Reverse Chain Rule (Cambridge (CIE) IGCSE Additional Maths): Revision Note

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 $\int {\mathrm{e}}^{\mathrm{f}\left(x\right)}\mathrm{d}x\ne \frac{1}{\mathrm{f}\text{'}\left(x\right)}{\mathrm{e}}^{\mathrm{f}\left(x\right)}$

  • However, this result is true if$\mathrm{f}\left(x\right)$ is linear$(ax+b)$

• Formally, in function notation, the reverse chain rule is used for integrands of the form 

$I=\int g^{\text{'}}\left(x\right)f^{\text{'}}\left(g\left(x\right)\right)\mathrm{d}x$

• 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, ${\mathrm{e}}^{x^2}$ differentiates to $2x{\mathrm{e}}^{x^2}$ with the  chain rule

    • so$\int 2x{\mathrm{e}}^{x^2}\mathrm{d}x={\mathrm{e}}^{x^2}+c$ with the reverse chain rule

  • But to do $\int 5x{\mathrm{e}}^{x^2}\mathrm{d}x$  we need to:

    • Take out the five: $5\int x{\mathrm{e}}^{x^2}\mathrm{d}x$

    • Force a 2 inside (adjust) and divide the outside by a 2 (compensate): $\frac{5}{2}\int 2x{\mathrm{e}}^{x^2}\mathrm{d}x$

    • The bit inside the integral is now a reverse chain rule

    • The answer is $\frac{5}{2}{\mathrm{e}}^{x^2}+c$

• A particularly useful instance of the reverse chain rule to recognise is

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

• 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.  $I=\int \frac{x^2+1}{x^3+3x}\mathrm{d}x=\frac{1}{3}\int 3\frac{x^2+1}{x^3+3x}\mathrm{d}x=\frac{1}{3}\int \frac{3x^2+3}{x^3+3x}\mathrm{d}x=\frac{1}{3}\ln |x^3+3x|+c$

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$\mathit{a}\mathit{x}\mathbf{+}\mathit{b}$

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 $\int {(ax+b)}^{n}dx$(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

    • $\int {(ax + b)}^n\mathrm{ }\mathrm{d}x=\frac{{\left(ax+b\right)}^{n+1}}{n+1}\times \frac{1}{a}+c$

• 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

  •  $\int \sin (ax+b)\mathrm{d}x=-\frac{1}{a}\cos (ax+b)+c$

  •  $\int \cos (ax+b)\mathrm{d}x=\frac{1}{a}\sin (ax+b)+c$

  • $\int {\mathrm{e}}^{ax+b}\mathrm{d}x=\frac{1}{a}{\mathrm{e}}^{ax+b}+c$

  • $\int \frac{1}{ax+b}\mathrm{d}x=\frac{1}{a}\ln \left|ax+b\right|+c$

• $c$, 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