The Chain Rule

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 20 July 2012

The Chain Rule

There are different kinds of derivatives:

$\begin{array}{cc}\text{Derivative with respect to }x& \text{Derivative with respect to }g\\ f⟶\frac{d}{dx}⟶\frac{df}{dx}& f⟶\frac{d}{dg}⟶\frac{df}{dg}\\ \text{This one satisfies}& \text{This one satisfies}\\ {\displaystyle \frac{dx}{dx}=1\text{,}}& {\displaystyle \frac{dg}{dg}=1\text{,}}\\ {\displaystyle \frac{d\left(cf\right)}{dx}=c\frac{df}{dx}\text{,}\text{if }c\text{ is a constant,}}& {\displaystyle \frac{d\left(cf\right)}{dg}=c\frac{df}{dg}\text{,}\text{if }c\text{ is a constant,}}\\ {\displaystyle \frac{d\left(y+z\right)}{dx}=\frac{dy}{dx}+\frac{dz}{dx}\text{,}}& {\displaystyle \frac{d\left(y+z\right)}{dg}=\frac{dy}{dg}+\frac{dz}{dg}\text{,}}\\ {\displaystyle \frac{d\left(yz\right)}{dx}=y\frac{dz}{dx}+\frac{dy}{dx}z\text{,}}& {\displaystyle \frac{d\left(yz\right)}{dg}=y\frac{dz}{dg}+\frac{dy}{dg}z\text{,}}\end{array}$

${\displaystyle \text{<strong>What is the relation between</strong>}\frac{df}{dx}\text{<strong>and</strong>}\frac{df}{dg}\text{<strong>?</strong>}}$

$\begin{array}{cc}{\displaystyle ⟶\frac{d}{dg}⟶}& {\displaystyle ⟶\frac{d}{dx}⟶}\\ \\ {\displaystyle \frac{d{g}^0}{dg}=\frac{d1}{dg}=0\text{,}}& {\displaystyle \frac{d{g}^0}{dx}=\frac{d1}{dx}=0\text{,}}\\ \\ {\displaystyle \frac{dg}{dg}=1\text{,}}& {\displaystyle \frac{dg}{dx}=\frac{dg}{dx}\text{,}}\\ \\ {\displaystyle \frac{d{g}^2}{dg}=\frac{dg·g}{dg}}& {\displaystyle \frac{d{g}^2}{dx}=\frac{dg·g}{dx}}\\ {\displaystyle =g\frac{dg}{dg}+\frac{dg}{dg}g}& {\displaystyle =g\frac{dg}{dx}+\frac{dg}{dx}g}\\ {\displaystyle =g+g=2g\text{,}}& {\displaystyle =2g\frac{dg}{dx}\text{,}}\\ \\ {\displaystyle \frac{d{g}^3}{dg}=\frac{d{g}^2·g}{dg}}& {\displaystyle \frac{d{g}^3}{dx}=\frac{d{g}^2·g}{dx}}\\ {\displaystyle =g^2\frac{dg}{dg}+\frac{d{g}^2}{dg}g}& {\displaystyle =g^2\frac{dg}{dx}+\frac{d{g}^2}{dx}g}\\ {\displaystyle =g^2+2g·g=3g^2\text{,}}& {\displaystyle =g^2\frac{dg}{dx}+2g\frac{dg}{dx}g}\\ & {\displaystyle =g^2\frac{dg}{dx}+2g^2\frac{dg}{dx}}\\ & {\displaystyle =3g^2\frac{dg}{dx}\text{,}}\\ \\ {\displaystyle \frac{d{g}^4}{dg}=\frac{d{g}^3·g}{dg}}& {\displaystyle \frac{d{g}^4}{dx}=\frac{d{g}^3·g}{dx}}\\ {\displaystyle =g^3\frac{dg}{dg}+\frac{d{g}^3}{dg}g}& {\displaystyle =g^3\frac{dg}{dx}+\frac{d{g}^3}{dx}g}\\ {\displaystyle =g^3+3g^2·g=4g^3\text{,}}& {\displaystyle =g^3\frac{dg}{dx}+3g^2\frac{dg}{dx}g}\\ & {\displaystyle =g^3\frac{dg}{dx}+3g^3\frac{dg}{dx}}\\ & {\displaystyle =4g^3\frac{dg}{dx}\text{,}}\\ \\ ⋮& ⋮\\ \\ {\displaystyle \frac{d{g}^{6342}}{dg}=6342g^{6341}\text{,}}& {\displaystyle \frac{d{g}^{6342}}{dx}=6342g^{6341}\frac{dg}{dx}\text{,}}\\ \\ {\displaystyle \frac{d\left(3g^2+2g+7\right)}{dg}=\frac{d\left(3g^2\right)}{dg}+\frac{d\left(2g\right)}{dg}+\frac{d7}{dg}}& {\displaystyle \frac{d\left(3g^2+2g+7\right)}{dx}=\frac{d\left(3g^2\right)}{dx}+\frac{d\left(2g\right)}{dx}+\frac{d7}{dx}}\\ {\displaystyle =3\frac{d{g}^2}{dg}+2\frac{dg}{dg}+0}& {\displaystyle =3\frac{d{g}^2}{dx}+2\frac{dg}{dx}+0}\\ {\displaystyle =3·2g+2·1}& {\displaystyle =3·2g\frac{dg}{dx}+2\frac{dg}{dx}}\\ {\displaystyle =6g+2\text{,}}& {\displaystyle =\left(6g+2\right)\frac{dg}{dx}\text{,}}\end{array}$

