3.4: Chain Rule

$\newenvironment {prompt}{}{} \newcommand {\DeclareMathOperator }[2]{\@OldDeclareMathOperator {##1}{##2}\immediate \write \myfile {\unexpanded {\DeclareMathOperator }{\unexpanded {##1}}{\unexpanded {##2}}}} \newcommand {\reserved@a }[2]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{} \newcommand {\AppendGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Append \grfext@Check }{\grfext@Append \grfext@@Add }} \newcommand {\PrependGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Prepend \grfext@Check }{\grfext@Prepend \grfext@@Add }} \newcommand {\RemoveGraphicsExtensions }[1]{\@ifundefined {Gin@extensions}{\def \Gin@extensions {}}{\edef \grfext@tmp {\zap@space ##1 \@empty }\@for \grfext@ext :=\grfext@tmp \do {\def \grfext@next {\let \grfext@tmp \Gin@extensions \@expandtwoargs \@removeelement \grfext@ext \Gin@extensions \Gin@extensions \ifx \grfext@tmp \Gin@extensions \let \grfext@next \relax \fi \grfext@next }\grfext@next }}\grfext@Print \RemoveGraphicsExtensions } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} \newcommand {\epstopdfcall }[1]{\ifETE@InsideSetfile \expandafter \@firstoftwo \else \expandafter \@secondoftwo \fi {‘##1}{\Gin@base \Gin@ext }} \newcommand {\epstopdfDeclareGraphicsRule }[4]{\ifx \\##4\\\@PackageError {epstopdf-base}{Conversion command is missing}\@ehc \else \begingroup \@ifundefined {Gin@rule@##1}{}{\@PackageInfo {epstopdf-base}{Redefining graphics rule for ‘##1’}}\endgroup \@namedef {Gin@rule@##1}####1{{##2}{##3}{\epstopdfcall {##4}}}\fi } \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\GetTitleString }[0]{\ifGTS@expand \expandafter \GetTitleStringExpand \else \expandafter \GetTitleStringNonExpand \fi } \newcommand {\GetTitleStringExpand }[1]{\def \GetTitleStringResult {##1}\begingroup \GTS@DisablePredefinedCmds \GTS@DisableHook \edef \x {\endgroup \noexpand \def \noexpand \GetTitleStringResult {\GetTitleStringResult }}\x } \newcommand {\GetTitleStringNonExpand }[1]{\def \GetTitleStringResult {##1}\global \let \GTS@GlobalString \GetTitleStringResult \begingroup \GTS@RemoveLeft \GTS@RemoveRight \endgroup \let \GetTitleStringResult \GTS@GlobalString } \newcommand {\GTS@DisableHook }[0]{} \newcommand {\label@hook }[0]{} \newcommand {\@currentlabelname }[0]{} \newcommand {\reserved@a }[0]{} \newcommand {\@safe@activestrue }[0]{} \newcommand {\@safe@activesfalse }[0]{} \newcommand {\nameref }[0]{\@ifstar \T@nameref \T@nameref } \newcommand {\Sectionformat }[2]{##1} \newcommand {\vnameref }[1]{\unskip ~\nameref {##1}\@vpageref [\unskip ]{##1}} \newcommand {\ref }[0]{\@ifstar \@refstar \T@ref } \newcommand {\pageref }[0]{\@ifstar \@pagerefstar \T@pageref } \newcommand {\nameref }[0]{\@ifstar \@namerefstar \T@nameref } \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{}$

Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Problems

Here are the main derivative rules of the section.

Assume $f(x)$ and $g(x)$ are differentiable functions.

(Chain Rule) $\displaystyle \frac {d}{dx}f(g(x)) = f'(g(x))g'(x)$ .
(Exponentials) $\displaystyle \frac {d}{dx}b^x = b^x \ln (b)$ .

If $f(x) = e^{x^2}$ , find $f'(x)$ .

We can write $f(x)$ as a composition of two functions $g$ and $h$ , say $f(x) = g(h(x))$ , where $g(x) = e^x, \quad h(x) = \answer {x^2}.$ Notice $g'(x) = \answer {e^x}, \quad h'(x) = \answer {2x},$ and $g'(h(x)) = \answer {e^{x^2}}.$ Therefore, the chain rule says $f'(x) = g'(h(x))h'(x) = \answer {2xe^{x^2}}.$

If $f(x) = \sin (x\cos (x))$ , find $f'(x)$ .

