Differential Calculus

Average vs. Instantaneous Rate of Change

In calculus, one of the biggest ideas is understanding how a quantity changes. Sometimes we look at change over a whole interval, and sometimes we want the change at one exact moment. These two ideas are called the average rate of change and the instantaneous rate of change.

This lesson explains the difference between them, shows how they are connected, and teaches how to compute instantaneous rates of change using the limit definition. This is the foundation of differentiation.

1. What is a rate of change?

A rate of change compares how much one quantity changes compared to another. In functions, we usually compare the change in output to the change in input.

If a function is written as \(y=f(x)\), then the rate of change tells us how much \(f(x)\) changes when \(x\) changes.

For example, if distance depends on time, then the rate of change of distance with respect to time is speed.

2. Average rate of change

The average rate of change of a function over an interval \([a,b]\) is the slope of the secant line joining the points \((a,f(a))\) and \((b,f(b))\).

The formula is

$$\text{Average rate of change} = \frac{f(b)-f(a)}{b-a}$$

This tells us the function's overall change over the interval, not what happens at a single point.

Geometrically:

• Choose two points on the graph.
• Draw the line through them.
• The slope of that line is the average rate of change.

3. Instantaneous rate of change

The instantaneous rate of change is the rate of change at one exact value of \(x\). It is the slope of the tangent line to the graph at that point.

This is more subtle, because one point by itself does not form a line. So we begin with average rate of change over a very small interval and then let that interval shrink toward zero.

That is why instantaneous rate of change is defined using a limit.

At \(x=a\), the instantaneous rate of change is

$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$

This is called the limit definition of the derivative.

Here:

• \(h\) is a very small change in \(x\),
• \(f(a+h)-f(a)\) is the corresponding change in the output,
• \(\frac{f(a+h)-f(a)}{h}\) is the average rate of change over the tiny interval,
• taking the limit as \(h\to 0\) gives the instantaneous rate of change.

4. Connecting average and instantaneous rate of change

The average rate of change uses two separate points. The instantaneous rate of change is what happens when the second point moves closer and closer to the first point.

So you can think of the instantaneous rate of change as the limiting value of the average rates of change.

In symbols, over the interval from \(a\) to \(a+h\), the average rate of change is

$$\frac{f(a+h)-f(a)}{h}$$

Then the instantaneous rate of change at \(a\) is

$$\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$

This transition from interval-based slope to exact-point slope is one of the central ideas of calculus.

5. Physical meaning: average speed vs instantaneous speed

Suppose a car travels \(120\) km in \(2\) hours. Its average speed is

$$\frac{120}{2}=60\text{ km/h}$$

But this does not mean the car was traveling exactly \(60\) km/h at every moment. It may have sped up or slowed down.

If we ask, “How fast was the car going at exactly 1.5 hours?” we are asking for the instantaneous speed. That is an instantaneous rate of change.

6. Notation for instantaneous rate of change

If \(y=f(x)\), then the instantaneous rate of change at \(x=a\) may be written as:

• \(f'(a)\)
• \(\left.\frac{dy}{dx}\right|_{x=a}\)
• the derivative of \(f\) at \(a\)

If we find a formula for the derivative for all \(x\), we write \(f'(x)\).

7. Worked Example 1: Average rate of change

Find the average rate of change of \(f(x)=x^2+1\) on the interval \([1,4]\).

Step 1: Use the formula

$$\frac{f(4)-f(1)}{4-1}$$

Step 2: Evaluate the function values

$$f(4)=4^2+1=17$$ $$f(1)=1^2+1=2$$

Step 3: Substitute

$$\frac{17-2}{3}=\frac{15}{3}=5$$

Answer: The average rate of change is \(5\).

This means that over the interval from \(x=1\) to \(x=4\), the function increases on average by \(5\) units of \(y\) for each 1 unit increase in \(x\).

8. Worked Example 2: Instantaneous rate of change from the limit definition

Find the instantaneous rate of change of \(f(x)=x^2\) at \(x=3\) using the limit definition.

Step 1: Write the definition

$$f'(3)=\lim_{h\to 0}\frac{f(3+h)-f(3)}{h}$$

Step 2: Compute each part

$$f(3+h)=(3+h)^2=9+6h+h^2$$ $$f(3)=9$$

Step 3: Substitute into the quotient

$$f'(3)=\lim_{h\to 0}\frac{(9+6h+h^2)-9}{h}$$ $$=\lim_{h\to 0}\frac{6h+h^2}{h}$$

Step 4: Simplify

$$=\lim_{h\to 0}(6+h)$$

Step 5: Take the limit

$$f'(3)=6$$

Answer: The instantaneous rate of change at \(x=3\) is \(6\).

This is the slope of the tangent line to \(y=x^2\) at \(x=3\).

9. Seeing the transition numerically

Before using the limit formally, it helps to see what happens numerically for \(f(x)=x^2\) at \(x=3\).

The average rate of change from \(x=3\) to \(x=3+h\) is

$$\frac{(3+h)^2-3^2}{h}=\frac{6h+h^2}{h}=6+h$$

If \(h\) is small:

• if \(h=1\), rate \(=7\)
• if \(h=0.1\), rate \(=6.1\)
• if \(h=0.01\), rate \(=6.01\)
• if \(h=-0.1\), rate \(=5.9\)
• if \(h=-0.01\), rate \(=5.99\)

From both sides, the value gets closer to \(6\). That is why the instantaneous rate of change is \(6\).

10. Worked Example 3: Find the derivative formula using the limit definition

Use the limit definition to find \(f'(x)\) for \(f(x)=x^2+3x\).

Step 1: Write the definition

$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

Step 2: Find \(f(x+h)\)

$$f(x+h)=(x+h)^2+3(x+h)$$ $$=x^2+2xh+h^2+3x+3h$$

Step 3: Substitute into the difference quotient

$$f'(x)=\lim_{h\to 0}\frac{(x^2+2xh+h^2+3x+3h)-(x^2+3x)}{h}$$

Step 4: Simplify the numerator

$$f'(x)=\lim_{h\to 0}\frac{2xh+h^2+3h}{h}$$

Step 5: Factor out \(h\)

$$=\lim_{h\to 0}\frac{h(2x+h+3)}{h}$$ $$=\lim_{h\to 0}(2x+h+3)$$

Step 6: Take the limit

$$f'(x)=2x+3$$

Answer: The derivative is

$$f'(x)=2x+3$$

This means the instantaneous rate of change of the function at any value of \(x\) is \(2x+3\).

For example, at \(x=2\), the instantaneous rate of change is

$$f'(2)=2(2)+3=7$$

11. Worked Example 4: Comparing average and instantaneous rates

Let \(f(x)=x^3\). Compare the average rate of change on \([1,2]\) with the instantaneous rate of change at \(x=1\).

Part A: Average rate of change on \([1,2]\)

$$\frac{f(2)-f(1)}{2-1}$$ $$=\frac{8-1}{1}=7$$

So the average rate of change is \(7\).

Part B: Instantaneous rate of change at \(x=1\)

Use the limit definition:

$$f'(1)=\lim_{h\to 0}\frac{(1+h)^3-1^3}{h}$$

Expand:

$$ (1+h)^3=1+3h+3h^2+h^3 $$

Substitute:

$$f'(1)=\lim_{h\to 0}\frac{1+3h+3h^2+h^3-1}{h}$$ $$=\lim_{h\to 0}\frac{3h+3h^2+h^3}{h}$$ $$=\lim_{h\to 0}(3+3h+h^2)$$ $$=3$$

Answer:

• Average rate of change on \([1,2]\): \(7\)
• Instantaneous rate of change at \(x=1\): \(3\)

This example shows that the average rate over an interval can be very different from the rate at one exact point.

12. Differentiation rules and instantaneous rate of change

The limit definition is the foundation of the derivative, but once the idea is understood, we usually use differentiation rules to find instantaneous rates more efficiently.

Some common rules are:

• If \(f(x)=c\), then \(f'(x)=0\)
• If \(f(x)=x^n\), then \(f'(x)=nx^{n-1}\)
• If \(f(x)=af(x)+bg(x)\), then derivative is found term-by-term

For example, if \(f(x)=4x^3-2x+5\), then

$$f'(x)=12x^2-2$$

This derivative gives the instantaneous rate of change at any value of \(x\).

At \(x=2\), the instantaneous rate of change is

$$f'(2)=12(2)^2-2=48-2=46$$

13. Common mistakes to avoid

• Mixing up average and instantaneous rate of change. Average uses two endpoints. Instantaneous uses a limit at one point.
• Forgetting the formula. Average rate of change is \(\frac{f(b)-f(a)}{b-a}\), not just \(f(b)-f(a)\).
• Substituting too early in the limit definition. If you plug in \(h=0\) before simplifying, you often get \(\frac{0}{0}\), which is undefined.
• Making algebra errors when expanding. Be careful with brackets, especially in expressions like \((x+h)^2\) or \((x+h)^3\).
• Thinking the derivative is only a formula. It also has meaning: slope of the tangent line and instantaneous rate of change.

14. How to decide which method to use

If the question asks for change over an interval, use average rate of change:

$$\frac{f(b)-f(a)}{b-a}$$

If the question asks for change at a point, use the instantaneous rate of change:

• either the limit definition, or
• a derivative rule if you already know the derivative.

Key words can help:

• Average, between, over the interval \(\rightarrow\) average rate of change
• At, exactly when, instantaneous, tangent \(\rightarrow\) instantaneous rate of change

15. Quick comparison table

• Average rate of change: over an interval
• Formula: \(\frac{f(b)-f(a)}{b-a}\)
• Graph meaning: slope of a secant line
• Instantaneous rate of change: at a single point
• Formula: \(f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\)
• Graph meaning: slope of a tangent line

16. Final summary

The average rate of change describes how a function changes across an interval, and it is found using the slope between two points. The instantaneous rate of change describes how a function changes at one exact point, and it is found by taking the limit of average rates of change as the interval becomes extremely small.

This idea leads directly to the derivative. The derivative is one of the most important tools in calculus because it lets us describe motion, slope, growth, and change precisely.

Once you understand that instantaneous rate of change comes from the limit of average rate of change, the whole idea of differentiation becomes much more meaningful.

Lesson: The Derivative as a Limit of the Difference Quotient

In calculus, one of the most important ideas is the derivative. The derivative tells us how fast a function is changing at a single point. You can think of it as the instantaneous rate of change or the slope of the tangent line to a graph.

