2.7 Chain Rule

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\newrobustcmd }[0]{} \newcommand {\csshow }[1]{\begingroup \expandafter \endgroup \expandafter \show \csname #1\endcsname } \newcommand {\csmeaning }[1]{\ifcsname #1\endcsname \expandafter \meaning \csname #1\endcsname \else \detokenize {undefined}\fi } \newcommand {\ifdefmacro }[0]{} \newcommand {\ifdefparam }[0]{} \newcommand {\ifdefprotected }[0]{} \newcommand {\ifnumequal }[1]{\ifnumcomp {#1}=} \newcommand {\ifnumgreater }[1]{\ifnumcomp {#1}>} \newcommand {\ifnumless }[1]{\ifnumcomp {#1}<} \newcommand {\ifdimequal }[1]{\ifdimcomp {#1}=} \newcommand {\ifdimgreater }[1]{\ifdimcomp {#1}>} \newcommand {\ifdimless }[1]{\ifdimcomp {#1}<} \newcommand {\expandonce }[1]{\unexpanded \expandafter {#1}} \newcommand {\csexpandonce }[1]{\expandafter \expandonce \csname #1\endcsname } \newcommand {\protecting }[0]{} \newcommand {\csdef }[1]{\expandafter \def \csname #1\endcsname } \newcommand {\csedef }[1]{\expandafter \edef \csname #1\endcsname } \newcommand {\csgdef }[1]{\expandafter \gdef \csname #1\endcsname } \newcommand {\csxdef }[1]{\expandafter \xdef \csname #1\endcsname } \newcommand {\cslet }[2]{\expandafter \let \csname #1\endcsname #2} \newcommand {\letcs }[2]{\ifcsdef {#2} {\expandafter \let \expandafter #1\csname #2\endcsname } {\undef #1}} \newcommand {\csletcs }[2]{\ifcsdef {#2} {\expandafter \let \csname #1\expandafter \endcsname \csname #2\endcsname } {\csundef {#1}}} \newcommand {\csuse }[1]{\ifcsname #1\endcsname \csname #1\expandafter \endcsname \fi } \newcommand {\appto }[2]{\ifundef {#1} {\edef #1{\unexpanded {#2}}} {\edef #1{\expandonce #1\unexpanded {#2}}}} \newcommand {\eappto }[2]{\ifundef {#1} {\edef #1{#2}} {\edef #1{\expandonce #1#2}}} \newcommand {\gappto }[2]{\ifundef {#1} {\xdef #1{\unexpanded {#2}}} {\xdef #1{\expandonce #1\unexpanded {#2}}}} \newcommand {\xappto }[2]{\ifundef {#1} {\xdef #1{#2}} {\xdef #1{\expandonce #1#2}}} \newcommand {\preto }[2]{\ifundef {#1} {\edef #1{\unexpanded {#2}}} {\edef #1{\unexpanded {#2}\expandonce #1}}} \newcommand {\epreto }[2]{\ifundef {#1} {\edef #1{#2}} {\edef #1{#2\expandonce #1}}} \newcommand {\gpreto }[2]{\ifundef {#1} {\xdef #1{\unexpanded {#2}}} {\xdef #1{\unexpanded {#2}\expandonce #1}}} \newcommand {\xpreto }[2]{\ifundef {#1} {\xdef #1{#2}} {\xdef #1{#2\expandonce #1}}} \newcommand {\csappto }[1]{\expandafter \appto \csname #1\endcsname } \newcommand {\cseappto }[1]{\expandafter \eappto \csname #1\endcsname } \newcommand {\csgappto }[1]{\expandafter \gappto \csname #1\endcsname } \newcommand {\csxappto }[1]{\expandafter \xappto \csname #1\endcsname } \newcommand {\cspreto }[1]{\expandafter \preto \csname #1\endcsname } \newcommand {\csepreto }[1]{\expandafter \epreto \csname #1\endcsname } \newcommand {\csgpreto }[1]{\expandafter \gpreto \csname #1\endcsname } \newcommand {\csxpreto }[1]{\expandafter \xpreto \csname #1\endcsname } \newcommand {\csnumdef }[1]{\expandafter \numdef \csname #1\endcsname } \newcommand {\csnumgdef }[1]{\expandafter \numgdef \csname #1\endcsname } \newcommand {\csdimdef }[1]{\expandafter \dimdef \csname #1\endcsname } \newcommand {\csdimgdef }[1]{\expandafter \dimgdef \csname #1\endcsname } \newcommand {\csgluedef }[1]{\expandafter \gluedef \csname #1\endcsname } \newcommand {\csgluegdef }[1]{\expandafter \gluegdef \csname #1\endcsname } \newcommand {\mudef }[2]{\ifundef #1{\def #1{0mu}}{}\edef #1{\the \muexpr #2}} \newcommand {\csmudef }[1]{\expandafter \mudef \csname #1\endcsname } \newcommand {\csmugdef }[1]{\expandafter \mugdef \csname #1\endcsname } \newcommand {\listbreak }[0]{} \newcommand {\listadd }[2]{\ifblank {#2}{}{\appto #1{#2|}}} \newcommand {\listgadd }[2]{\ifblank {#2}{}{\gappto #1{#2|}}} \newcommand {\listcsadd }[1]{\expandafter \listadd \csname #1\endcsname } \newcommand {\listcseadd }[1]{\expandafter \listeadd \csname #1\endcsname } \newcommand {\listcsgadd }[1]{\expandafter \listgadd \csname #1\endcsname } \newcommand {\listcsxadd }[1]{\expandafter \listxadd \csname #1\endcsname } \newcommand {\dolistloop }[0]{\forlistloop \do } \newcommand {\dolistcsloop }[0]{\forlistcsloop \do } \newcommand {\AfterPreamble }[0]{\AtBeginDocument } \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\C }[0]{\mathbb C} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\dis }[0]{\displaystyle } \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ppx }[0]{\frac {\partial }{\partial x}} \newcommand {\ppy }[0]{\frac {\partial }{\partial y}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{ #1 \bigg |} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{proj} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\cis }{cis} \DeclareMathOperator {\Arg }{Arg} \DeclareMathOperator {\Rep }{Re} \DeclareMathOperator {\Imp }{Im} \DeclareMathOperator {\sech }{sech} \DeclareMathOperator {\csch }{csch} \DeclareMathOperator {\Log }{Log} \newcommand {\tightoverset }[2]{\mathop {#2}\limits ^{\vbox to -.5ex{\kern -0.75ex\hbox {$#1$}\vss }}} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\utan }[0]{\vec {\hat {t}}} \newcommand {\unormal }[0]{\vec {\hat {n}}} \newcommand {\ubinormal }[0]{\vec {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

