The Chain Rule

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Our goal is to differentiate functions such as

\[ y = (3x + 1)^{10} \nonumber \]

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 \[\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} \nonumber \]

In our example we have

\[ y = u^{10} \nonumber \]

and

\[ u = 3x + 1 \nonumber \]

so that

\[\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} \nonumber \]

= (10u⁹)(3) = 30(3x+1)⁹

Proof: Chain Rule
Recall an alternate definition of the derivative: $chain.1.gif$

Proof: Chain Rule

Recall an alternate definition of the derivative:

$chain.1.gif$

Example 2
Find f '(x) if \[ f(x) = (x^4 - 3x^3 + x)^5 \nonumber \] Solution Here \[ f(u) = u^5 \nonumber \] and \[ u(x) = x^4 - 3x^3 + x \nonumber \] So that the derivative is (5u⁴)(4x³ - 9x² + 1) = [5(x⁴ - 3x³ + x)⁴](4x³ - 9x² + 1)

Example 2

Find f '(x) if

\[ f(x) = (x^4 - 3x^3 + x)^5 \nonumber \]

Solution

Here

\[ f(u) = u^5 \nonumber \]

and

\[ u(x) = x^4 - 3x^3 + x \nonumber \]

So that the derivative is

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

Example 3
Find f '(x) if \[ f(x) = (x^3 - x + 1)^{20} \nonumber \] Solution Here \[ f(u) = u^{20} \nonumber \] and \[ u(x) = x^3 - x + 1 \nonumber \] So that the derivative is \[ (20u^{19})(3x^2 - 1) = \left[20(x^3 - x + 1)^{19}\right](3x^2 - 1) \nonumber \]

Example 3

Find f '(x) if

\[ f(x) = (x^3 - x + 1)^{20} \nonumber \]

Solution

Here

\[ f(u) = u^{20} \nonumber \]

and

\[ u(x) = x^3 - x + 1 \nonumber \]

So that the derivative is

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

Example 4
Find f '(x) if \[ f(x) = (1 - x)^9 (1-x^2)^4 \nonumber \] 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 4

Find f '(x) if

\[ f(x) = (1 - x)^9 (1-x^2)^4 \nonumber \]

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 \[ f(x)= \dfrac{ (x^3 + 4x - 3)^7}{ (2x - 1)^3} \nonumber \] Solution Here we need both the quotient and the chain rule. \[ f'(x) = \dfrac{(2x - 1)^3\left[(x^3 + 4x - 3)^7\right]' - (x^3 + 4x - 3)^7 \left[(2x - 1)^3\right]'}{(2x - 1)^6} \nonumber \] 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 \[ f](x) = \dfrac{7(2x - 1)^3(x^3 + 4x - 3)^6(3x^2 + 4) + 6(x^3 + 4x - 3)^7 (2x - 1)^2}{ (2x - 1)^6 } \nonumber \]

Example 5

Find f '(x) if

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

Solution

Here we need both the quotient and the chain rule.

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

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

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

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\[ f(x)= \dfrac{x^2(5 - x^3)^4 }{3 - x} \nonumber \]