Thus, we are seeing that

${\displaystyle \frac{df}{dx}=\frac{df}{dg}\frac{dg}{dx}\text{,}\text{which is the chain rule.}}$

 

$\begin{array}{c}{\displaystyle \text{<strong>Example: </strong>Find }\frac{dy}{dx}\text{ when }y={\left(2x-5\right)}^2\text{.}}\\ \text{If }g=2x-5\text{ then }y=g^2\text{.}\\ \begin{array}{c}{\displaystyle \frac{dy}{dx}=\frac{dy}{dg}\frac{dg}{dx}=\frac{d{g}^2}{dg}\frac{d\left(2x-5\right)g}{dx}=2g\left(2-0\right)=2\left(2x-5\right)·2}\\ =4\left(2x-5\right)=8x-20\text{.}\end{array}\end{array}$

 

$\begin{array}{c}{\displaystyle \text{<strong>Example: </strong>Find }\frac{dy}{dx}\text{ when }y={\left(3x-4\right)}^3\text{.}}\\ \text{If }g=3x-4\text{ then }y=g^3\text{.}\\ \begin{array}{c}{\displaystyle \frac{dy}{dx}=\frac{dy}{dg}\frac{dg}{dx}=\frac{d{g}^3}{dg}\frac{d\left(3x-4\right)g}{dx}=3g^2\left(3-0\right)=9{\left(3x-4\right)}^2}\\ =9\left(9x^2-24x+16\right)=81x^2-72x+144\text{.}\end{array}\end{array}$

 

$\begin{array}{c}{\displaystyle \text{<strong>Example: </strong>Find }\frac{dy}{dx}\text{ when }y={\left(2x-5\right)}^2{\left(3x-4\right)}^3\text{.}}\\ \begin{array}{c}{\displaystyle \frac{dy}{dx}=\frac{d{\left(2x-5\right)}^2{\left(3x-4\right)}^3}{dx}={\left(2x-5\right)}^2\frac{d{\left(3x-4\right)}^3}{dx}+\frac{d{\left(2x-5\right)}^2}{dx}{\left(3x-4\right)}^3}\\ ={\left(2x-5\right)}^2·3{\left(3x-4\right)}^2·3+2\left(2x-5\right)·2{\left(3x-4\right)}^3\\ =\left(2x-5\right){\left(3x-4\right)}^2\left(9\left(2x-5\right)+4\left(3x-4\right)\right)=\left(2x-5\right){\left(3x-4\right)}^2\left(30x-61\right)\text{.}\end{array}\end{array}$

 

