The Chain Rule

Last updated

Oct 27, 2024
Save as PDF
- Implicit Differentiation
- The Product and Quotient Rules

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

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

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

( \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)=x2(5−x3)43−x f(x)= \dfrac{x^2(5 - x^3)^4 }{3 - x} \nonumber

Support Center

How can we help?

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