The Chain Rule

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The Chain Rule

Our goal is to differentiate functions such as

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

The Chain Rule

If
$y = y(u)$

is a function of $u$, and

$u = u(x)$

is a function of $ x $ then

\[ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} \]

In our example we have

$y = u^{10}$

and

$ u = 3x + 1 $

so that

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

$= (10u^9) (3) = 30u^9 = 30 (3x+1)^9 $

Proof of the Chain Rule

Recall an alternate definition of the derivative:

$ \displaystyle \frac{d}{dx}f(g(x)) = \lim_{x \to c} \frac{f(g(x)) - f(g(c))}{x - c} $

$ \displaystyle = \lim_{x \to c} \frac{f(g(x)) - f(g(c))}{g(x) - g(c)} \frac{g(x) - g(c)}{x - c} $

$ = f'(g(c))g(c) $

Examples

Find $f '(x)$ if

$ f(x) = (x^3 - x + 1)^20 $
$ f(x) = (x^4 - 3x^3 +x)^5 $
$ f(x) = (1 - x)^9 (1 - x^2)^4 $
$ f(x) = \frac{(x^3 + 4x - 3)^7}{(2x - 1)^3} $

Solution:

Here

        $ f(u) = u^{20} $

and

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

So that the derivative is

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

$ f(u) = u^5 $

and

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

So that the derivative is

$ [5u^4] [4x^3 - 9x^2 + 1] = [5(x^4 - 3x^3 + x)^4] [4x^3 - 9x^2 + 1] $
Here we need both the product and the chain rule.

    $ f'(x) = [(1 - x)^9] [(1 - x^2)^4]' + [(1 - x)^9] ' [(1 - x^2)^4] $

We first compute

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

and

$ [(1 - x)^9] ' = [9(1 - x)^8] [-1] $

Putting this all together gives

   $ f'(x) = [(1 - x)^9] [4(1 - x^2)^3] [-2x] - [9(1 - x)^8] [(1 - x^2)^4] $
Here we need both the quotient and the chain rule.

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

We first compute

$ [(x^3 + 4x - 3)^7]' = [7(x^3 + 4x - 3)^6 ][3x^2 + 4] $

and

       $[(2x - 1)^3] ' = [3(2x - 1)^2] [2] $

Putting this all together gives

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

Application

Suppose that you put $1000 into a bank at an interest rate r compounded monthly for 3 years. Then the amount $A$ that will be in the account at the end of the three years will be

$ A = 1000(1 + \frac{r}{12})^{36} $

Find the rate at which $ A $ rises with respect to a rise in the interest rate when the interest rate is 6%.

Solution

We are asked to find a derivative. We use the chain rule with

$u = 1 + \frac{r}{12}$ and $ A(u) = 1000u^{36} $

The two derivatives are

$u' = \frac{1}{12} $ and $A' = 36000u^35 $

The chain rule gives

$ \frac{dA}{dr} = \frac{dA}{du}\frac{du}{dr} = \frac{1}{12} 36000 u^{35} $

$ = 3000 u^{35} = 3000(1 + \frac{r}{12})^{35} $

Now plug in $ r = 6% = 0.06 $ to get

$ 3000(1 + \frac{0.06}{12})^{35} = 3572.18 $

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In our example we have

\(y = u^{10}\)

and

\( u = 3x + 1 \)

so that

\( \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}\)

\(= (10u^9) (3) = 30u^9 = 30 (3x+1)^9 \)

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

In our example we have \(y = u^{10}\) and \( u = 3x + 1 \) so that \( \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}\) \(= (10u^9) (3) = 30u^9 = 30 (3x+1)^9 \)

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

In our example we have

\(y = u^{10}\)

and

\( u = 3x + 1 \)

so that

\( \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}\)

\(= (10u^9) (3) = 30u^9 = 30 (3x+1)^9 \)