$\begin{array}{c}{\displaystyle \text{<strong>Example: </strong>Find }\frac{d{x}^{m/n}}{dx}\text{ when }m\text{ and }n\text{ are integers, }n\ne 0\text{.}}\\ {\displaystyle \frac{d{\left(x^{m/n}\right)}^n}{dx}=\frac{d{x}^m}{dx}=m{x}^{m-1}\text{.}\text{On the other hand}\frac{d{\left(x^{m/n}\right)}^n}{dx}=n{\left(x^{m/n}\right)}^{n-1}\frac{d{x}^{m/n}}{dx}\text{.}}\\ {\displaystyle \text{So}m{x}^{m-1}=n{\left(x^{m/n}\right)}^{n-1}\frac{d{x}^{m/n}}{dx}\text{and we can solve for}\frac{d{x}^{m/n}}{dx}\text{.}}\\ \begin{array}{c}{\displaystyle \frac{d{x}^{m/n}}{dx}=\frac{m{x}^{m-1}}{n{\left(x^{m/n}\right)}^{n-1}}=\frac{m{x}^{m-1}}{n{\left(x^{m/n}\right)}^n{\left(x^{m/n}\right)}^{-1}}}\\ {\displaystyle =\frac{m{x}^{m-1}}{n{x}^m\frac{1}{x^{m/n}}}=\left(\frac{m}{n}\right)x^{-1}{x}^{m/n}=\left(\frac{m}{n}\right)x^{\left(m/n\right)-1}\text{.}}\end{array}\end{array}$

 

$\begin{array}{c}{\displaystyle \text{<strong>Example: </strong>Find }\frac{dy}{dx}\text{ when }y=\frac{x}{\sqrt{1-2x}}\text{.}}\\ \\ \begin{array}{c}{\displaystyle \frac{dy}{dx}=\frac{d\frac{x}{\sqrt{1-2x}}}{dx}=\frac{dx{\left(\sqrt{1-2x}\right)}^{-1}}{dx}=\frac{dx{\left({\left(1-2x\right)}^{1/2}\right)}^{-1}}{dx}}\\ {\displaystyle =\frac{dx{\left(1-2x\right)}^{-\left(1/2\right)}}{dx}=x\frac{d{\left(1-2x\right)}^{-\left(1/2\right)}}{dx}+\frac{dx}{dx}{\left(1-2x\right)}^{-\left(1/2\right)}}\\ {\displaystyle =x}\left(-\frac{1}{2}\right){\displaystyle {\left(1-2x\right)}^{-3/2}\frac{d\left(1-2x\right)}{dx}+1·\frac{1}{\sqrt{1-2x}}}\\ {\displaystyle =\frac{-x}{2{\left(1-2x\right)}^{-3/2}}·\left(-2\right)+\frac{1}{{\left(1-2x\right)}^{1/2}}=\frac{x+1-2x}{{\left(1-2x\right)}^{3/2}}=\frac{1-x}{{\left(1-2x\right)}^{3/2}}\text{.}}\end{array}\end{array}$

 

