The Chain Rule

Last updated
Save as PDF

Page ID: 625

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

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

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

Search

Text Color

Text Size

Margin Size

Font Type

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

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