
# The Chain Rule

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Our goal is to differentiate functions such as

$y = (3x + 1)^{10}$

The last thing that we would want to do is FOIL this out ten times. We now look for a better way.

Definition: The Chain Rule

If  $$y = y(u)$$ is a function of  $$u$$, and $$u = u(x)$$ is a function of $$x$$ then $\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx}$

In our example we have

$y = u^{10}$

and

$u = 3x + 1$

so that

$\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx}$

= (10u9)(3) = 30(3x+1)9

Proof: Chain Rule

Recall an alternate definition of the derivative:

Example 2

Find f '(x) if

$f(x) = (x^4 - 3x^3 + x)^5$

Solution

Here

$f(u) = u^20$

and

$u(x) = x^3 - x + 1$

So that the derivative is

$(20u^{19})(3x^2 - 1) = \left[20(x^3 - x + 1)^{19}\right](3x^2 - 1)$

Example 3

Find f '(x) if

$f(x) = (x^3 - x + 1)^{20}$

Solution

Here

$f(u) = u^5$

and

$u(x) = x^4 - 3x^3 + x$

So that the derivative is

(5u4)(4x3 - 9x2 + 1)  =  [5(x4 - 3x3 + x)4](4x3 - 9x2 + 1)

Example 4

Find f '(x) if

$f(x) = (1 - x)^9 (1-x^2)^4$

Solution

Here we need both the product and the chain rule.  First the product rule

f '(x) = [(1 - x)9][(1 - x2)4] ' + [(1 - x)9]' [(1 - x2)4]

Now compute

[(1 - x2)4]' = [4(1 - x2)3](-2x)

and

[(1 - x)9]'  = [9(1 - x)8](-1)

Putting this all together gives

f '(x) = [(1 - x)9][4(1 - x2)3](-2x) - [9(1 - x)8] [(1 - x2)4]

Example 5

Find f '(x) if

$f(x)= \dfrac{ (x^3 + 4x - 3)^7}{ (2x - 1)^3}$

Solution

Here we need both the quotient and the chain rule.

$f'(x) = \dfrac{(2x - 1)^3\left[(x^3 + 4x - 3)^7\right]' - (x^3 + 4x - 3)^7 \left[(2x - 1)^3\right]'}{(2x - 1)^6}$

We first compute

[(x3 + 4x - 3)7]' = [7(x3 + 4x - 3)6](3x2 + 4)

and

[(2x - 1)3]'  = [3(2x - 1)2](2)

Putting this all together gives

$f](x) = \dfrac{7(2x - 1)^3(x^3 + 4x - 3)^6(3x^2 + 4) + 6(x^3 + 4x - 3)^7 (2x - 1)^2}{ (2x - 1)^6 }$

Exercise

Find the derivative of

$f(x)= \dfrac{x^2(5 - x^3)^4 }{3 - x}$