Before learning shortcut differentiation rules, it is important to understand where the derivative comes from. The derivative is built from the idea of the slope between two nearby points on a graph. This slope is called the difference quotient.

This lesson explains how to compute the derivative from first principles, using limits and the generalized slope formula.

1. From average rate of change to instantaneous rate of change

Suppose we have a function \(f(x)\). If we choose two points on the graph, at \(x\) and \(x+h\), then the change in the output is

$$f(x+h)-f(x)$$

and the change in the input is

$$h$$

So the slope between these two points is

$$\frac{f(x+h)-f(x)}{h}$$

This expression is called the difference quotient. It gives the average rate of change of the function over the interval from \(x\) to \(x+h\).

But what if we want the rate of change at exactly one point? Then we make the second point move closer and closer to the first point. In algebraic terms, we let \(h \to 0\).

That leads to the definition of the derivative:

$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

This is called the limit definition of the derivative.

2. What this definition means

• \(f(x+h)-f(x)\) measures how much the function changes.
• \(h\) measures how much the input changes.
• \(\frac{f(x+h)-f(x)}{h}\) gives the average rate of change over a small interval.
• \(\lim_{h\to 0}\) makes that interval shrink to zero.
• The result is the exact rate of change at \(x\).

You should notice that we do not simply substitute \(h=0\) right away, because that would cause division by zero. Instead, we simplify first and then take the limit.

3. A step-by-step method for finding a derivative from first principles

1. Write the definition: $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$
2. Find \(f(x+h)\) by replacing every \(x\) in the function with \(x+h\).
3. Substitute into the difference quotient.
4. Simplify the numerator carefully.
5. Factor and cancel \(h\), if possible.
6. Take the limit as \(h\to 0\).

This process may feel long at first, but it helps you understand how differentiation works.

4. Worked Example 1: A basic quadratic

Find the derivative of \(f(x)=x^2\) using first principles.

Step 1: Write the definition

$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

Step 2: Find \(f(x+h)\)

$$f(x+h)=(x+h)^2$$

Expand:

$$f(x+h)=x^2+2xh+h^2$$

Step 3: Substitute into the difference quotient

$$f'(x)=\lim_{h\to 0}\frac{(x^2+2xh+h^2)-x^2}{h}$$

Step 4: Simplify the numerator

$$f'(x)=\lim_{h\to 0}\frac{2xh+h^2}{h}$$

Step 5: Factor out \(h\)

$$f'(x)=\lim_{h\to 0}\frac{h(2x+h)}{h}$$

Cancel \(h\):

$$f'(x)=\lim_{h\to 0}(2x+h)$$

Step 6: Take the limit

$$f'(x)=2x$$

So the derivative of \(x^2\) is

$$f'(x)=2x$$

This means the instantaneous rate of change of \(x^2\) at any value of \(x\) is \(2x\).

5. Worked Example 2: A cubic function

Find the derivative of \(f(x)=x^3\) using the limit definition.

Step 1: Write \(f(x+h)\)

$$f(x+h)=(x+h)^3$$

Expand:

$$f(x+h)=x^3+3x^2h+3xh^2+h^3$$

Step 2: Substitute into the definition

$$f'(x)=\lim_{h\to 0}\frac{(x^3+3x^2h+3xh^2+h^3)-x^3}{h}$$

Step 3: Simplify

$$f'(x)=\lim_{h\to 0}\frac{3x^2h+3xh^2+h^3}{h}$$

Factor out \(h\):

$$f'(x)=\lim_{h\to 0}\frac{h(3x^2+3xh+h^2)}{h}$$

Cancel \(h\):

$$f'(x)=\lim_{h\to 0}(3x^2+3xh+h^2)$$

Step 4: Take the limit

$$f'(x)=3x^2$$

So,

$$\frac{d}{dx}(x^3)=3x^2$$

This example shows why derivative rules later become so useful, but the limit definition is the foundation.

6. Worked Example 3: A linear function

Find the derivative of \(f(x)=5x-4\) using first principles.

Step 1: Find \(f(x+h)\)

$$f(x+h)=5(x+h)-4=5x+5h-4$$

Step 2: Substitute into the difference quotient

$$f'(x)=\lim_{h\to 0}\frac{(5x+5h-4)-(5x-4)}{h}$$

Step 3: Simplify

$$f'(x)=\lim_{h\to 0}\frac{5h}{h}$$ $$f'(x)=\lim_{h\to 0}5$$

Step 4: Take the limit

$$f'(x)=5$$

This makes sense because a straight line has the same slope everywhere. Its instantaneous rate of change is constant.

7. Worked Example 4: A function with a fraction

Find the derivative of \(f(x)=\frac{1}{x}\) using first principles.

Step 1: Find \(f(x+h)\)

$$f(x+h)=\frac{1}{x+h}$$

Step 2: Substitute into the definition

$$f'(x)=\lim_{h\to 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$$

Step 3: Combine the fractions in the numerator

$$\frac{1}{x+h}-\frac{1}{x}=\frac{x-(x+h)}{x(x+h)}=\frac{-h}{x(x+h)}$$

So:

$$f'(x)=\lim_{h\to 0}\frac{\frac{-h}{x(x+h)}}{h}$$

This is the same as

$$f'(x)=\lim_{h\to 0}\frac{-h}{hx(x+h)}$$

Cancel \(h\):

$$f'(x)=\lim_{h\to 0}\frac{-1}{x(x+h)}$$

Step 4: Take the limit

$$f'(x)=\frac{-1}{x^2}$$

So,

$$\frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}$$

This example shows that first principles also works for non-polynomial expressions, as long as you simplify carefully.

8. Interpreting the derivative at a point

Once you have a formula for \(f'(x)\), you can find the instantaneous rate of change at a specific value.

For example, if \(f(x)=x^2\), then we found that

$$f'(x)=2x$$

At \(x=3\),

$$f'(3)=2(3)=6$$

This means:

• The function is changing at a rate of 6 units of output per 1 unit of input when \(x=3\).
• The slope of the tangent line to the graph at \(x=3\) is 6.

9. Common mistakes to avoid

• Substituting \(h=0\) too early: This gives division by zero. Always simplify first.
• Forgetting brackets: When subtracting \(f(x)\), use brackets so negative signs are handled correctly.
• Expanding incorrectly: Be careful with expressions like \((x+h)^2\) and \((x+h)^3\).
• Not factoring out \(h\): In many problems, the key step is finding and canceling a common factor of \(h\).
• Mixing up average and instantaneous rate of change: The difference quotient alone gives an average rate; the limit gives the instantaneous rate.

10. Why this idea matters

The limit definition is the starting point of differential calculus. Even though later you will use faster differentiation rules, this definition explains why derivatives work.

It connects three big ideas:

• slope
• rate of change
• limits

Understanding this connection makes the rest of calculus much easier.

11. Brief summary

The derivative of a function is defined by

$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

This formula comes from the slope between two nearby points on a graph. The difference quotient gives the average rate of change, and the limit turns it into the instantaneous rate of change.

To find a derivative from first principles, calculate \(f(x+h)\), substitute into the formula, simplify, cancel \(h\), and then take the limit. This method helps you understand the meaning of the derivative before using shortcut rules.

Differentiability and Continuity are two closely related ideas in calculus. Both help us understand how a function behaves at a point. Continuity tells us whether the graph has a break, hole, or jump. Differentiability tells us whether the function has a well-defined tangent slope, or instantaneous rate of change, at that point.

In this lesson, you will learn what continuity and differentiability mean, why differentiability implies continuity, and how to recognize points such as corners, cusps, and vertical tangents where a derivative does not exist.

This idea is very important because many functions are continuous but not differentiable at some points. So it is essential to know the difference.

1. Continuity at a Point

A function \(f(x)\) is continuous at \(x=a\) if all three of the following are true:

1. \(f(a)\) is defined.
2. \(\lim_{x \to a} f(x)\) exists.
3. \(\lim_{x \to a} f(x) = f(a)\).

If these conditions hold, the graph has no break at \(x=a\). Informally, you can draw the graph through that point without lifting your pencil.

Examples of discontinuity include:

• Hole: the limit exists, but the function value is missing or different.
• Jump: the left-hand and right-hand limits are different.
• Infinite discontinuity: the function grows without bound near the point.

2. Differentiability at a Point

A function \(f(x)\) is differentiable at \(x=a\) if the derivative exists there. Using the limit definition, this means:

$$f'(a)=\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$$

If this limit exists and is finite, then the function has a definite instantaneous rate of change at \(x=a\).

Geometrically, differentiability means the graph has a clear tangent line with one definite slope at that point.

3. Relationship Between Differentiability and Continuity

The key result is:

If a function is differentiable at a point, then it is continuous at that point.

This means:

• Differentiable \(\Rightarrow\) Continuous
• But Continuous \(\nRightarrow\) Differentiable in general

So differentiability is a stronger condition than continuity.

Why does differentiability imply continuity?

Suppose \(f\) is differentiable at \(x=a\). Then

$$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$$

exists. Let this limit be some finite number \(L\).

Then we can write

$$f(a+h)-f(a)=\left(\frac{f(a+h)-f(a)}{h}\right)h$$

Now as \(h \to 0\), the first factor approaches \(L\), and the second factor approaches \(0\). So their product approaches \(0\):

$$\lim_{h \to 0} [f(a+h)-f(a)] = 0$$

Therefore,

$$\lim_{h \to 0} f(a+h) = f(a)$$

which means \(f\) is continuous at \(x=a\).

Important: The reverse is not always true. A function can be continuous but still fail to be differentiable.

4. When Does a Derivative Fail to Exist?

A function may fail to be differentiable at a point for several common reasons:

• The function is not continuous there.
• The graph has a corner.
• The graph has a cusp.
• The graph has a vertical tangent.

Let us understand each one.

5. Non-differentiability Due to Discontinuity

If a function is not continuous at \(x=a\), then it cannot be differentiable there. This follows directly from the theorem above.

So if you are asked whether a function is differentiable at a point, it is often smart to check continuity first.

6. Corner Points

A corner occurs when the graph changes direction sharply, and the left-hand slope and right-hand slope are different.

A common example is \(f(x)=|x|\) at \(x=0\).

For \(x<0\), \(|x|=-x\), whose slope is \(-1\). For \(x>0\), \(|x|=x\), whose slope is \(1\).

