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Chain Rule Simplified Revision Notes

Revision notes with simplified explanations to understand Chain Rule quickly and effectively.

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Chain Rule

The chain rule is useful for differentiating complicated functions. Up until now we've only dealt with polynomial expressions where each term is in index form. We've also seen examples where we can easily convert an expression to index form, for example x    x12\sqrt{x} \implies x^{\frac{1}{2}}. In many cases as we will see, this isn't always possible.

Chain rule assume there is an inner function and outer function. Formally :

f(x)=u(v(x))    f(x)=dudvdvdxf(x)=u(v(x)) \implies f'(x)=\frac{du}{dv} \cdot \frac{dv}{dx}

Example

infoNote

Differentiate x22x2\sqrt{x^2-2x-2}.

First, identify an inner and outer function. We can set the inner function to be the expression under the square root :

v(x)=x22x2v(x)=x^2-2x-2

Then the outer function can be written as :

u(x)=v(x)u(x)=\sqrt{v(x)}

According to chain rule, we differentiate both functions :

v(x)=2x2v'(x)=2x-2u(x)=12v12=12vu'(x)=\tfrac{1}{2}v^{-\tfrac{1}{2}}=\frac{1}{2\sqrt{v}}

Next, multiply both derivatives :

u(x)v(x)=(2x2)12v=2x22v=x1v\begin{align*} u'(x) \cdot v'(x)&=(2x-2) \cdot \frac{1}{2\sqrt{v}} \\\\ &= \frac{2x-2}{2\sqrt{v}} \\\\ &= \frac{x-1}{\sqrt{v}} \end{align*}

Lastly, resubstitute vv to have your answer in terms of xx.

x1v=x1x22x2\frac{x-1}{\sqrt{v}}=\frac{x-1}{\sqrt{x^2-2x-2}}

Example

infoNote

Differentiate (5x+7)100(5x+7)^{100}.

Identify an inner and outer function :

v=5x+7v=5x+7u=v100u=v^{100}

Differentiate both functions :

v=5v'=5u=100v99u'=100v^{99}

Multiply derivatives :

uv=5100v99=500v99=500(5x+7)99\begin{align*} u' \cdot v'&=5 \cdot 100v^{99} \\\\ &= 500v^{99} \\\\ &= 500(5x+7)^{99} \end{align*}
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