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Reverse chain rule

Reverse chain rule

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Reverse chain rule

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Explainer Video

Tutor: Bilal

Summary

Reverse chain rule

​​In a nutshell

The reverse chain rule makes use of the chain rule for differentiation to integrate functions that are of particular forms, such has k[f(x)]nf′(x)k [f(x)]^n f'(x) k[f(x)]nf′(x) and kf′(x)f(x)k\dfrac{f'(x)}{f(x)}kf(x)f′(x)​.



Functions of the form kf′(x)f(x)k\dfrac{f'(x)}{f(x)}kf(x)f′(x)​​

​​​​​​​When integrating functions of the form kf′(x)f(x)k\dfrac{f'(x)}{f(x)}kf(x)f′(x)​, try differentiating y=ln⁡(f(x))y=\ln(f(x))y=ln(f(x)) and adjust for any constants.


Example 1

Evaluate ∫4x+2x2+x+5 dx\int \dfrac{4x+2}{x^2+x+5}\, dx∫x2+x+54x+2​dx​.


Try to differentiate the denominator to see if it matches the numerator:

​ddx(x2+x+5)=2x+1\dfrac{d}{dx}(x^2+x+5)=2x+1dxd​(x2+x+5)=2x+1​​


​Note that this is very similar to the numerator - it differs by a constant factor. Therefore, try differentiating y=ln⁡(x2+x+2)y=\ln(x^2+x+2)y=ln(x2+x+2) and adjust for any constants:

​ddxln⁡(x2+x+5)=2x+1x2+x+5\dfrac{d}{dx}\ln(x^2+x+5)=\dfrac{2x+1}{x^2+x+5}dxd​ln(x2+x+5)=x2+x+52x+1​​​


​⇒∫2x+1x2+x+5 dx=ln⁡(x2+x+5)+C\Rightarrow \int \dfrac{2x+1}{x^2+x+5}\, dx = \ln(x^2+x+5)+C⇒∫x2+x+52x+1​dx=ln(x2+x+5)+C​​


Use the fact that 4x+2=2×(2x+1)4x+2=2\times(2x+1)4x+2=2×(2x+1):

​∫4x+2x2+x+5 dx=∫2×2x+1x2+x+5 dx=2×∫2x+1x2+x+5 dx=2×ln⁡(x2+x+5)+C\begin{aligned} \int \dfrac{4x+2}{x^2+x+5}\, dx &= \int 2\times \dfrac{2x+1}{x^2+x+5}\, dx\\&= 2\times \int \dfrac{2x+1}{x^2+x+5}\, dx\\&=2\times \ln(x^2+x+5)+C\end{aligned}∫x2+x+54x+2​dx​=∫2×x2+x+52x+1​dx=2×∫x2+x+52x+1​dx=2×ln(x2+x+5)+C​​​


​∫4x+2x2+x+5 dx=2ln⁡(x2+x+5)+C‾\int \dfrac{4x+2}{x^2+x+5}\,dx = \underline{2 \ln(x^2+x+5)+C}∫x2+x+54x+2​dx=2ln(x2+x+5)+C​​​



Functions of the form k[f(x)]nf′(x)k [f(x)]^n f'(x) k[f(x)]nf′(x)​​

For functions of the form k[f(x)]nf′(x)k [f(x)]^n f'(x) k[f(x)]nf′(x), try differentiating y=[f(x)]n+1y=[f(x)]^{n+1}y=[f(x)]n+1 and adjust for any constants.


Example 2

Evaluate ∫12sin⁡(x)cos⁡3(x) dx\int \dfrac12\sin(x)\cos^3(x)\,dx∫21​sin(x)cos3(x)dx.


​Try to differentiate the function inside the power to see if it matches the other function:

ddx(cos⁡(x))=−sin⁡(x)\dfrac{d}{dx}(\cos(x))=-\sin(x)dxd​(cos(x))=−sin(x)​​


This differs from the numerator by only a constant factor. Therefore, try differentiating y=[cos⁡(x)]4y=[\cos(x)]^4y=[cos(x)]4:

​ddx(cos⁡4(x))=−4cos⁡3(x)sin⁡(x)\dfrac{d}{dx} (\cos^4(x))=-4\cos^3(x)\sin(x)dxd​(cos4(x))=−4cos3(x)sin(x)​​


​⇒∫−4cos⁡3(x)sin⁡(x) dx=cos⁡4(x)+C\Rightarrow \int -4\cos^3(x)\sin(x) \, dx= \cos^4(x)+C⇒∫−4cos3(x)sin(x)dx=cos4(x)+C​​


Adjust for the constant factor:

​∫12sin⁡(x)cos⁡3(x) dx=−18×∫−4sin⁡(x)cos⁡3(x) dx=−18×cos⁡4(x)+C\begin{aligned} \int \dfrac12\sin(x)\cos^3(x)\,dx &= -\dfrac{1}{8}\times \int -4\sin(x)\cos^3(x)\,dx\\&=-\dfrac18 \times \cos^4(x)+C \end{aligned}∫21​sin(x)cos3(x)dx​=−81​×∫−4sin(x)cos3(x)dx=−81​×cos4(x)+C​​​


​∫12sin⁡(x)cos⁡3(x) dx=−18cos⁡4(x)+C‾ \int \dfrac12\sin(x)\cos^3(x)\,dx =\underline{-\dfrac18 \cos^4(x)+C}∫21​sin(x)cos3(x)dx=−81​cos4(x)+C​​​


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Differentiating functions with multiple terms

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Differentiating functions with multiple terms

The chain rule

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The chain rule

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Reverse chain rule

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FAQs - Frequently Asked Questions

How do you integrate functions of the form k[f(x)]^n f'(x)?

Try differentiating y=[f(x)]^(n+1) and adjust for any constants.

How do you integrate functions of the form kf'(x)/f(x)?

Try differentiating y=ln(f(x)) and adjust for any constants.

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