Since the left-hand derivative and right-hand derivative are not equal, the derivative at \(x=0\) does not exist.

Graphically, the tangent slope is not unique, so the function is not differentiable there.

7. Cusps

A cusp is a sharper point than a corner. At a cusp, the slopes on the two sides become very large in magnitude, often with opposite signs.

A standard example is \(f(x)=x^{2/3}\) at \(x=0\).

Its derivative is

$$f'(x)=\frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$$

As \(x \to 0^+\), \(f'(x) \to +\infty\). As \(x \to 0^-\), \(f'(x) \to -\infty\).

Because the slopes do not approach one finite value, the derivative at \(x=0\) does not exist.

8. Vertical Tangents

A vertical tangent occurs when the slope becomes infinite in the same direction from both sides.

A common example is \(f(x)=\sqrt[3]{x}\).

Its derivative is

$$f'(x)=\frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}$$

As \(x \to 0\), \(f'(x) \to +\infty\). The graph is continuous at \(x=0\), but the tangent is vertical, so the derivative is not defined as a finite number.

This is another example showing that continuity does not guarantee differentiability.

9. One-Sided Derivatives

Sometimes it is helpful to compare the derivative from the left and from the right.

The left-hand derivative at \(x=a\) is

$$\lim_{h \to 0^-} \frac{f(a+h)-f(a)}{h}$$

and the right-hand derivative at \(x=a\) is

$$\lim_{h \to 0^+} \frac{f(a+h)-f(a)}{h}$$

A function is differentiable at \(x=a\) only if both one-sided derivatives exist and are equal.

This is especially useful for piecewise functions and functions with sharp points.

10. Worked Example 1: Showing Differentiability Implies Continuity

Suppose a function \(f\) is differentiable at \(x=2\). Show that it must be continuous at \(x=2\).

Solution:

If \(f\) is differentiable at \(x=2\), then

$$\lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$$

exists.

Now write

$$f(2+h)-f(2)=\left(\frac{f(2+h)-f(2)}{h}\right)h$$

As \(h \to 0\), the first factor approaches the finite value \(f'(2)\), and \(h \to 0\). So

$$\lim_{h \to 0} [f(2+h)-f(2)] = 0$$

Thus,

$$\lim_{h \to 0} f(2+h)=f(2)$$

So \(f\) is continuous at \(x=2\).

Conclusion: differentiability at a point guarantees continuity there.

11. Worked Example 2: Continuous but Not Differentiable at a Corner

Determine whether \(f(x)=|x|\) is continuous and differentiable at \(x=0\).

Step 1: Check continuity.

We know \(f(0)=|0|=0\).

Also,

$$\lim_{x \to 0} |x| = 0$$

Since

$$\lim_{x \to 0} |x| = f(0)$$

the function is continuous at \(x=0\).

Step 2: Check differentiability.

Using one-sided derivatives:

• For \(x<0\), \(f(x)=-x\), so slope \(=-1\).
• For \(x>0\), \(f(x)=x\), so slope \(=1\).

Thus,

$$f'_-(0)=-1, \qquad f'_+(0)=1$$

Since these are not equal, \(f'(0)\) does not exist.

Conclusion: \(f(x)=|x|\) is continuous at \(0\), but not differentiable there because of a corner.

12. Worked Example 3: Discontinuous Functions Are Not Differentiable

Consider

$$f(x)= \begin{cases} 1, & x<0 \\ 2, & x\ge 0 \end{cases}$$

Determine whether \(f\) is continuous and differentiable at \(x=0\).

Step 1: Check continuity.

The left-hand limit is

$$\lim_{x \to 0^-} f(x)=1$$

and the right-hand limit is

$$\lim_{x \to 0^+} f(x)=2$$

Since these are not equal, \(\lim_{x \to 0} f(x)\) does not exist.

So the function is not continuous at \(x=0\).

Step 2: Check differentiability.

Because differentiability implies continuity, and the function is not continuous at \(x=0\), it cannot be differentiable there.

Conclusion: discontinuity always prevents differentiability.

13. Worked Example 4: Vertical Tangent / Cusp Type Behavior

Determine whether \(f(x)=\sqrt[3]{x}\) is differentiable at \(x=0\).

Step 1: Check continuity.

Since \(\sqrt[3]{x}\) is defined for all real \(x\), and

$$\lim_{x \to 0} \sqrt[3]{x} = 0 = f(0)$$

the function is continuous at \(x=0\).

Step 2: Check differentiability using the derivative.

$$f'(x)=\frac{1}{3}x^{-2/3}$$

At \(x=0\), this expression is undefined.

As \(x\) gets very close to \(0\), \(x^{-2/3}\) becomes very large and positive, so the slope tends to \(+\infty\).

This means the graph has a vertical tangent at \(x=0\), so the derivative does not exist there as a finite number.

Conclusion: \(f(x)=\sqrt[3]{x}\) is continuous at \(0\), but not differentiable there.

14. Quick Comparison Table

• Continuous and differentiable: smooth curves like \(x^2\), \(\sin x\), \(3x+1\)
• Continuous but not differentiable: \(|x|\) at \(0\), \(\sqrt[3]{x}\) at \(0\), \(x^{2/3}\) at \(0\)
• Not continuous and not differentiable: jump or hole functions at the point of discontinuity

15. How to Test a Point Systematically

When asked whether a function is differentiable at a point \(x=a\), follow these steps:

1. Check whether \(f(a)\) is defined.
2. Check whether the function is continuous at \(x=a\).
3. If it is continuous, check the derivative using:

• the limit definition, or
• left-hand and right-hand derivatives, or
• known derivative rules if the expression is smooth.

Then ask:

• Is there a corner?
• Is there a cusp?
• Is there a vertical tangent?

If any of these appear, the derivative does not exist at that point.

16. Common Mistakes to Avoid

• Assuming continuity automatically means differentiability.
• Forgetting to check left-hand and right-hand derivatives for piecewise functions.
• Thinking an undefined derivative formula always means discontinuity. A function may still be continuous.
• Ignoring special graph shapes such as corners, cusps, and vertical tangents.

17. Final Summary

Continuity means a function has no break at a point, while differentiability means it has a well-defined tangent slope there. Every differentiable function is continuous, but not every continuous function is differentiable.

The derivative fails to exist at points of discontinuity and also at certain special continuous points such as corners, cusps, and vertical tangents. To decide whether a function is differentiable at a point, first test continuity, then compare the behavior of slopes from both sides.

Lesson: Power Rule and Basic Derivative Properties

In calculus, the derivative tells us the instantaneous rate of change of a function. In simpler words, it tells us how fast a quantity is changing at a specific moment.

You may already know that the derivative can be found from the limit definition. That definition is very important because it explains what a derivative means. However, using the limit definition every time can be slow. For polynomial functions, we use faster rules such as the power rule, the constant multiple rule, and the sum and difference rules.

In this lesson, you will learn how these rules work and how to use them to differentiate polynomial expressions quickly and accurately.

1. Reminder: What is a derivative?

For a function \(f(x)\), the derivative is written as \(f'(x)\) or \(\dfrac{dy}{dx}\). It measures the slope of the tangent line to the graph at a point.

The limit definition of the derivative is

$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

This formula is the foundation of differentiation. The rules in this lesson are shortcuts that come from this definition.

2. The Power Rule

The most important rule for differentiating polynomials is the power rule.

If

$$f(x)=x^n$$

then

$$f'(x)=nx^{n-1}$$

This means:

• Bring the exponent down to the front.
• Then reduce the exponent by 1.

Examples of the power rule:

• \(\dfrac{d}{dx}(x^5)=5x^4\)
• \(\dfrac{d}{dx}(x^3)=3x^2\)
• \(\dfrac{d}{dx}(x^1)=1x^0=1\)
• \(\dfrac{d}{dx}(x^2)=2x\)

Important special case: constants

A constant does not change, so its rate of change is 0.

$$\frac{d}{dx}(c)=0$$

For example:

• \(\dfrac{d}{dx}(7)=0\)
• \(\dfrac{d}{dx}(-12)=0\)

Why does this fit the power rule?

A constant can be thought of as \(cx^0\), since \(x^0=1\). Then the derivative becomes

$$\frac{d}{dx}(cx^0)=c\cdot 0\cdot x^{-1}=0$$

3. Constant Multiple Rule

If a function is multiplied by a constant, the derivative is multiplied by the same constant.

If

$$f(x)=c\,g(x)$$

then

$$f'(x)=c\,g'(x)$$

This means you can keep the constant in front and differentiate the rest.

Examples:

• \(\dfrac{d}{dx}(4x^3)=4\cdot \dfrac{d}{dx}(x^3)=4(3x^2)=12x^2\)
• \(\dfrac{d}{dx}(-2x^5)=-2\cdot 5x^4=-10x^4\)

4. Sum and Difference Rules

Many polynomial expressions are made of several terms added or subtracted. The derivative of a sum is the sum of the derivatives, and the derivative of a difference is the difference of the derivatives.

If

$$f(x)=g(x)+h(x)$$

then

$$f'(x)=g'(x)+h'(x)$$

If

$$f(x)=g(x)-h(x)$$

then

$$f'(x)=g'(x)-h'(x)$$

This lets us differentiate each term separately.

Example:

$$\frac{d}{dx}(x^4+3x^2-7)=\frac{d}{dx}(x^4)+\frac{d}{dx}(3x^2)-\frac{d}{dx}(7)$$ $$=4x^3+3(2x)-0$$ $$=4x^3+6x$$

5. How to Differentiate a Polynomial Step by Step

When differentiating a polynomial, follow this process:

1. Look at each term separately.
2. Use the constant multiple rule if there is a number in front.
3. Use the power rule on the variable part.
4. Use the sum or difference rule to combine the results.
5. Simplify your final answer.

