The Reverse Chain Rule Revision Notes for HSC SSCE Mathematics Advanced

The Reverse Chain Rule

Introduction

When we differentiate a composite function using the chain rule, the result is a product of two terms. For example:

$\frac{d}{dx}(x^3 + 5)^4 = 4(x^3 + 5)^3 \times 3x^2$

The reverse chain rule allows us to work backwards from this product to find the original function. This technique is essential for finding primitives (antiderivatives) of more complex functions.

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The chain rule produces a characteristic pattern: the derivative of the outer function multiplied by the derivative of the inner function. Recognizing this pattern is the key to applying the reverse chain rule successfully.

Reversing chain-rule differentiation

Finding primitives is the reverse process of differentiation. Every differentiation result can be written in reverse as an indefinite integral.

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Key principle: Once a chain-rule differentiation has been performed, the result can be written down in reverse as an indefinite integral.

Basic example

Let's see how this works with a complete example:

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Worked Example: Understanding the Reverse Process

Part a: Differentiate $(x^3 + 5)^4$ using the chain rule

Let $y = (x^3 + 5)^4$

Let $u = x^3 + 5$

Then $y = u^4$

Hence $\frac{du}{dx} = 3x^2$ and $\frac{dy}{du} = 4u^3$

Using the chain rule:

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

$= 4(x^3 + 5)^3 \times 3x^2$

$= 12x^2(x^3 + 5)^3$

Part b: Find a primitive of $12x^2(x^3 + 5)^3$

From part a, we know that:

$\frac{d}{dx}(x^3 + 5)^4 = 12x^2(x^3 + 5)^3$

Reversing this differentiation:

$\int 12x^2(x^3 + 5)^3 dx = (x^3 + 5)^4 + C$

Part c: Find the primitive of $x^2(x^3 + 5)^3$

Notice that $x^2(x^3 + 5)^3$ is $\frac{1}{12}$ of the expression in part b.

Dividing by $12$ :

$\int x^2(x^3 + 5)^3 dx = \frac{1}{12}(x^3 + 5)^4 + C$

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Important note: It's best not to add the arbitrary constant $C$ until the last line. Dividing the constant would be pointless since it remains arbitrary.

The reverse chain rule formula

The chain rule for differentiation can be written in two equivalent forms:

$\frac{d}{dx}(u^n) = nu^{n-1}\frac{du}{dx}$

$\frac{d}{dx}(f(x))^n = n(f(x))^{n-1}f'(x)$

These can both be reversed to give integration formulas:

$\int u^{n-1} \frac{du}{dx} dx = \frac{u^n}{n} + C$

$\int (f(x))^{n-1} f'(x) dx = \frac{(f(x))^n}{n} + C$

By replacing $n-1$ with $n$ and $n$ with $n+1$ , we get the standard forms:

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Standard Forms of the Reverse Chain Rule

Form 1: $\int u^n \frac{du}{dx} dx = \frac{u^{n+1}}{n+1} + C$

Form 2: $\int (f(x))^n f'(x) dx = \frac{(f(x))^{n+1}}{n+1} + C$

You can use either formula - they give the same result. The key challenge is identifying which part of the integrand should be $u$ or $f(x)$ .

Essential steps for using the formula

Follow these important steps when applying the reverse chain rule:

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Recommended Working Method:

Show working that identifies $u$ or $f(x)$ and its derivative on the right-hand side
Write down the standard form, substituting the particular value of $n$
Complete the integration using the formula

Worked examples

Example 1: Using the reverse chain rule formula

Find $\int x^2(x^3 + 5)^3 dx$

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Worked Example: Applying the Formula

The crucial insight is recognizing that $x^2$ is a multiple of the derivative of $x^3 + 5$ .

Since $\frac{d}{dx}(x^3 + 5) = 3x^2$ , we can see that $x^2 = \frac{1}{3} \times 3x^2$ .

Using the first formula:

Let $u = x^3 + 5$

Then $\frac{du}{dx} = 3x^2$

Here we use: $\int u^3 \frac{du}{dx} dx = \frac{1}{4}u^4 + C$

Working:

$\int x^2(x^3 + 5)^3 dx = \frac{1}{3}\int 3x^2(x^3 + 5)^3 dx$

$= \frac{1}{3} \times \frac{1}{4}(x^3 + 5)^4 + C$

$= \frac{1}{12}(x^3 + 5)^4 + C$

Using the second formula:

Let $f(x) = x^3 + 5$

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

Here we use: $\int (f(x))^3 f'(x) dx = \frac{1}{4}(f(x))^4 + C$

Working:

$\int x^2(x^3 + 5)^3 dx = \frac{1}{3}\int 3x^2(x^3 + 5)^3 dx$

$= \frac{1}{3} \times \frac{1}{4}(x^3 + 5)^4 + C$

$= \frac{1}{12}(x^3 + 5)^4 + C$

Notice that both approaches give identical results - they differ only in the notation used.

Example 2: Integrating with square roots

Find $\int x\sqrt{1 - x^2} dx$

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Worked Example: Working with Square Roots

This integral is based on recognizing that $\frac{d}{dx}(1 - x^2) = -2x$ .

We can rewrite the square root as a power: $\sqrt{1 - x^2} = (1 - x^2)^{\frac{1}{2}}$

Using the first formula:

Let $u = 1 - x^2$

Then $\frac{du}{dx} = -2x$

Here we use: $\int u^{\frac{1}{2}} \frac{du}{dx} dx = \frac{2}{3}u^{\frac{3}{2}} + C$

Working:

$\int x\sqrt{1 - x^2} dx = -\frac{1}{2}\int(-2x) \times (1 - x^2)^{\frac{1}{2}} dx$

$= -\frac{1}{2} \times \frac{2}{3}(1 - x^2)^{\frac{3}{2}} + C$

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

Using the second formula:

Let $f(x) = 1 - x^2$

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

Here we use: $\int (f(x))^{\frac{1}{2}} f'(x) dx = \frac{2}{3}(f(x))^{\frac{3}{2}} + C$

Working:

$\int x\sqrt{1 - x^2} dx = -\frac{1}{2}\int(-2x) \times (1 - x^2)^{\frac{1}{2}} dx$

$= -\frac{1}{2} \times \frac{2}{3}(1 - x^2)^{\frac{3}{2}} + C$

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

Important note about definite integrals

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When evaluating definite integrals, always check the domain. For example, $\int_0^2 x\sqrt{1 - x^2} dx$ is meaningless because $\sqrt{1 - x^2}$ is undefined for $x > 1$ (as this would require taking the square root of a negative number).

Key points for success

To use the reverse chain rule formula effectively, you must write the integrand as a product with two factors:

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Pattern Recognition is Key

One factor is a power of a function $f(x)$ or $u$
The other factor is the derivative of that function

The pattern to look for is:

$\text{(derivative)} \times \text{(power of inner function)}$

You may need to manipulate the integral by factoring out constants to match this pattern exactly.

Remember!

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Key Points to Remember:

The reverse chain rule undoes chain-rule differentiation to find primitives
Always identify the inner function ( $u$ or $f(x)$ ) and check that its derivative appears in the integrand
Use the formula: $\int u^n \frac{du}{dx} dx = \frac{u^{n+1}}{n+1} + C$ or $\int (f(x))^n f'(x) dx = \frac{(f(x))^{n+1}}{n+1} + C$
Show your working clearly: identify $u$ or $f(x)$ , find its derivative, and state the value of $n$
You may need to factor out constants to match the derivative exactly
Add the constant of integration $C$ only in your final answer

The Reverse Chain Rule (HSC SSCE Mathematics Advanced): Revision Notes