06 1chainrule

# The Chain Rule

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The Chain Rule provides a method of differentiating composite functions.
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Consider the composite function: … $y = f \big( g(x) \big)$

… … Let $u = g(x) \quad \text{ so } \quad y = f(u)$

… … Then the Chain Rule is:
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… … $\dfrac{dy}{dx}=\dfrac{dy}{du} \times \dfrac{du}{dx}$

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Note: if the "dy" "dx" and "du" were treated as terms in a fraction form,

• the "du" would cancel out of the right side, leaving "dy" over "dx"

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In function notation, the chain rule can be written as

… … If $y(x)=f \big( g(x) \big)$
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… …then $y'(x)=f ' \big( g(x) \big) \times g'(x)$

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### Example 1

Differentiate $y=(4x^3-5x)^{-2}$

Solution:

… … $y=(4x^3-5x)^{-2}$

… … Let $u=4x^3-5x \quad \text{ then } \quad y=u^{-2}$

… … $\dfrac{du}{dx}=12x^2-5 \quad \quad \dfrac{dy}{du}=-2u^{-3}$

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… … Use: $\dfrac{dy}{dx}=\dfrac{dy}{du} \times \dfrac{du}{dx}$
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… … $\dfrac{dy}{dx}=-2u^{-3} \times \left( 12x^2-5 \right)$

… … … .. $=-2 \left( 4x^3-5x \right)^{-3} \left( 12x^2-5 \right)$

… … … .. $=\dfrac{-2 \left( 12x^2-5 \right) } { \left( 4x^3-5x \right)^3}$

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## Chain Rule Shortcut

A quick method of applying the chain rule can be used to differentiate expressions in terms of x involving brackets with high powers or rational powers or negative powers.

… … If $y=mu^n$

… … Then $\dfrac{dy}{dx}= n \times m(u)^{n-1} \times\dfrac{du}{dx}$
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ie take the derivative of the outside function (leaving the contents of the bracket untouched)
then multiply by the derivative of the contents of the bracket.

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For the example above,

… … $y= \left( 4x^3-5x \right)^{-2}$ … so
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… … $\dfrac{dy}{dx}=-2 \left ( 4x^3-5x \right)^{-3} \times \left( 12x^2-5 \right)$

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### Example 2

Find the gradient of the curve with the equation: … $y = \dfrac{16}{3x^2+1}$
at the point … $(1, \;4)$

Solution:

… … Rewrite $y=\dfrac{16}{3x^2+1} \text{ as } y=16(3x^2+1)^{-1}$

… … $\dfrac{dy}{dx}=-1 \times 16 \big( 3x^2+1 \big)^{-2} \times \big( 6x \big)$

… … … .. $=\dfrac{-96x}{(3x^2+1)^2}$
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… … when $x = 1$

… … $\dfrac{dy}{dx}=\dfrac{-96\times1}{(3\times1^2+1)^2}$

… … … .. $=-6$

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### Example 3

The table gives information about functions f(x) and g(x) and their derivatives f'(x) and g'(x).

Given that $h(x) = f \big( g(x) \big)$, find the value of $h' (1)$.

Solution:

… … Use the chain rule with function notation:

… … $h'(x)=f' \big( g(x) \big) \times g'(x)$

… … $h'(1)=f' \big( g(1) \big) \times g'(1)$

… … … … $=f'(3) \times g'(1)$ … … … as $g(1)=3$

… … … … $=(-1)\times(-3)$ … … … as $f'(3)=-1 \text{ and } g'(1)=-3$

… … … … $=3$

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