$\begin{array}{c}{\displaystyle \text{<strong>Example: </strong>Find }\frac{dy}{dx}\text{ when }y=\frac{\sqrt{1+x^2}}{\sqrt{1-x^2}}\text{.}}\\ \\ \begin{array}{c}{\displaystyle \frac{dy}{dx}=\frac{d\frac{\sqrt{1+x^2}}{\sqrt{1-x^2}}}{dx}=\frac{d\frac{{\left(1+x^2\right)}^{1/2}}{{\left(1-x^2\right)}^{1/2}}}{dx}=\frac{d{\left(\frac{1+x^2}{1-x^2}\right)}^{1/2}}{dx}}\\ {\displaystyle =\frac{1}{2}·{\left(\frac{1+x^2}{1-x^2}\right)}^{\left(1/2\right)-1}\frac{d\left(\frac{1+x^2}{1-x^2}\right)}{dx}}\\ {\displaystyle =\frac{1}{2}·{\left(\frac{1+x^2}{1-x^2}\right)}^{-\left(1/2\right)}\frac{d\left(1+x^2\right){\left(1-x^2\right)}^{-1}}{dx}}\\ {\displaystyle =\frac{1}{2}·{\left(\frac{1-x^2}{1+x^2}\right)}^{1/2}\left(\left(1+x^2\right)\frac{d{\left(1-x^2\right)}^{-1}}{dx}+\frac{d\left(1+x^2\right)}{dx}{\left(1-x^2\right)}^{-1}\right)}\\ {\displaystyle =\frac{1}{2}·{\left(\frac{1-x^2}{1+x^2}\right)}^{1/2}\left(\left(1+x^2\right)\left(-1\right){\left(1-x^2\right)}^{-2}\frac{d{\left(1-x^2\right)}^{-1}}{dx}+2x{\left(1-x^2\right)}^{-1}\right)}\\ {\displaystyle =\frac{1}{2}·{\left(\frac{1-x^2}{1+x^2}\right)}^{1/2}\left(\frac{\left(-1\right)\left(1+x^2\right)\left(-2x\right)}{{\left(1-x^2\right)}^2}+\frac{2x}{1-x^2}\right)}\\ {\displaystyle =\frac{1}{2}·{\left(\frac{1-x^2}{1+x^2}\right)}^{1/2}\left(\frac{2x\left(1+x^2\right)}{{\left(1-x^2\right)}^2}+\frac{2x\left(1-x^2\right)}{{\left(1-x^2\right)}^2}\right)}\\ {\displaystyle =\frac{1}{2}·{\left(\frac{1-x^2}{1+x^2}\right)}^{1/2}\left(\frac{2x\left(1+x^2+1-x^2\right)}{{\left(1-x^2\right)}^2}\right)}\\ {\displaystyle =\frac{1}{2}·\frac{{\left(1-x^2\right)}^{1/2}}{{\left(1+x^2\right)}^{1/2}}·\frac{4x}{{\left(1-x^2\right)}^2}=\frac{2x}{{\left(1+x^2\right)}^{1/2}{\left(1-x^2\right)}^{3/2}}\text{.}}\end{array}\end{array}$

 

$\begin{array}{c}{\displaystyle \text{<strong>Example: </strong>Differentiate }\frac{x^2}{1+x^2}\text{ with respect to }{x}^2\text{.}}\\ \\ \text{This is the same problem as:}\\ {\displaystyle \text{Find }\frac{dz}{dp}\text{ when }z=\frac{x^2}{1+x^2}\text{ and }p=x^2\text{.}}\\ {\displaystyle \text{Since }\frac{dz}{dx}=\frac{dz}{dp}\frac{dp}{dx}\text{,}\frac{dz}{dp}=\frac{\left(dz/dx\right)}{\left(dp/dx\right)}\text{.}}\\ \text{So}\\ \begin{array}{c}{\displaystyle \frac{dz}{dp}=\frac{\frac{d}{dx}\left(\frac{x^2}{1+x^2}\right)}{\frac{d}{dx}\left(x^2\right)}=\frac{\frac{d{x}^2{\left(1+x^2\right)}^{-1}}{dx}}{\frac{d{x}^2}{dx}}=\frac{x^2\frac{d{\left(1+x^2\right)}^{-1}}{dx}+\frac{d{x}^2}{dx}{\left(1+x^2\right)}^{-1}}{2x}}\\ {\displaystyle =\frac{x^2\left(-1\right){\left(1+x^2\right)}^{-2}\frac{d\left(1+x^2\right)}{dx}+2x{\left(1+x^2\right)}^{-1}}{2x}}\\ {\displaystyle =\frac{\frac{{-x}^2}{{\left(1+x^2\right)}^2}·2x+\frac{2x}{1+x^2}}{2x}=\frac{{-x}^2}{{\left(1+x^2\right)}^2}+\frac{1}{1+x^2}}\\ {\displaystyle =\frac{-x^2+1+x^2}{{\left(1+x^2\right)}^2}=\frac{1}{{\left(1+x^2\right)}^2}\text{.}}\end{array}\end{array}$

 

