# Reverse Chain Rule

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Reverse Chain Rule

The chain rule allows us to differentiate in terms of something other than $x$, and we end up with a product of two derivatives.

We can do this in reverse to integrate complicated functions where a function and its derivative both appear in that which is to be integrated.

A Level

## The Reverse Chain Rule

Recall: The chain rule.

$\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}$

Now the reverse chain rule:

${\LARGE \int}\dfrac{du}{dx}f'(u)dx=f(u)+c$

The easiest way to spot when to use this is to look for a function and its derivative.

A Level

## Look for a Function and Its Derivative

Some general results from the reverse chain rule appear so frequently that they are worth remembering.

${\LARGE \int}(n+1)f'(x)(f(x))^{n}dx=(f(x))^{n+1}+c$

The multiple in the integral is $n+1$, not $n$, which is very important to note.

A Level
A Level

## Example 1: Reverse Chain Rule

Find the integral of $2\cos(2x)e^{\sin(2x)}$

[2 marks]

$\dfrac{d}{dx}(\sin(2x))=2\cos(2x)$ so our integral is of the form ${\LARGE \int}\dfrac{du}{dx}f'(u)dx$ where $f'(u)=e^{u}$. Hence:

$\int2\cos(2x)e^{\sin(2x)}dx=e^{\sin(2x)}+c$

A Level

## Example 2: A Function and its Derivative

Find $\int\dfrac{4}{x}(\ln(x))^{3}dx$

[2 marks]

Since $\dfrac{d}{dx}(\ln(x))=\dfrac{1}{x}$, this is of the form $\int(n+1)f'(x)(f(x))^{n}dx$ where $n=3$ and $f(x)=\ln(x)$. Hence:

${\LARGE \int}\dfrac{4}{x}(\ln(x))^{3}dx=(\ln(x))^{4}+c$

A Level

## Example Questions

i) $\dfrac{d}{dx}(x^{3})=3x^{2}$ so this is in reverse chain rule form. Hence:

\begin{aligned}\int3x^{2}e^{x^{3}}dx=e^{x^{3}}+c\end{aligned}

ii) $\dfrac{d}{dx}\left(\dfrac{1}{2}x^{8}+4\right)=4x^{7}$ so this is in reverse chain rule form. Hence:

\begin{aligned}\int4x^{7}\left(\dfrac{1}{2}x^{8}+4\right)^{2}dx=\dfrac{1}{3}\left(\dfrac{1}{2}x^{8}+4\right)^{3}+c\end{aligned}

iii) $\dfrac{d}{dx}\left(\dfrac{1}{3}x^{3}\right)=x^{2}$ so this is in reverse chain rule form. Hence:

\begin{aligned}\int x^{2}\cos\left(\dfrac{1}{3}x^{3}\right)dx=\sin\left(\dfrac{1}{3}x^{3}\right)+c\end{aligned}

i) $\dfrac{d}{dx}(x^{4})=4x^{3}$ so this is in reverse chain rule form. Hence:

\begin{aligned}\int12x^{3}e^{x^{4}}dx&=3\int4x^{3}e^{x^{4}}dx\\[1.2em]&=3e^{x^{4}}+c\end{aligned}

ii) $\dfrac{d}{dx}(\log(x))=\dfrac{1}{x}$ so this is in reverse chain rule form. Hence:

\begin{aligned}\int\dfrac{2}{x}(\log(x))^{5}dx&=2\times\dfrac{1}{6}(\log(x))^{6}+c\\[1.2em]&=\dfrac{1}{3}(\log(x))^{6}+c\end{aligned}

iii) $\dfrac{d}{dx}(\tan(x))=\sec^{2}(x)$ so this is in reverse chain rule form. Hence:

\begin{aligned}\int4sec^{2}(x)e^{\tan(x)}dx=4e^{\tan(x)}+c\end{aligned}

i) $\dfrac{d}{dx}(x^{3}+3x^{2}+x+5)=3x^{2}+6x+1$ so this is in reverse chain rule form. Hence:

\begin{aligned}&\int(6x^{2}+12x+2)e^{x^{3}+3x^{2}+x+5}dx\\[1.2em]&=2\int(3x^{2}+6x+1)e^{x^{3}+3x^{2}+x+5}dx\\[1.2em]&=e^{x^{3}+3x^{2}+x+5}+c\end{aligned}

ii) $\dfrac{d}{dx}(3x^{2}+2x+9)=6x+2$ so this is in reverse chain rule form. Hence:

\begin{aligned}&\int(12x+4)(3x^{2}+2x+9)^{6}dx\\[1.2em]&=2\int(6x+2)(3x^{2}+2x+9)^{6}dx\\[1.2em]&=2\times\dfrac{1}{7}(3x^{2}+2x+9)^{7}+c\\[1.2em]&=\dfrac{2}{7}(3x^{2}+2x+9)^{7}+c\end{aligned}

iii) $\dfrac{d}{dx}(\cot(x))=-\cosec^{2}(x)$ so this is in reverse chain rule form. Hence:

\begin{aligned}&\int19\cosec^{2}(3x)\cot^{12}(3x)dx\\[1.2em]&=-19\int-\cosec^{2}(3x)\cot^{12}(3x)dx\\[1.2em]&=-19\times\dfrac{1}{3}\times\dfrac{1}{13}\cot^{13}(3x)+c\\[1.2em]&=-\dfrac{19}{39}\cot^{13}(3x)+c\end{aligned}

A Level

A Level

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