Again, $f(x)$ is a composition of two functions, say $f(x) = g(h(x))$ , where $g(x) = \sin (x), \quad h(x) = \answer {x\cos (x)}.$ Then $g'(x) = \answer {\cos (x)}$ , and to differentiate $h(x)$ , we need to use the productquotientpower rule. Doing so gives $h'(x) = \answer {-x\sin (x)+\cos (x)}.$ Finally, notice $g'(h(x)) = \answer {\cos (x\cos (x))}$ . Therefore, $f'(x) = g'(h(x))h'(x) = \answer {\cos (x\cos (x))(-x\sin (x)+\cos (x))}.$

Note that the chain rule applies even when you have more than two functions composed, the pattern remains the same: peel the function from the outside in, taking derivatives and leaving the core intact at each step.

If $f(x) = \sqrt {1+\sin (x^2)}$ , then $f'(x) = \answer {\frac {x\cos (x^2)}{\sqrt {1+\sin (x^2)}}}.$

Write $f(x) = g(h(k(x)))$ , where $g(x) = \sqrt {x},\quad h(x) = 1+\sin (x),\quad k(x) = \answer {x^2}.$ Then $g'(x) = \answer {\frac {1}{2\sqrt {x}}},\quad h'(x) = \answer {\cos (x)},\quad k'(x) = \answer {2x}.$ Then $g'(h(k(x))) = \answer {\frac {1}{2\sqrt {1+\sin (x^2)}}},$ and $h'(k(x)) = \answer {\cos (x^2)}$ . Now put everything together to get $f'(x)$ .

For the following functions, find $f'(x)$ .

If $f(x) = \tan (x^3)$ , then $f'(x) = \answer {3x^2\sec ^2(x^3)}$ .
If $f(x) = (x^4+x)^{100}$ , then $f'(x) = \answer {100(x^4+x)^{99}(4x^3+1)}$ .
If $\displaystyle f(x) = \frac {\sin (x)}{\cos (x)+1}$ , then $f'(x) = \answer {\frac {\cos (x)(\cos (x)+1)+\sin (x)\sin (x)}{(\cos (x)+1)^2}}$ .
If $f(x) = \sin (x)\cos (2x+1)$ , then $f'(x) = \answer {-2\sin (x)\sin (2x+1)+\cos (x)\cos (2x+1)}$ .
If $f(x) = e^{1-x}$ , then $f'(x) = \answer {-e^{1-x}}$ .
If $\displaystyle f(x) = \frac {xe^{1-2x}}{x^2+1}$ , then $f'(x) = \answer {\frac {(x^2+1)(e^{1-2x}-2xe^{1-2x})-xe^{1-2x}(2x)}{(x^2+1)^2}}$ .

Give examples to show $\displaystyle \frac {d}{dx}f(g(x)) \neq f'(g'(x))$ .

Consider $f(x) = x$ and $g(x) = 1$ . Then $f(g(x)) = \answer {1}$ , so $\frac {d}{dx}f(g(x)) = \answer {0}.$ However, notice $f'(x) = \answer {1}$ and $g'(x) = \answer {0}$ , so $f'(g'(x)) = \answer {1} \neq 0$ . Therefore, the two cannot be equal.

Consider $f(x) = 5^{-(x+1)^2}+1$ . How many horizontal asymptotes does $f(x)$ have? $\answer {1}$ .

The equation of the horizontal asymptote is $y = \answer {1}$ .

The derivative is $f'(x) = \answer {5^{-(x+1)^2}\ln (5)(-2x-2)}$ .

How many horizontal tangent lines does $f(x)$ have? $\answer {1}$ .

The horizontal tangent line happens at the point $(\answer {-1},\answer {2})$ .

Consider the following table giving the values of $f(x)$ , $g(x)$ , $f'(x)$ , and $g'(x)$ . $\begin {array}{c|cccc} x & f(x) & g(x) & f'(x) & g'(x) \\ \hline 0 & 1 & 3 & 1 & 2 \\ 1 & 0 & 1 & 2 & -1 \\ 2 & 2 & -1 & 3 & 0 \\ \end {array}$ Find:

$f(1) = \answer {0}$ .
$f(f(2)) = \answer {2}$ .
$g(f(1)) = \answer {3}$ .
$\displaystyle \frac {d}{dx}f(g(x))\Big |_{x=1} = \answer {-2}$ .
The equation of the tangent line to $y = f(x^2g(x))$ at $x=1$ is $y = \answer {2x-2}.$