6. Worked Examples

Example 1: Differentiate \(f(x)=x^6\)

Use the power rule. The exponent is 6.

$$f'(x)=6x^{5}$$

Answer: \(f'(x)=6x^5\)

Example 2: Differentiate \(f(x)=5x^4\)

Use the constant multiple rule and the power rule.

$$f'(x)=5\cdot \frac{d}{dx}(x^4)=5(4x^3)=20x^3$$

Answer: \(f'(x)=20x^3\)

Example 3: Differentiate \(f(x)=3x^5-2x^3+4x-9\)

Differentiate each term one at a time.

$$\frac{d}{dx}(3x^5)=15x^4$$ $$\frac{d}{dx}(-2x^3)=-6x^2$$ $$\frac{d}{dx}(4x)=4$$ $$\frac{d}{dx}(-9)=0$$

Now combine the results:

$$f'(x)=15x^4-6x^2+4$$

Answer: \(f'(x)=15x^4-6x^2+4\)

Example 4: Differentiate \(y=-7x^8+x^2-5\)

Apply the rules term by term.

$$\frac{dy}{dx}=-7\cdot 8x^7+2x^{1}-0$$ $$\frac{dy}{dx}=-56x^7+2x$$

Answer: \(\dfrac{dy}{dx}=-56x^7+2x\)

7. Common Mistakes to Avoid

• Forgetting to subtract 1 from the exponent
  For example, \(\dfrac{d}{dx}(x^4)\) is \(4x^3\), not \(4x^4\).
• Forgetting that constants have derivative 0
  For example, \(\dfrac{d}{dx}(6)=0\).
• Not differentiating every term
  In a polynomial, every term must be handled separately.
• Sign errors
  Be careful with negative signs. For example, \(\dfrac{d}{dx}(-3x^2)=-6x\), not \(6x\).

8. Quick Practice Ideas

Try differentiating these on your own:

• \(f(x)=x^7\)
• \(g(x)=6x^3\)
• \(h(x)=2x^4-5x^2+8\)
• \(p(x)=-4x^5+x-1\)

You should get:

• \(f'(x)=7x^6\)
• \(g'(x)=18x^2\)
• \(h'(x)=8x^3-10x\)
• \(p'(x)=-20x^4+1\)

9. Why these rules matter

These basic derivative rules make calculus much faster. Instead of going back to the limit definition each time, you can quickly find slopes, rates of change, and equations of tangent lines for polynomial functions.

They are also the foundation for more advanced differentiation rules you will learn later. If you are confident with the power rule and the basic properties, you will be much more successful in the rest of differential calculus.

Summary

The power rule says that \(\dfrac{d}{dx}(x^n)=nx^{n-1}\). The derivative of a constant is 0, the constant multiple rule lets you keep constants out front, and the sum and difference rules let you differentiate each term of a polynomial separately.

Using these rules together, you can differentiate polynomial expressions quickly and correctly. The key idea is to work term by term, apply the right rule, and simplify your final answer.

Lesson: The Product Rule and Quotient Rule

When finding derivatives, many functions can be differentiated using basic rules such as the power rule. However, some functions are built by multiplying or dividing two simpler functions. In these cases, we need special rules called the Product Rule and the Quotient Rule.

These rules help us compute the instantaneous rate of change of more complicated functions without going back to the full limit definition every time. They are important because many real problems involve expressions like a polynomial times an exponential function, or one algebraic expression divided by another.

In this lesson, you will learn:

• when to use the Product Rule,
• when to use the Quotient Rule,
• how to apply each rule step by step,
• and how to avoid common mistakes.

1. When do we need these rules?

If a function is the product of two functions, such as

$$f(x)=u(x)v(x),$$

then we use the Product Rule.

If a function is the quotient of two functions, such as

$$f(x)=\frac{u(x)}{v(x)},$$

then we use the Quotient Rule, as long as \(v(x)\neq 0\).

Here, \(u(x)\) and \(v(x)\) are just two simpler parts of the function. A very useful habit is to first identify these two parts before differentiating.

2. The Product Rule

If

$$f(x)=u(x)v(x),$$

then

$$f'(x)=u'(x)v(x)+u(x)v'(x).$$

In words: differentiate the first function and leave the second alone, then add the first function unchanged times the derivative of the second.

A common memory aid is:

$$\text{Product Rule: } (uv)'=u'v+uv'$$

Important note: The derivative of a product is not just the product of the derivatives. In general,

$$\frac{d}{dx}[u(x)v(x)]\neq u'(x)v'(x).$$

Steps for using the Product Rule

1. Identify the two factors: \(u(x)\) and \(v(x)\).
2. Find \(u'(x)\) and \(v'(x)\).
3. Substitute into \((uv)'=u'v+uv'\).
4. Simplify if possible.

3. Worked Example 1: Product of a polynomial and a linear function

Find the derivative of

$$f(x)=x^2(x+5).$$

Step 1: Identify the two parts.

Let

$$u(x)=x^2, \qquad v(x)=x+5.$$

Step 2: Differentiate each part.

$$u'(x)=2x, \qquad v'(x)=1.$$

Step 3: Apply the Product Rule.

$$f'(x)=u'v+uv'$$

$$f'(x)=2x(x+5)+x^2(1).$$

Step 4: Simplify.

$$f'(x)=2x^2+10x+x^2=3x^2+10x.$$

So the derivative is

$$f'(x)=3x^2+10x.$$

Check: Since this function can also be expanded first, we get

$$f(x)=x^3+5x^2,$$

and differentiating gives

$$f'(x)=3x^2+10x,$$

which matches our answer.

4. Worked Example 2: Product involving a square root

Find the derivative of

$$g(x)=x^3\sqrt{x}.$$

This is a product, so we use the Product Rule.

Step 1: Identify the parts.

$$u(x)=x^3, \qquad v(x)=\sqrt{x}=x^{1/2}.$$

Step 2: Differentiate each part.

$$u'(x)=3x^2, \qquad v'(x)=\frac{1}{2}x^{-1/2}.$$

Step 3: Apply the Product Rule.

$$g'(x)=u'v+uv'$$

$$g'(x)=3x^2(x^{1/2})+x^3\left(\frac{1}{2}x^{-1/2}\right).$$

Step 4: Simplify.

$$g'(x)=3x^{5/2}+\frac{1}{2}x^{5/2}$$

$$g'(x)=\frac{7}{2}x^{5/2}.$$

So,

$$g'(x)=\frac{7}{2}x^{5/2}.$$

Note: In this example, you could also simplify the original function first:

$$x^3\sqrt{x}=x^3\cdot x^{1/2}=x^{7/2},$$

so

$$g'(x)=\frac{7}{2}x^{5/2}.$$

This confirms the result. Even when a function can be simplified first, practicing the Product Rule is still useful.

5. The Quotient Rule

If

$$f(x)=\frac{u(x)}{v(x)},$$

then

$$f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}.$$

In words: derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.

A common memory aid is:

$$\text{Quotient Rule: } \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}$$

Important note: Order matters. The numerator must be

$$u'v-uv',$$

not the other way around. Switching the order changes the sign of the answer.

Steps for using the Quotient Rule

1. Identify the numerator \(u(x)\) and denominator \(v(x)\).
2. Find \(u'(x)\) and \(v'(x)\).
3. Substitute into \(\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\).
4. Simplify carefully.

6. Worked Example 3: Quotient of two polynomials

Find the derivative of

$$h(x)=\frac{x^2+1}{x-3}.$$

Step 1: Identify the top and bottom.

$$u(x)=x^2+1, \qquad v(x)=x-3.$$

Step 2: Differentiate each part.

$$u'(x)=2x, \qquad v'(x)=1.$$

Step 3: Apply the Quotient Rule.

$$h'(x)=\frac{u'v-uv'}{v^2}$$

$$h'(x)=\frac{2x(x-3)-(x^2+1)(1)}{(x-3)^2}.$$

Step 4: Simplify the numerator.

$$h'(x)=\frac{2x^2-6x-x^2-1}{(x-3)^2}$$

$$h'(x)=\frac{x^2-6x-1}{(x-3)^2}.$$

So the derivative is

$$h'(x)=\frac{x^2-6x-1}{(x-3)^2}.$$

7. Worked Example 4: Quotient with a power in the denominator

Find the derivative of

$$k(x)=\frac{3x+2}{x^2}.$$

You can use the Quotient Rule directly.

Step 1: Identify the parts.

$$u(x)=3x+2, \qquad v(x)=x^2.$$

Step 2: Differentiate each part.

$$u'(x)=3, \qquad v'(x)=2x.$$

Step 3: Apply the Quotient Rule.

$$k'(x)=\frac{u'v-uv'}{v^2}$$

$$k'(x)=\frac{3(x^2)-(3x+2)(2x)}{(x^2)^2}.$$

Step 4: Simplify.

$$k'(x)=\frac{3x^2-(6x^2+4x)}{x^4}$$

$$k'(x)=\frac{-3x^2-4x}{x^4}.$$

We can simplify further by factoring out \(x\):

$$k'(x)=\frac{-x(3x+4)}{x^4}=\frac{-(3x+4)}{x^3}.$$

So,

$$k'(x)=\frac{-(3x+4)}{x^3}.$$

Alternative method: Rewrite first.

$$k(x)=\frac{3x+2}{x^2}=\frac{3x}{x^2}+\frac{2}{x^2}=3x^{-1}+2x^{-2}.$$

Then differentiate term by term:

$$k'(x)=-3x^{-2}-4x^{-3}=\frac{-3x-4}{x^3}.$$

This matches the quotient rule result.

8. Common mistakes to avoid

• Using the product rule for addition: If the function is a sum like \(x^2+\sin x\), use the sum rule, not the product rule.
• Forgetting one term in the Product Rule: \((uv)'\) has two terms: \(u'v+uv'\).
• Getting the Quotient Rule in the wrong order: It is \(u'v-uv'\), not \(uv'-u'v\).
• Forgetting to square the denominator: In the Quotient Rule, the denominator becomes \(v^2\).
• Not simplifying carefully: Sign errors often happen when expanding brackets.

9. How to decide which rule to use

Ask yourself:

• Are two expressions being multiplied? Use the Product Rule.
• Is one expression being divided by another? Use the Quotient Rule.
• Can the expression be simplified first? Sometimes rewriting the function makes differentiation easier.

For example:

• \(x^2(x+1)\) → Product Rule
• \(\frac{x^2+1}{x-1}\) → Quotient Rule
• \(x^2\sqrt{x}\) → Product Rule, though it can also be rewritten as a single power
• \(\frac{5}{x^3}\) → often easier to rewrite as \(5x^{-3}\)

10. Final summary

The Product Rule is used when a function is made by multiplying two functions:

$$\frac{d}{dx}[u(x)v(x)]=u'(x)v(x)+u(x)v'(x).$$

The Quotient Rule is used when a function is made by dividing two functions:

$$\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right]=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}. $$

To use these rules well, first split the function into two parts, differentiate each part, substitute carefully, and simplify. With practice, these rules become a reliable way to find instantaneous rates of change for more complicated functions.

