07Dchainrule

toc = The Chain Rule =

The Chain Rule provides a method of differentiating composite functions.

Consider the composite function: math . \qquad y = f \big( g(x) \big) \qquad. math

math . \qquad \text{Let } u = g(x) \quad \text{ so } \quad y = f(u) \text{ then the Chain Rule is:} math

math . \qquad \dfrac{dy}{dx}=\dfrac{dy}{du} \times \dfrac{du}{dx} \qquad. math

Notice that 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"

In function notation, the chain rule can be written as

math \\ . \qquad \text{If } y(x)=f \big( g(x) \big) \qquad. \\ . \\ . \qquad \text{then }y'(x)=f ' \big( g(x) \big) \times g'(x) \qquad. math

**Example 1**
math \text{Differentiate }y=(4x^3-5x)^{-2} \qquad. math

__Solution:__

math . \qquad y=(4x^3-5x)^{-2} \qquad. math

math \\ . \qquad \text{Let } u=4x^3-5x \quad \text{ then } \quad y=u^{-2} \qquad. \\ . \\ . \quad \dfrac{du}{dx}=12x^2-5 \quad \quad \dfrac{dy}{du}=-2u^{-3} \qquad. math

Use: math \dfrac{dy}{dx}=\dfrac{dy}{du} \times \dfrac{du}{dx} \qquad. math

math \\ \dfrac{dy}{dx}=-2u^{-3} \times \left( 12x^2-5 \right) \qquad .\\. \\ .\quad=-2 \left( 4x^3-5x \right)^{-3} \left( 12x^2-5 \right) \qquad .\\. \\ .\quad=\dfrac{-2 \left( 12x^2-5 \right) } { \left( 4x^3-5x \right)^3} math

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.

math \\ \text{If } y=mu^n \text{ then } \qquad. \\ . \\ \dfrac{dy}{dx}= n \times m(u)^{n-1} \times\dfrac{du}{dx} \qquad. math

{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.}

For the example above, math \\ y= \left( 4x^3-5x \right)^{-2} \quad \text{so} \qquad. \\ . \\ \dfrac{dy}{dx}=-2 \left ( 4x^3-5x \right)^{-3} \times \left( 12x^2-5 \right) \qquad. math

**Example 2**
Find the gradient of the curve with the equation: math . \qquad y = \dfrac{16}{3x^2+1} \text{ at the point } (1,4) \qquad. math

__Solution: __

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

math \\ \dfrac{dy}{dx}=-1 \times16 \big( 3x^2+1 \big)^{-2} \times \big( 6x \big) \qquad. \\ . \\ .\quad=\dfrac{-96x}{(3x^2+1)^2} \qquad. math

when x = 1 math \\ \dfrac{dy}{dx}=\dfrac{-96\times1}{(3\times1^2+1)^2} \qquad. \\ . \\ .\quad=-6 math

**Example 3**
The table gives information about functions //f// and //g// and their derivatives //f'// and //g'//. Given that //h (x) = f// [//g (x)//], find the value of //h' (//1//)//.



__Solution:__

math \text{Use the chain rule with function notation, }h'(x)=f' \big( g(x) \big) \times g'(x) math

math \\ h'(1)=f' \big( g(1) \big) \times g'(1) \\ \\ .\qquad=f'(3) \times g'(1) \quad\quad\quad \big( \text{as } g(1)=3 \big) math

math \\ .\qquad=(-1)\times(-3) \quad\quad \big( \text{as } f'(3)=-1 \text{ and } g'(1)=-3 \big) \\ \\ .\qquad=3 math Go to top of page flat

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