Reverse Chain Rule (Cambridge O Level Additional Maths)

Revision Note

Author

Paul

Last updated

10 July 2024

Reverse Chain Rule

What is the reverse chain rule?

The Chain Rule is a way of differentiating two (or more) functions
The Reverse Chain Rule (RCR) refers to integrating by inspection
- Spotting that chain rule would be used in the reverse (differentiating) process

How do I know when to use the reverse chain rule?

The reverse chain rule is used when we have the product of a composite function and the derivative of its second function
Integration is trickier than differentiation; many of the shortcuts do not work
- For example, in general $\int e^{f (x)} d x \neq \frac{1}{f' (x)} e^{f (x)}$
- However, this result is true if $f (x)$ is linear $(a x + b)$
Formally, in function notation, the reverse chain rule is used for integrands of the form

$I = \int g^{'} (x) f^{'} (g (x)) d x$

- This does not have to be strictly true, but ‘algebraically’ it should be
If the coefficients do not match ‘adjust and compensate’ can be used
- For example, differentiates to with the chain rule
  - so $\int 2 x e^{x^{2}} d x = e^{x^{2}} + c$ with the reverse chain rule
- But to do we need to:
  - Take out the five: $5 \int x e^{x^{2}} d x$
  - Force a 2 inside (adjust) and divide the outside by a 2 (compensate): $\frac{5}{2} \int 2 x e^{x^{2}} d x$
  - The bit inside the integral is now a reverse chain rule
  - The answer is $\frac{5}{2} e^{x^{2}} + c$
A particularly useful instance of the reverse chain rule to recognise is

$I = \int \frac{f' (x)}{f (x)} d x = \ln | f (x) | + c$

- i.e. the numerator is (almost) the derivative of the denominator
- 'adjust and compensate' may need to be used to deal with any coefficients
  - e.g. $I = \int \frac{x^{2} + 1}{x^{3} + 3 x} d x = \frac{1}{3} \int 3 \frac{x^{2} + 1}{x^{3} + 3 x} d x = \frac{1}{3} \int \frac{3 x^{2} + 3}{x^{3} + 3 x} d x = \frac{1}{3} \ln | x^{3} + 3 x | + c$

Examiner Tip

You can always check your work by differentiating, if you have time

Worked example

A curve has the gradient function $f' (x) = 5 x^{2} \sin (2 x^{3})$ .

Given that the curve passes through the point $(0, 1)$ , find an expression for $f (x)$ .

Write f(x) as an integral.

$f (x) = \int 5 x^{2} \sin (2 x^{3}) d x$

Take 5 out of the integral as a factor.

$f (x) = 5 \int x^{2} \sin (2 x^{3}) d x$

The main function is sin(...), which would have come from -cos(...).

Adjust and compensate the coefficients.
2x³would differentiate to 6x²so -cos(2x³) would differentiate to (6x²)sin(2x³)

$f (x) = 5 \times (\frac{1}{6}) \times \int (6 x^{2}) \sin (2 x^{3}) d x$

Integrate.

$f (x) = (\frac{5}{6}) \times - \cos (2 x^{3}) + c$

$f (x) = - \frac{5}{6} \cos (2 x^{3}) + c$

Integrating Composite Functions (ax+b)

What is a composite function?

A composite function involves one function being applied after another
A composite function may be described as a “function of a function”
This Revision Note focuses on one of the functions being linear – i.e. of the form $a x + b$

How do I integrate linear (ax+b) functions?

The reverse chain rule can be used for integrating functions in the form y = (ax + b)ⁿ
- Make sure you are confident using the chain rule to differentiate functions in the form y = (ax + b)ⁿ
- The reverse chain rule works backwards
For n = 2 you will most likely expand the brackets and integrate each term separately
If n > 2 this becomes time-consuming and if n is not a positive integer we need a different method completely
To use the reverse chain rule $\int {(a x + b)}^{n} d x$ (provided n is not -1)
- Raise the power of n by 1
- Divide by this new power
- Divide this whole function by the coefficient of x
  - $\int {(a x + b)}^{n} d x = \frac{{(a x + b)}^{n + 1}}{n + 1} \times \frac{1}{a} + c$
You can check your answer by differentiating it
- You should get the original function when you differentiate your answer
Note that this method only works when the function in the brackets is linear (ax + b)
The special cases for trigonometric functions and exponential and logarithmic functions are
- $\int \sin (a x + b) d x = - \frac{1}{a} \cos (a x + b) + c$
- $\int \cos (a x + b) d x = \frac{1}{a} \sin (a x + b) + c$
- $\int e^{a x + b} d x = \frac{1}{a} e^{a x + b} + c$
- $\int \frac{1}{a x + b} d x = \frac{1}{a} \ln |a x + b| + c$
$c$ , in all cases, is the constant of integration
All the above can be deduced using reverse chain rule
- However, spotting them can make solutions more efficient

Worked example

Find the following integrals

\int 3 {(7 - 2 x)}^{\frac{5}{3}} d x

Name the integral.

$I = \int 3 {(7 - 2 x)}^{\frac{5}{3}} d x = 3 \int {(7 - 2 x)}^{\frac{5}{3}} d x$

Using the rule 'raise the power by one, divide by the new power and then multiply by the reciprocal of the derivative' integrate the expression.

$I = 3 [\frac{1}{\frac{8}{3}} {(7 - 2 x)}^{\frac{8}{3}} \times \frac{1}{- 2}] + c$

Simplify.

$I = 3 [- \frac{3}{16} {(7 - 2 x)}^{\frac{8}{3}}] + c$

$I = - \frac{9}{16} {(7 - 2 x)}^{\frac{8}{3}} + c$

\int \frac{1}{2} \cos (3 x - 2) d x

Name the integral.

$I = \int \frac{1}{2} \cos (3 x - 2) d x = \frac{1}{2} \int \cos (3 x - 2) d x$

Using the rule $\int \cos (a x + b) d x = \frac{1}{a} \sin (a x + b) + c$ , integrate the expression.

$I = \frac{1}{2} [\frac{1}{3} \sin (3 x - 2)] + c$

Simplify.

$I = \frac{1}{6} \sin (3 x - 2) + c$

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Reverse Chain Rule (Cambridge O Level Additional Maths)

Revision Note

What is the reverse chain rule?

How do I know when to use the reverse chain rule?

What is a composite function?

How do I integrate linear (ax+b) functions?

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

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