The Chain Rule is one of the most important differentiation rules in calculus. It is used when a function is built inside another function. These are called composite functions, or nested functions.

For example, in the function \((3x+1)^5\), the expression \(3x+1\) is inside the power of 5. In \(\sqrt{2x^2-7}\), the expression \(2x^2-7\) is inside the square root. In cases like these, we cannot just differentiate the outside part and ignore the inside. We need the Chain Rule.

The Chain Rule helps us find the derivative of a composite function by working from the outside inward. In simple words: differentiate the outer function, keep the inner part the same, then multiply by the derivative of the inner function.

General form of the Chain Rule:

If \(y=f(g(x))\), then

$$\frac{dy}{dx}=f'(g(x))\cdot g'(x)$$

This means:

• Differentiate the outer function.
• Leave the inner function unchanged for the moment.
• Multiply by the derivative of the inner function.

Another common way to write it is:

If \(u=g(x)\) and \(y=f(u)\), then

$$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$$

This notation makes the chain idea easier to see: one rate of change is multiplied by another.

Why does this rule make sense?

Suppose \(y\) depends on \(u\), and \(u\) depends on \(x\). Then changing \(x\) changes \(u\), and that changes \(y\). So the effect of \(x\) on \(y\) happens through the intermediate variable \(u\). That is why the derivatives are multiplied.

For example, if

$$u=3x+1 \quad \text{and} \quad y=u^5,$$

then

$$\frac{dy}{du}=5u^4 \quad \text{and} \quad \frac{du}{dx}=3.$$

So

$$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=5u^4\cdot 3=15u^4=15(3x+1)^4.$$

This is exactly the Chain Rule in action.

How to recognize when to use the Chain Rule

You usually need the Chain Rule when you see one function applied to another. Common signs include:

• A power of an expression, such as \((2x-5)^7\)
• A square root or cube root of an expression, such as \(\sqrt{x^2+4}\)
• An exponential function with a non-trivial exponent, such as \(e^{x^2}\)
• A trigonometric function of an expression, such as \(\sin(4x^3)\)
• A logarithm of an expression, such as \(\ln(5x+2)\)

If the input to a function is not just plain \(x\), but another expression in \(x\), the Chain Rule is very likely needed.

A step-by-step method

1. Identify the outer function and the inner function.
2. Differentiate the outer function.
3. Keep the inner expression unchanged while doing that.
4. Multiply by the derivative of the inner expression.
5. Simplify if needed.

Worked Example 1: A basic power of a linear expression

Differentiate

$$y=(2x+3)^4$$

Step 1: Identify outer and inner functions.

• Outer function: \(u^4\)
• Inner function: \(u=2x+3\)

Step 2: Differentiate the outer function.

The derivative of \(u^4\) is \(4u^3\).

Step 3: Multiply by the derivative of the inner function.

The derivative of \(2x+3\) is \(2\).

So,

$$\frac{dy}{dx}=4(2x+3)^3\cdot 2$$ $$\frac{dy}{dx}=8(2x+3)^3$$

Answer: \(\boxed{8(2x+3)^3}\)

Notice that if we had simply used the power rule and written \(4(2x+3)^3\), we would have missed the extra factor of 2 from the inside function.

Worked Example 2: A square root of a quadratic expression

Differentiate

$$y=\sqrt{x^2+1}$$

First rewrite the square root using an exponent:

$$y=(x^2+1)^{1/2}$$

Outer function: \(u^{1/2}\)

Inner function: \(u=x^2+1\)

Differentiate the outer function:

$$\frac{d}{du}(u^{1/2})=\frac{1}{2}u^{-1/2}$$

Differentiate the inner function:

$$\frac{d}{dx}(x^2+1)=2x$$

Apply the Chain Rule:

$$\frac{dy}{dx}=\frac{1}{2}(x^2+1)^{-1/2}\cdot 2x$$

Simplify:

$$\frac{dy}{dx}=x(x^2+1)^{-1/2}$$ $$\frac{dy}{dx}=\frac{x}{\sqrt{x^2+1}}$$

Answer: \(\boxed{\dfrac{x}{\sqrt{x^2+1}}}\)

Worked Example 3: A trigonometric composite function

Differentiate

$$y=\sin(3x^2-4x)$$

Outer function: \(\sin(u)\)

Inner function: \(u=3x^2-4x\)

The derivative of the outer function \(\sin(u)\) is \(\cos(u)\).

The derivative of the inner function is

$$\frac{d}{dx}(3x^2-4x)=6x-4$$

Now apply the Chain Rule:

$$\frac{dy}{dx}=\cos(3x^2-4x)(6x-4)$$

Answer: \(\boxed{(6x-4)\cos(3x^2-4x)}\)

This example shows that the Chain Rule works not only with powers, but also with trigonometric functions.

Worked Example 4: A function with more than one layer

Differentiate

$$y=(5x^2+1)^3$$

This has two layers:

• Outer layer: cube
• Inner layer: \(5x^2+1\)

Apply the Chain Rule:

$$\frac{dy}{dx}=3(5x^2+1)^2\cdot \frac{d}{dx}(5x^2+1)$$ $$\frac{dy}{dx}=3(5x^2+1)^2\cdot 10x$$ $$\frac{dy}{dx}=30x(5x^2+1)^2$$

Answer: \(\boxed{30x(5x^2+1)^2}\)

If you like, you can think of it as:

• Derivative of the outside: \(3(\text{inside})^2\)
• Times derivative of the inside: \(10x\)

Common patterns to remember

• $$\frac{d}{dx}(ax+b)^n=n(ax+b)^{n-1}\cdot a$$
• $$\frac{d}{dx}\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}}$$
• $$\frac{d}{dx}\sin(f(x))=\cos(f(x))\cdot f'(x)$$
• $$\frac{d}{dx}\cos(f(x))=-\sin(f(x))\cdot f'(x)$$
• $$\frac{d}{dx}e^{f(x)}=e^{f(x)}\cdot f'(x)$$
• $$\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}$$

These are all direct uses of the Chain Rule.

Common mistakes students make

• Forgetting to multiply by the derivative of the inside.
  This is the most common mistake.
• Changing the inside too early.
  When differentiating the outer function, keep the entire inner expression unchanged at first.
• Using the Chain Rule when it is not needed.
  For example, \(x^4\) is not composite in the same way as \((x^2+1)^4\).
• Missing multiple layers.
  Some functions have an inside inside another inside. In these cases, apply the Chain Rule more than once.

A quick comparison

Compare these two derivatives:

• \(\dfrac{d}{dx}(x^5)=5x^4\)
• \(\dfrac{d}{dx}(x^2+1)^5=5(x^2+1)^4\cdot 2x=10x(x^2+1)^4\)

The second one needs the Chain Rule because the quantity being raised to the fifth power is not just \(x\); it is the whole expression \(x^2+1\).

Practice check

Try these on your own:

• \(y=(4x-1)^6\)
• \(y=\cos(x^3)\)
• \(y=\ln(2x^2+5)\)

Answers:

• \(\dfrac{dy}{dx}=24(4x-1)^5\)
• \(\dfrac{dy}{dx}=-\sin(x^3)\cdot 3x^2\)
• \(\dfrac{dy}{dx}=\dfrac{4x}{2x^2+5}\)

Summary

The Chain Rule is used to differentiate composite functions, where one function is inside another. The key idea is: differentiate the outer function, leave the inner function unchanged, then multiply by the derivative of the inner function.

In symbols, if \(y=f(g(x))\), then

$$\frac{dy}{dx}=f'(g(x))\cdot g'(x)$$

As you practice, always ask yourself: What is the outer function? What is the inner function? If you can identify those clearly, the Chain Rule becomes much easier to apply correctly.

Derivatives of Transcendental Functions

In calculus, the derivative tells us the instantaneous rate of change of a function. You have already seen how derivatives work for powers such as \(x^2\), \(x^3\), and other algebraic functions. In this lesson, we extend differentiation to transcendental functions, especially trigonometric, exponential, and logarithmic functions.

These functions appear in many real situations. Trigonometric functions model waves and oscillations, exponential functions model growth and decay, and logarithmic functions help describe inverse relationships and scales such as pH and sound intensity. Knowing their derivatives is an essential part of differential calculus.

We will learn the main derivative formulas, understand where some of them come from using the limit definition, and practise applying them carefully.

1. Reminder: The limit definition of the derivative

The derivative of a function \(f(x)\) at a point is defined by

$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

This formula measures how fast the function changes at a specific value of \(x\). For many transcendental functions, the derivative formulas are usually memorized, but it is also helpful to know how they are connected to this limit definition.

2. Derivatives of trigonometric functions

The basic trigonometric derivative formulas are:

• \(\dfrac{d}{dx}(\sin x)=\cos x\)
• \(\dfrac{d}{dx}(\cos x)=-\sin x\)
• \(\dfrac{d}{dx}(\tan x)=\sec^2 x\)
• \(\dfrac{d}{dx}(\cot x)=-\csc^2 x\)
• \(\dfrac{d}{dx}(\sec x)=\sec x\tan x\)
• \(\dfrac{d}{dx}(\csc x)=-\csc x\cot x\)

These are true when angles are measured in radians. This is very important. The derivative formulas for sine and cosine work in this simple form only when the input is in radians.

How to derive \(\dfrac{d}{dx}(\sin x)=\cos x\)

Start with the limit definition:

$$\frac{d}{dx}(\sin x)=\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}$$

Use the angle addition identity:

$$\sin(x+h)=\sin x\cos h+\cos x\sin h$$

Substitute this into the limit:

$$\lim_{h\to 0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$$

Group terms:

$$\lim_{h\to 0}\left[\sin x\cdot \frac{\cos h-1}{h}+\cos x\cdot \frac{\sin h}{h}\right]$$

Now use the standard trigonometric limits:

$$\lim_{h\to 0}\frac{\sin h}{h}=1 \qquad \text{and} \qquad \lim_{h\to 0}\frac{\cos h-1}{h}=0$$

So we get

$$\sin x\cdot 0+\cos x\cdot 1=\cos x$$

Therefore,

$$\frac{d}{dx}(\sin x)=\cos x$$

How to derive \(\dfrac{d}{dx}(\cos x)=-\sin x\)

Similarly, begin with

$$\frac{d}{dx}(\cos x)=\lim_{h\to 0}\frac{\cos(x+h)-\cos x}{h}$$

Use the identity

$$\cos(x+h)=\cos x\cos h-\sin x\sin h$$

Then

$$\lim_{h\to 0}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}$$ $$=\lim_{h\to 0}\left[\cos x\cdot \frac{\cos h-1}{h}-\sin x\cdot \frac{\sin h}{h}\right]$$

Applying the same standard limits gives

$$\cos x\cdot 0-\sin x\cdot 1=-\sin x$$

So,

$$\frac{d}{dx}(\cos x)=-\sin x$$

Derivative of \(\tan x\)

Since \(\tan x=\dfrac{\sin x}{\cos x}\), we can use the quotient rule:

$$\frac{d}{dx}(\tan x)=\frac{\cos x(\cos x)-\sin x(-\sin x)}{\cos^2 x}$$ $$=\frac{\cos^2 x+\sin^2 x}{\cos^2 x}=\frac{1}{\cos^2 x}=\sec^2 x$$

3. Derivatives of exponential functions

The most important exponential derivative is:

• \(\dfrac{d}{dx}(e^x)=e^x\)

This is a special and very useful result. The function \(e^x\) is its own derivative, which means its rate of change is equal to its value at every point.

