The chain rule

Differentiation: Chain rule

The chain rule

We can also call a composition of functions a chain. The chain rule gives us a way to calculate the derivative of a composite function.

The chain rule for differentiation

For a composite function #f(x)=\blue{g(}\green{h(x)}\blue{)}#, the following applies:

\[\begin{array}{c}
f'(x)=\orange{g'(\green{h(x)})}\cdot \purple{h'(x)}
\end{array}\]

Example\[\begin{array}{rcl}f(x)&=&\blue{(\green{x^2-5x})^4} \\
f'(x) &=& \orange{4(\green{x^2-5x})^3} \cdot \purple{(2x-5)}
\end{array}\]

We will present a rough sketch of the proof of the chain rule. We write #\epsilon# for the difference, instead of #h#, in order to prevent confusion between #h# and #h(x)#.

\[\begin{array}{rcl}
f'(x)&=&\displaystyle\frac{f(x+\epsilon)-f(x)}{\epsilon} \\
&&\phantom{xx}\blue{\text{definition derivative}}\\
&=&\displaystyle\frac{g(h(x+\epsilon))-g(h(x))}{\epsilon}\\
&&\phantom{xx}\blue{f(x)=g(h(x))\text{ filled in}}\\
&=&\displaystyle\frac{g(h(x+\epsilon))-g(h(x))}{\epsilon}\cdot \frac{h(x+\epsilon)-h(x)}{h(x+\epsilon)-h(x)}\\
&&\phantom{xx}\blue{\text{multiplied with }1=\frac{h(x+\epsilon)-h(x)}{h(x+\epsilon)-h(x)}}\\
&=&\displaystyle\frac{g(h(x+\epsilon))-g(h(x))}{h(x+\epsilon)-h(x)}\cdot \frac{h(x+\epsilon)-h(x)}{\epsilon}\\
&&\phantom{xx}\blue{\text{denominators swapped}}\\
\end{array}\]

By definition, we have #h'(x)=\frac{h(x+\epsilon)-h(x)}{\epsilon}#. Showing that #\frac{g(h(x+\epsilon))-g(h(x))}{h(x+\epsilon)-h(x)}# equals #g'h(x)# is harder. We will use, without further explanation, the fact that if #\epsilon# gets close to zero, the difference #h(x+\epsilon)-h(x)# also gets close to zero. This means we can look at #h(x)+\epsilon# instead of #h(x+\epsilon)#, because if we let #\epsilon# get arbitrarly close to #0#, this does not really matter. We now have

\[\begin{array}{rcl}\dfrac{g(h(x+\epsilon))-g(h(x))}{h(x+\epsilon)-h(x)}&=&\dfrac{g(h(x)+\epsilon)-g(h(x))}{h(x)+\epsilon-h(x)}\\
&&\phantom{xx}\blue{h(x+\epsilon)\text{ replaced by }h(x)+\epsilon}\\
&=&\dfrac{g(h(x)+\epsilon)-g(h(x))}{\epsilon}\\
&&\phantom{xx}\blue{\text{simplified}}\\
&=&g'(h(x))\\
&&\phantom{xx}\blue{\text{definition of derivative, with }h(x)\text{ as variable}}
\end{array}\]

We now see that indeed #f'(x)=g'(h(x))\cdot h'(x)# if #f(x)=g(h(x))#.

To use the chain rule, we can use this step-by-step guide.

Step-by-step guide chain rule	Step-by-step	Example
	Determine the derivative of a function that is composed of multiple functions: #f(x)=\blue{g(}\green{h(x)}\blue{)}#.	#\qqquad \begin{array}{rcl} f(x)\phantom{'}&=&(x^2-1)^3\end{array}#
Step 1	Distinguish the simpler functions #\blue{g(x)}# and #\green{h(x)}# of which #f(x)# is composed.	#\qqquad\begin{array}{rcl}\blue{g(x)}\phantom{'}&=&\blue{x^3}\\ \green{h(x)}\phantom{'}&=&\green{x^2-1}\end{array}#
Step 2	Determine the derivatives #\orange{g'(x)}# and #\purple{h'(x)}#.	#\qqquad\begin{array}{rcl}\orange{g'(x)}&=&\orange{3x^2}\\ \purple{h'(x)}&=&\purple{2x}\end{array}#
Step 3	Calculate the derivative of #f# with the formula: \[\begin{array}{c} f'(x)=\orange{g'(\green{h(x)})}\cdot \purple{h'(x)} \end{array}\]	#\qqquad \begin{array}{rcl} f'(x)&=& \orange{3(\green{x^2-1})^2}\cdot \purple{2x}\\&=&(3x^4-6x^2+3)\cdot \purple{2x}\\&=& 6x^5-12x^3+6x\end{array}#

Calculate the derivative #f'(x)# for #f(x)=# #\sqrt{1-x^4}#.

#f'(x)=# #-{{2\cdot x^3}\over{\sqrt{1-x^4}}}#

Step 1	We distinguish the simpler functions #g(x)# and #h(x)# of which #f(x)# is composed. In other words, the functions for which #f(x)=g(h(x))#. #\begin{array}{rcl} g(x)&=&\displaystyle \sqrt{x}\\ h(x)&=& \displaystyle1-x^4\end{array}#
Step 2	We calculate the derivatives #g'(x)# and #h'(x)#. #\begin{array}{rcl} g'(x)&=& \displaystyle\dfrac{\dd}{\dd x}\left(\sqrt{x}\right)\\ &&\phantom{xxx}\blue{\text{definition derivative}}\\ &=& \displaystyle{{1}\over{2\cdot \sqrt{x}}}\\ &&\phantom{xxx}\blue{\text{power rule}}\end{array}# #\begin{array}{rcl} h'(x)&=& \displaystyle\dfrac{\dd}{\dd x}\left(1-x^4\right)\\ &&\phantom{xxx}\blue{\text{definition derivative}}\\ &=& \displaystyle-4\cdot x^3\\ &&\phantom{xxx}\blue{\text{sum rule and power rule}}\end{array}#
Step 3	We calculate the derivative #f'(x)#. #\begin{array}{rcl} f'\left(x\right)&=& \displaystyle g'(h(x))\cdot h'(x)\\ &&\phantom{xxx}\blue{\text{chain rule}}\\ &=& \displaystyle {{1}\over{2\cdot \sqrt{h\left(x\right)}}}\cdot -4\cdot x^3\\ &&\phantom{xxx}\blue{\text{substituting }g' \text{ and }h'}\\ &=& \displaystyle\displaystyle {{1}\over{2\cdot \sqrt{1-x^4}}} \cdot (-4\cdot x^3)\\ &&\phantom{xxx}\blue{h(x)\text{ substituted}}\\ &=& \displaystyle -{{2\cdot x^3}\over{\sqrt{1-x^4}}}\\ &&\phantom{xxx}\blue{\text{simplified}} \end{array}#

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