# Reverse Chain Rule

## Reverse Chain Rule Revision

**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**.

**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**.

**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**.

**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:

**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

## Reverse Chain Rule Example Questions

**Question 1: **Integrate:

i) 3x^{2}e^{x^{3}}

ii) 4x^{7}\left(\dfrac{1}{2}x^{8}+4\right)^{2}

iii) x^{2}\cos\left(\dfrac{1}{3}x^{3}\right)

**[9 marks]**

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}

**Question 2: **Integrate:

i) 12x^{3}e^{x^{4}}

ii) \dfrac{2}{x}(\log(x))^{5}

iii) 4\sec^{2}(x)e^{\tan(x)}

**[9 marks]**

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}

**Question 3: **Integrate:

i) (6x^{2}+12x+2)e^{x^{3}+3x^{2}+x+5}

ii) (12x+4)(3x^{2}+2x+9)^{6}

iii) 19\cosec^{2}(3x)\cot^{12}(3x)

**[9 marks]**

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}