For a general exponential function with base \(a\), where \(a>0\) and \(a\neq 1\),

• \(\dfrac{d}{dx}(a^x)=a^x\ln a\)

Why is \(e^x\) special?

If we use the limit definition for \(f(x)=a^x\), then

$$f'(x)=\lim_{h\to 0}\frac{a^{x+h}-a^x}{h}$$

Factor out \(a^x\):

$$f'(x)=a^x\lim_{h\to 0}\frac{a^h-1}{h}$$

The limit depends only on the base \(a\), and that value is called \(\ln a\). So,

$$\frac{d}{dx}(a^x)=a^x\ln a$$

When \(a=e\), we have \(\ln e=1\), so

$$\frac{d}{dx}(e^x)=e^x$$

4. Derivatives of logarithmic functions

The main logarithmic derivative formulas are:

• \(\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}\), for \(x>0\)
• \(\dfrac{d}{dx}(\log_a x)=\dfrac{1}{x\ln a}\), for \(x>0\)

Here, \(\ln x\) means logarithm base \(e\), also called the natural logarithm.

Why is the derivative of \(\ln x\) equal to \(\dfrac{1}{x}\)?

The functions \(y=e^x\) and \(y=\ln x\) are inverses of each other. Since the derivative of \(e^x\) is \(e^x\), the derivative of its inverse turns out to be \(\dfrac{1}{x}\).

You can also remember this result as one of the key basic derivatives in calculus:

$$\frac{d}{dx}(\ln x)=\frac{1}{x}$$

5. Using the chain rule with transcendental functions

In many questions, the function is not just \(\sin x\) or \(e^x\), but something like \(\sin(3x)\) or \(\ln(x^2+1)\). In these cases, we use the chain rule.

If \(y=f(g(x))\), then

$$\frac{dy}{dx}=f'(g(x))\cdot g'(x)$$

This means:

• Differentiate the outer function.
• Keep the inner function unchanged.
• Multiply by the derivative of the inner function.

Examples of chain rule results:

• \(\dfrac{d}{dx}(\sin 3x)=\cos 3x\cdot 3=3\cos 3x\)
• \(\dfrac{d}{dx}(e^{2x})=e^{2x}\cdot 2=2e^{2x}\)
• \(\dfrac{d}{dx}(\ln(x^2+1))=\dfrac{1}{x^2+1}\cdot 2x=\dfrac{2x}{x^2+1}\)

6. Worked Examples

Example 1: Differentiate \(y=\sin x+e^x\)

Differentiate each term separately:

$$\frac{d}{dx}(\sin x)=\cos x$$ $$\frac{d}{dx}(e^x)=e^x$$

So,

$$\frac{dy}{dx}=\cos x+e^x$$

Answer: \(y'=\cos x+e^x\)

Example 2: Differentiate \(y=\tan x+\ln x\)

Use the basic derivative formulas:

$$\frac{d}{dx}(\tan x)=\sec^2 x$$ $$\frac{d}{dx}(\ln x)=\frac{1}{x}$$

Therefore,

$$\frac{dy}{dx}=\sec^2 x+\frac{1}{x}$$

Answer: \(y'=\sec^2 x+\dfrac{1}{x}\)

Example 3: Differentiate \(y=e^{3x}\)

This is an exponential function with an inner function \(3x\). Use the chain rule.

Outer function: \(e^u\), whose derivative is \(e^u\)

Inner function: \(u=3x\), whose derivative is \(3\)

So,

$$\frac{dy}{dx}=e^{3x}\cdot 3=3e^{3x}$$

Answer: \(y'=3e^{3x}\)

Example 4: Differentiate \(y=\ln(2x^2+5)\)

Again use the chain rule.

Outer function: \(\ln u\), whose derivative is \(\dfrac{1}{u}\)

Inner function: \(u=2x^2+5\), whose derivative is \(4x\)

Therefore,

$$\frac{dy}{dx}=\frac{1}{2x^2+5}\cdot 4x$$ $$\frac{dy}{dx}=\frac{4x}{2x^2+5}$$

Answer: \(y'=\dfrac{4x}{2x^2+5}\)

7. Common mistakes to avoid

• Forgetting radians: The standard derivatives of trigonometric functions are based on radian measure.
• Missing the negative sign: \(\dfrac{d}{dx}(\cos x)=-\sin x\), not \(\sin x\).
• Forgetting the chain rule: For example, \(\dfrac{d}{dx}(\sin 5x)=5\cos 5x\), not just \(\cos 5x\).
• Mixing up exponential derivatives: \(\dfrac{d}{dx}(e^x)=e^x\), but \(\dfrac{d}{dx}(a^x)=a^x\ln a\).
• Ignoring the domain of logarithms: \(\ln x\) is only defined for \(x>0\).

8. Key formulas to memorize

• \(\dfrac{d}{dx}(\sin x)=\cos x\)
• \(\dfrac{d}{dx}(\cos x)=-\sin x\)
• \(\dfrac{d}{dx}(\tan x)=\sec^2 x\)
• \(\dfrac{d}{dx}(e^x)=e^x\)
• \(\dfrac{d}{dx}(a^x)=a^x\ln a\)
• \(\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}\)
• \(\dfrac{d}{dx}(\log_a x)=\dfrac{1}{x\ln a}\)

9. Brief summary

Transcendental functions include trigonometric, exponential, and logarithmic functions. Their derivatives are fundamental in calculus and should be known well. Some of the most important results are \(\dfrac{d}{dx}(\sin x)=\cos x\), \(\dfrac{d}{dx}(e^x)=e^x\), and \(\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}\).

These formulas often combine with the chain rule, product rule, or quotient rule in more complicated questions. If you remember the core derivative rules and apply them carefully, you can differentiate a wide range of transcendental functions successfully.

Implicit Differentiation is a method used to find \\(\dfrac{dy}{dx}\\) when \\(x\\) and \\(y\\) are connected by an equation, but \\(y\\) is not written clearly as a function of \\(x\\).

In many problems, we can solve for \\(y\\) first and then differentiate. For example, if \\(y=x^2+3x\\), we can directly find the derivative. But sometimes the equation is mixed, such as \\(x^2+y^2=25\\). In this case, solving for \\(y\\) first is possible but not always convenient, and for more complicated equations it may be very difficult.

Implicit differentiation lets us differentiate both sides of the equation with respect to \\(x\\), treating \\(y\\) as a function of \\(x\\). That means whenever we differentiate a term involving \\(y\\), we must multiply by \\(\dfrac{dy}{dx}\\).

This idea comes from the chain rule. Since \\(y\\) depends on \\(x\\), differentiating \\(y\\) with respect to \\(x\\) gives

$$\frac{d}{dx}(y)=\frac{dy}{dx}.$$

More generally,

$$\frac{d}{dx}(y^n)=n y^{n-1}\frac{dy}{dx}.$$

This extra factor of \\(\dfrac{dy}{dx}\\) is the key feature of implicit differentiation.

When do we use implicit differentiation?

• When \\(y\\) cannot be easily isolated.
• When solving for \\(y\\) would be messy or time-consuming.
• When the equation describes a curve that is not a function in the usual sense.

For example, the equation \\(x^2+y^2=25\\) represents a circle. A circle is not a single function of \\(x\\), because one \\(x\\)-value can give two different \\(y\\)-values. Even so, we can still find the slope of the tangent line at points on the circle using implicit differentiation.

Main idea: Differentiate every term with respect to \\(x\\), remembering that terms with \\(y\\) need \\(\dfrac{dy}{dx}\\).

Basic derivative rules used in implicit differentiation

• \\(\dfrac{d}{dx}(x^n)=nx^{n-1}\\)
• \\(\dfrac{d}{dx}(y)=\dfrac{dy}{dx}\\)
• \\(\dfrac{d}{dx}(y^n)=ny^{n-1}\dfrac{dy}{dx}\\)
• \\(\dfrac{d}{dx}(xy)=x\dfrac{dy}{dx}+y\\) using the product rule
• Constants differentiate to 0

General steps for implicit differentiation

1. Differentiate both sides of the equation with respect to \\(x\\).
2. Whenever you differentiate a \\(y\\)-term, include \\(\dfrac{dy}{dx}\\).
3. Collect all terms containing \\(\dfrac{dy}{dx}\\) on one side.
4. Factor out \\(\dfrac{dy}{dx}\\).
5. Solve for \\(\dfrac{dy}{dx}\\).

Worked Example 1: A circle

Find \\(\dfrac{dy}{dx}\\) if

$$x^2+y^2=25.$$

Differentiate both sides with respect to \\(x\\):

$$\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(25).$$

This gives

$$2x+2y\frac{dy}{dx}=0.$$

Now solve for \\(\dfrac{dy}{dx}\\):

$$2y\frac{dy}{dx}=-2x$$

$$\frac{dy}{dx}=-\frac{2x}{2y}=-\frac{x}{y}.$$

So the derivative is

$$\frac{dy}{dx}=-\frac{x}{y}.$$

This tells us the slope of the tangent line at any point \\((x,y)\\) on the circle, as long as \\(y\neq 0\\).