$\begin{array}{c}{\displaystyle \text{<strong>Example: </strong>Find }\frac{dy}{dx}\text{ when }{x}^4+y^4=4a^2{x}^2{y}^2\text{.}}\\ {\displaystyle \frac{d\left(x^4+y^4\right)}{dx}=\frac{d\left(4a^2{x}^2{y}^2\right)}{dx}\text{.}\text{So}\frac{d{x}^4}{dx}+\frac{d{y}^4}{dy}=4a^2\frac{d{x}^2{y}^2}{dy}\text{.}}\\ {\displaystyle \text{So}4x^3+4y^3\frac{dy}{dx}=4a^2\left(x^2\frac{d{y}^2}{dx}+\frac{d{x}^2}{dx}{y}^2\right)\text{.}}\\ {\displaystyle \text{So}4x^3+4y^3\frac{dy}{dx}=4a^2\left(x^{2}2y\frac{dy}{dx}+2x{y}^2\right)}\\ {\displaystyle =4a^2{x}^{2}2y\frac{dy}{dx}+4a^{2}2x{y}^2\text{.}}\\ {\displaystyle \text{So}4x^3-4a^{2}2x{y}^2=4a^2{x}^{2}2y\frac{dy}{dx}-4y^3\frac{dy}{dx}\text{.}}\\ {\displaystyle \text{So}4x^3-4a^{2}2x{y}^2=\left(4a^2{x}^{2}2y-4y^3\right)\frac{dy}{dx}\text{.}}\\ {\displaystyle \text{So}\frac{4x^3-4a^{2}2x{y}^2}{4a^2{x}^{2}2y-4y^3}=\frac{dy}{dx}\text{.}}\\ {\displaystyle \text{So}\frac{dy}{dx}=\frac{x^3-2a^{2}x{y}^2}{2a^2{x}^{2}y-y^3}\text{.}}\\ {\displaystyle \text{All we did is take the derivative of both sides and then solve for }\frac{dy}{dx}\text{.}}\end{array}$

 

$\begin{array}{c}{\displaystyle \text{<strong>Example: </strong>Find }\frac{dy}{dx}\text{ when }x=\frac{3at}{1+t^3}\text{and}y=\frac{3a{t}^2}{1+t^3}}\\ {\displaystyle \text{Since }y=\frac{3a{t}^2}{1+t^3}=\left(\frac{3at}{1+t^3}\right)t=xt\text{,}\frac{dy}{dx}=x\frac{dt}{dx}+\frac{dx}{dx}·t=x\frac{dt}{dx}+t\text{.}}\\ {\displaystyle \text{What is }\frac{dt}{dx}\text{??}}\\ {\displaystyle \text{Since }\frac{dx}{dx}=\frac{dx}{dt}\frac{dt}{dx}\text{,}\frac{dt}{dx}=\frac{\left(dx/dx\right)}{\left(dx/dt\right)}=\frac{1}{dx/dt}}\\ \text{So}\\ {\displaystyle \begin{array}{c}\frac{dt}{dx}=\frac{1}{dx/dt}=\frac{1}{\frac{d}{dt}\left(\frac{3at}{1+t^3}\right)}=\frac{1}{\frac{d\left(3at\right){\left(1+t^3\right)}^{-1}}{dt}}\\ =\frac{1}{3at\left(-1\right){\left(1+t^3\right)}^{-2}\frac{d\left(1+t^3\right)}{dt}+3a{\left(1+t^3\right)}^{-1}}\\ =\frac{1}{\frac{-3at}{{\left(1+t^3\right)}^2}3t^2+\frac{3a}{1+t^3}}=\frac{1}{\frac{-9a{t}^3+3a\left(1+t^3\right)}{{\left(1+t^3\right)}^2}}\\ =\frac{{\left(1+t^3\right)}^2}{-9a{t}^3+3a\left(1+t^3\right)}=\frac{{\left(1+t^3\right)}^2}{3a-6a{t}^3}\text{.}\end{array}}\\ \text{So}\\ {\displaystyle \begin{array}{c}\frac{dy}{dx}=x\frac{dt}{dx}+t=\frac{3at}{1+t^3}\frac{{\left(1+t^3\right)}^2}{3a\left(1-2t^3\right)}+t\\ =\frac{t\left(1+t^3\right)}{1-2t^3}+\frac{t\left(1-2t^3\right)}{1-2t^3}=\frac{t+t^4+t-2t^4}{1-2t^3}=\frac{2t-t^4}{1-2t^3}\end{array}}\end{array}$

Notes

These notes are from lecture notes of Arun Ram from May 20, 2004.

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