Worked Example 2: An equation with an \\(xy\\) term

Find \\(\dfrac{dy}{dx}\\) if

$$x^2+xy+y^2=7.$$

Differentiate each term with respect to \\(x\\):

$$\frac{d}{dx}(x^2)+\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=\frac{d}{dx}(7).$$

Now apply the rules:

• \\(\dfrac{d}{dx}(x^2)=2x\\)
• For \\(xy\\), use the product rule: \\(\dfrac{d}{dx}(xy)=x\dfrac{dy}{dx}+y\\)
• \\(\dfrac{d}{dx}(y^2)=2y\dfrac{dy}{dx}\\)

So we get

$$2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0.$$

Group the terms with \\(\dfrac{dy}{dx}\\):

$$x\frac{dy}{dx}+2y\frac{dy}{dx}=-(2x+y).$$

Factor out \\(\dfrac{dy}{dx}\\):

$$\left(x+2y\right)\frac{dy}{dx}=-(2x+y).$$

Now solve:

$$\frac{dy}{dx}=\frac{-(2x+y)}{x+2y}.$$

So,

$$\frac{dy}{dx}=-\frac{2x+y}{x+2y}.$$

Worked Example 3: Higher powers

Find \\(\dfrac{dy}{dx}\\) if

$$x^3+y^3=6xy.$$

Differentiate both sides:

$$\frac{d}{dx}(x^3)+\frac{d}{dx}(y^3)=\frac{d}{dx}(6xy).$$

Now differentiate term by term:

$$3x^2+3y^2\frac{dy}{dx}=6\left(x\frac{dy}{dx}+y\right).$$

Expand the right side:

$$3x^2+3y^2\frac{dy}{dx}=6x\frac{dy}{dx}+6y.$$

Move all \\(\dfrac{dy}{dx}\\) terms to one side:

$$3y^2\frac{dy}{dx}-6x\frac{dy}{dx}=6y-3x^2.$$

Factor out \\(\dfrac{dy}{dx}\\):

$$\left(3y^2-6x\right)\frac{dy}{dx}=6y-3x^2.$$

Solve for the derivative:

$$\frac{dy}{dx}=\frac{6y-3x^2}{3y^2-6x}.$$

Simplify by dividing top and bottom by 3:

$$\frac{dy}{dx}=\frac{2y-x^2}{y^2-2x}.$$

Worked Example 4: Finding the slope at a specific point

For the curve

$$x^2+xy+y^2=12,$$

find the slope of the tangent line at the point \\((2,2)\\).

First, differentiate implicitly:

$$2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0.$$

Group the \\(\dfrac{dy}{dx}\\) terms:

$$\left(x+2y\right)\frac{dy}{dx}=-(2x+y).$$

So,

$$\frac{dy}{dx}=-\frac{2x+y}{x+2y}.$$

Now substitute \\(x=2\\) and \\(y=2\\):

$$\frac{dy}{dx}=-\frac{2(2)+2}{2+2(2)}=-\frac{6}{6}=-1.$$

The slope of the tangent line at \\((2,2)\\) is

$$-1.$$

Why does \\(\dfrac{dy}{dx}\\) appear?

Suppose we have \\(y^2\\). Since \\(y\\) depends on \\(x\\), we cannot treat \\(y\\) like a constant. We use the chain rule:

$$\frac{d}{dx}(y^2)=2y\cdot\frac{dy}{dx}.$$

This is similar to differentiating \\((u)^2\\), where \\(u\\) is a function of \\(x\\):

$$\frac{d}{dx}(u^2)=2u\frac{du}{dx}.$$

Here, \\(u=y\\), so \\(\dfrac{du}{dx}=\dfrac{dy}{dx}\\).

Common mistakes to avoid

• Forgetting \\(\dfrac{dy}{dx}\\) when differentiating a \\(y\\)-term. For example, \\(\dfrac{d}{dx}(y^3)\\) is not just \\(3y^2\\); it is \\(3y^2\dfrac{dy}{dx}\\).
• Not using the product rule for terms like \\(xy\\).
• Substituting a point too early. Usually, find \\(\dfrac{dy}{dx}\\) first, then substitute the point.
• Algebra errors when collecting and factoring the \\(\dfrac{dy}{dx}\\) terms.

Quick check of understanding

If

$$x^2+y^2=10,$$

then differentiating gives

$$2x+2y\frac{dy}{dx}=0.$$

So,

$$\frac{dy}{dx}=-\frac{x}{y}.$$

This means the slope changes depending on where you are on the curve. Implicit differentiation helps us find that slope directly from the relation.

Summary

• Use implicit differentiation when \\(x\\) and \\(y\\) are mixed in one equation and \\(y\\) is not easily isolated.
• Differentiate both sides with respect to \\(x\\).
• Every time you differentiate a \\(y\\)-term, include \\(\dfrac{dy}{dx}\\).
• Collect the \\(\dfrac{dy}{dx}\\) terms and solve for \\(\dfrac{dy}{dx}\\).
• This method is especially useful for curves like circles and other relations that are not easy to write as \\(y=f(x)\\).

Once you are comfortable with the chain rule, implicit differentiation becomes a very powerful and systematic tool for finding instantaneous rates of change on curves defined by equations.

Derivatives of Inverse Functions help us find the slope of a function like \(f^{-1}(x)\) by using what we already know about the original function \(f(x)\).

This idea is very useful because inverse functions can be hard to differentiate directly. Instead, we use a relationship between the slopes of a function and its inverse.

In this lesson, you will learn:

• what an inverse function is,
• how the derivative of an inverse function is connected to the original function,
• how to use the formula correctly,
• and how this applies to inverse trigonometric functions.

1. What is an inverse function?

A function \(f\) and its inverse \(f^{-1}\) undo each other.

If \(f(a)=b\), then \(f^{-1}(b)=a\).

For example, if \(f(x)=3x+2\), then solving \(y=3x+2\) for \(x\) gives

$$f^{-1}(x)=\frac{x-2}{3}$$

The graphs of a function and its inverse are reflections of each other across the line \(y=x\).

This reflection idea helps explain why their slopes are closely related.

2. The key derivative formula

If \(y=f^{-1}(x)\), then \(f(y)=x\).

Differentiate both sides with respect to \(x\):

$$\frac{d}{dx}[f(y)] = \frac{d}{dx}[x]$$

Using the chain rule,

$$f'(y)\cdot \frac{dy}{dx}=1$$

So,

$$\frac{dy}{dx}=\frac{1}{f'(y)}$$

Since \(y=f^{-1}(x)\), we get the standard formula:

$$\left(f^{-1}\right)'(x)=\frac{1}{f'\left(f^{-1}(x)\right)}$$

This is the main formula for derivatives of inverse functions.

It says:

• first evaluate the inverse function,
• then plug that result into \(f'\),
• then take the reciprocal.

3. A second way to think about it

If a point \((a,b)\) is on the graph of \(f\), then \((b,a)\) is on the graph of \(f^{-1}\).

If the slope of \(f\) at \((a,b)\) is \(f'(a)\), then the slope of \(f^{-1}\) at \((b,a)\) is the reciprocal:

$$\left(f^{-1}\right)'(b)=\frac{1}{f'(a)}$$

This only works when \(f'(a)\neq 0\). If the original slope is zero, its reciprocal is undefined.

4. Conditions for using the formula

The derivative formula for inverse functions works when:

• the function \(f\) is one-to-one, so an inverse exists,
• \(f\) is differentiable near the point,
• and \(f'(a)\neq 0\).

A function must be one-to-one because every input must match exactly one output, and every output must come from exactly one input for the inverse to be a function.

For example, \(f(x)=x^2\) is not one-to-one on all real numbers, but if we restrict the domain to \(x\ge 0\), then it does have an inverse: \(f^{-1}(x)=\sqrt{x}\).

5. How to use the formula step by step

1. Identify the original function \(f(x)\).
2. Find the input value \(a\) such that \(f(a)=x\), or write \(a=f^{-1}(x)\).
3. Compute \(f'(a)\).
4. Take the reciprocal.

At a specific point, a very efficient version is:

• if \(f(a)=b\), then \(\left(f^{-1}\right)'(b)=\dfrac{1}{f'(a)}\).

Worked Example 1: A simple algebraic function

Let \(f(x)=2x-5\). Find \(\left(f^{-1}\right)'(7)\).

Step 1: Find the input \(a\) such that \(f(a)=7\).

$$2a-5=7$$ $$2a=12$$ $$a=6$$

So \(f(6)=7\).

Step 2: Differentiate the original function.

$$f'(x)=2$$

Therefore,

$$f'(6)=2$$

Step 3: Use the reciprocal relationship.

$$\left(f^{-1}\right)'(7)=\frac{1}{f'(6)}=\frac{1}{2}$$

Answer: \(\boxed{\frac{1}{2}}\)

This makes sense because the inverse of \(2x-5\) is \(\dfrac{x+5}{2}\), whose derivative is indeed \(\dfrac{1}{2}\).

Worked Example 2: A nonlinear function

Let \(f(x)=x^3+1\). Find \(\left(f^{-1}\right)'(9)\).

Step 1: Find \(a\) such that \(f(a)=9\).

$$a^3+1=9$$ $$a^3=8$$ $$a=2$$

So \(f(2)=9\).

Step 2: Differentiate.

$$f'(x)=3x^2$$

Then

$$f'(2)=3(2^2)=12$$

Step 3: Take the reciprocal.

$$\left(f^{-1}\right)'(9)=\frac{1}{12}$$

Answer: \(\boxed{\frac{1}{12}}\)

6. Inverse trigonometric functions

Inverse trigonometric functions are very important applications of this idea.

For example, if

$$y=\sin^{-1}(x)$$

then

$$\sin(y)=x$$

Differentiate both sides:

$$\cos(y)\cdot \frac{dy}{dx}=1$$

So,

$$\frac{dy}{dx}=\frac{1}{\cos(y)}$$

Now we need to rewrite \(\cos(y)\) in terms of \(x\).

Since \(\sin(y)=x\), imagine a right triangle where:

• opposite side = \(x\)
• hypotenuse = \(1\)

Then the adjacent side is

$$\sqrt{1-x^2}$$

So

$$\cos(y)=\sqrt{1-x^2}$$

Therefore,

$$\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}$$

for \(-1<x<1\).

In the same way, you can get other inverse trig derivatives:

• $$\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}$$
• $$\frac{d}{dx}(\cos^{-1}x)=\frac{-1}{\sqrt{1-x^2}}$$
• $$\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}$$

You do not need to memorize where these come from without understanding. They all follow from the derivative of an inverse function plus implicit differentiation.

Worked Example 3: Derivative of an inverse trigonometric function

Find the slope of \(y=\sin^{-1}(x)\) at \(x=\dfrac{1}{2}\).

Use the formula:

$$\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}$$

Substitute \(x=\dfrac{1}{2}\):

$$y'=\frac{1}{\sqrt{1-\left(\frac{1}{2}\right)^2}}$$ $$=\frac{1}{\sqrt{1-\frac{1}{4}}}$$ $$=\frac{1}{\sqrt{\frac{3}{4}}}$$ $$=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$$

Answer: \(\boxed{\frac{2}{\sqrt{3}}}\)

Worked Example 4: Using the reciprocal slope relationship at a point

Suppose \(f(3)=5\) and \(f'(3)=4\). Find \(\left(f^{-1}\right)'(5)\).

Because \(f(3)=5\), the point \((3,5)\) is on \(f\), so \((5,3)\) is on \(f^{-1}\).

Use the inverse derivative rule:

$$\left(f^{-1}\right)'(5)=\frac{1}{f'(3)}=\frac{1}{4}$$

Answer: \(\boxed{\frac{1}{4}}\)

7. Common mistakes to avoid

• Forgetting to use the matching input from the original function. If you want \(\left(f^{-1}\right)'(b)\), first find \(a\) so that \(f(a)=b\).
• Not taking the reciprocal. The slope of the inverse is not the same as the original slope; it is the reciprocal.
• Using a function that is not one-to-one. If the function does not have an inverse on its domain, the formula does not apply directly.
• Confusing \(f^{-1}(x)\) with \(\frac{1}{f(x)}\). These are completely different. \(f^{-1}(x)\) means inverse function, not reciprocal of the function.

8. Quick check idea

If the original function has a large slope at a point, then its inverse should have a small slope at the corresponding point.

If the original function has a slope of \(5\), the inverse has slope \(\dfrac{1}{5}\). If the original has slope \(\dfrac{1}{3}\), the inverse has slope \(3\).

This reciprocal pattern is a helpful way to check whether your answer is reasonable.

9. Summary

The derivative of an inverse function comes from the fact that the function and its inverse undo each other.

The main formula is

$$\left(f^{-1}\right)'(x)=\frac{1}{f'\left(f^{-1}(x)\right)}$$

At a point, if \(f(a)=b\), then

$$\left(f^{-1}\right)'(b)=\frac{1}{f'(a)}$$

This reciprocal slope relationship is especially useful for inverse trig functions, such as

$$\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}$$

When using this topic, always check that the function has an inverse, match the correct point, and remember to take the reciprocal of the original slope.

Higher-Order Derivatives are derivatives taken more than once. If the first derivative tells us the instantaneous rate of change of a function, then the second, third, and later derivatives tell us how that rate of change is itself changing.

In 12th Grade Maths, higher-order derivatives are important because they help us study velocity and acceleration in motion, and also help us understand the shape of graphs, especially concavity and changes in curvature.

This lesson will show you how to compute higher-order derivatives step by step, how to write them correctly, and how to use them in practical interpretation.

1. What is a higher-order derivative?

Suppose we start with a function \(y=f(x)\).

The first derivative is

$$f'(x)=\frac{dy}{dx}$$

This gives the rate of change of \(f(x)\) with respect to \(x\).

If we differentiate again, we get the second derivative:

$$f''(x)=\frac{d^2y}{dx^2}$$

If we differentiate a third time, we get the third derivative:

$$f'''(x)=\frac{d^3y}{dx^3}$$

Continuing this process gives the \(n\)th derivative:

$$f^{(n)}(x)=\frac{d^n y}{dx^n}$$

The key idea is simple: differentiate repeatedly.

2. Notation for higher-order derivatives

You may see several ways to write higher-order derivatives:

• \(f'(x), f''(x), f'''(x)\)
• \(y', y'', y'''\)
• \(\dfrac{d^2y}{dx^2}, \dfrac{d^3y}{dx^3}\)
• \(f^{(n)}(x)\) for the \(n\)th derivative

These all mean the same idea, but the notation with primes is most common for the first few derivatives.

3. How to find higher-order derivatives

To find higher-order derivatives:

1. Find the first derivative as usual.
2. Differentiate the result to get the second derivative.
3. Repeat as needed.

You use the same differentiation rules each time, such as the power rule, product rule, quotient rule, and chain rule.

For example, if \(f(x)=x^5\), then

$$f'(x)=5x^4$$ $$f''(x)=20x^3$$ $$f'''(x)=60x^2$$ $$f^{(4)}(x)=120x$$ $$f^{(5)}(x)=120$$ $$f^{(6)}(x)=0$$

This shows an important pattern for polynomials: after enough differentiations, the derivative becomes zero.

4. Meaning of the second derivative

The second derivative tells us how the first derivative is changing.

If \(f'(x)\) represents slope, then \(f''(x)\) tells us whether that slope is increasing or decreasing.

• If \(f''(x)>0\), the graph is concave up.
• If \(f''(x)<0\), the graph is concave down.
• If \(f''(x)=0\), there may be a possible change in concavity, but you must check the graph or sign of \(f''(x)\) around that point.

This is why the second derivative is very useful in graph analysis.

5. Higher-order derivatives in motion

When a function gives position \(s(t)\) at time \(t\):

• The first derivative \(s'(t)\) is velocity.
• The second derivative \(s''(t)\) is acceleration.
• The third derivative \(s'''(t)\) describes how acceleration changes.

At this level, the most important physical meaning is that the second derivative gives acceleration.

6. Worked Example 1: A basic polynomial

Find the first, second, and third derivatives of

$$f(x)=3x^4-2x^3+5x-7$$

Step 1: First derivative

Use the power rule term by term:

$$f'(x)=12x^3-6x^2+5$$

Step 2: Second derivative

$$f''(x)=36x^2-12x$$

Step 3: Third derivative

$$f'''(x)=72x-12$$

Answer:

$$f'(x)=12x^3-6x^2+5, \quad f''(x)=36x^2-12x, \quad f'''(x)=72x-12$$

Notice that each step is just another differentiation of the previous result.

7. Worked Example 2: Using the second derivative for concavity

For the function

$$f(x)=x^3-3x^2+2$$

find \(f''(x)\) and determine where the graph is concave up or concave down.

Step 1: First derivative

$$f'(x)=3x^2-6x$$

Step 2: Second derivative

$$f''(x)=6x-6$$

Step 3: Find where \(f''(x)=0\)

$$6x-6=0$$ $$x=1$$

Step 4: Test the sign of \(f''(x)\)

Choose a value less than 1, such as \(x=0\):

$$f''(0)=6(0)-6=-6$$

So \(f''(x)<0\) for \(x<1\), which means the graph is concave down there.

Choose a value greater than 1, such as \(x=2\):

$$f''(2)=6(2)-6=6$$

So \(f''(x)>0\) for \(x>1\), which means the graph is concave up there.

Conclusion:

• Concave down for \(x<1\)
• Concave up for \(x>1\)

The concavity changes at \(x=1\), so this is an inflection point.

8. Worked Example 3: Motion and acceleration

A particle moves along a line with position

$$s(t)=2t^3-9t^2+12t+1$$

Find the velocity and acceleration.

Step 1: Velocity is the first derivative

$$v(t)=s'(t)=6t^2-18t+12$$

Step 2: Acceleration is the second derivative

$$a(t)=s''(t)=12t-18$$

Answer:

$$v(t)=6t^2-18t+12$$ $$a(t)=12t-18$$

Interpretation: velocity tells how fast position is changing, while acceleration tells how fast velocity is changing.

For example, at \(t=2\):

$$a(2)=12(2)-18=6$$

This means the velocity is increasing at \(6\) units per second squared at that moment.

9. Worked Example 4: Finding an \(n\)th derivative pattern

Find the \(n\)th derivative of

$$f(x)=x^6$$

Differentiate repeatedly:

$$f'(x)=6x^5$$ $$f''(x)=30x^4$$ $$f'''(x)=120x^3$$ $$f^{(4)}(x)=360x^2$$ $$f^{(5)}(x)=720x$$ $$f^{(6)}(x)=720$$ $$f^{(7)}(x)=0$$

So the derivatives continue until the exponent drops to zero, and then the next derivative is zero.

In general, for powers of \(x\), each differentiation lowers the exponent by 1 and multiplies by the current exponent.

For this function:

• \(f^{(1)}(x)=6x^5\)
• \(f^{(2)}(x)=6\cdot 5x^4\)
• \(f^{(3)}(x)=6\cdot 5\cdot 4x^3\)

This pattern helps when finding higher derivatives of polynomial terms.

10. Useful patterns to remember

These patterns often appear in higher-order derivative questions:

• For \(f(x)=x^m\), repeated differentiation keeps reducing the power.
• For polynomial functions, higher derivatives eventually become zero.
• The second derivative is especially important for concavity and acceleration.

Here are some quick examples:

$$\frac{d}{dx}(x^4)=4x^3, \quad \frac{d^2}{dx^2}(x^4)=12x^2, \quad \frac{d^3}{dx^3}(x^4)=24x, \quad \frac{d^4}{dx^4}(x^4)=24$$ $$\frac{d}{dx}(7)=0 \Rightarrow \text{all later derivatives are also } 0$$

11. Common mistakes to avoid

• Forgetting to differentiate the derivative: the second derivative is the derivative of the first derivative, not of the original function again.
• Dropping constants incorrectly: a constant term disappears after one differentiation.
• Confusing \(f''(x)=0\) with an inflection point: it is only a possible inflection point. The concavity must actually change.
• Stopping too early: if the question asks for the third or \(n\)th derivative, keep going carefully.
• Sign errors: these are common, especially with negative terms.

12. Step-by-step strategy for exam questions

When solving a higher-order derivative question, follow this method:

1. Write the original function clearly.
2. Find the first derivative carefully.
3. Differentiate again for the second derivative.
4. Continue until you reach the required order.
5. If interpreting the result, connect it to the meaning of the derivative:
   • first derivative  rate of change or velocity
   • second derivative  acceleration or concavity

13. Brief recap

Higher-order derivatives are found by differentiating a function repeatedly. The first derivative gives rate of change, the second derivative gives the rate of change of that rate, and so on.

The second derivative is especially useful because it tells us about acceleration in motion and concavity in graphs. For polynomials, repeated differentiation follows a pattern and eventually gives zero.

If you stay organized and differentiate one step at a time, higher-order derivatives become much easier to handle.