We compute the derivative of a composition.

The Chain Rule

Compositions are common in typical looking functions. For example, the hypotensue of a right triangle with sides $x$ and $3$ is $\sqrt {x^2 + 9}$ . This function is a composition of the inside polynomial, $x^2 + 9,$ and the square root on the outside, $\sqrt {x}$ . Consider an exponential growth model with parameter $k$ , $f(t) = ae^{kt}$ , where $a>0$ . The expression $e^{kt}$ is the composition of the inner linear function $kt$ and the outer, exponential function $e^t$ . In this section, we will learn how to differentiate such functions.

Chain Rule If $f(x)$ and $g(x)$ are differentiable functions and if $h(x) = f(g(x))$ , then $h'(x) = f'(g(x))g'(x).$ This can also be written as $\frac {d}{dx}f(g(x)) = f'(g(x))g'(x),$

or in Leibniz notation as $\frac {dy}{dx} = \frac {dy}{du} \cdot \frac {du}{dx},$ where $u = g(x)$ and $y= f(g(x))$ .

In words, the derivative of a composition is the derivative of the outside (with the inside left in), times the derivative of the inside.

example 1 Find $h'(x)$ if $h(x) = \sin (2x)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = 2x \quad \text {and} \quad f(x) = \sin (x).$

To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) = 2; \quad \text {next}$ $f'(x) = \cos (x), \quad \text {so}$ $f'(g(x)) =\cos (g(x)) = \cos (2x).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \cos (2x) \cdot 2 = 2\cos (2x).$ We can also solve this problem using Leibniz notation. In that case, we let $u = 2x$ , and $y=\sin (2x)$ . Substituting $u$ for $2x$ in $y$ , we can write $y = \sin (u), \text { where } \, u = 2x.$ Then the derivatives are $\frac {dy}{du} = \cos (u) \,\,\,\text { and } \,\,\,\frac {du}{dx} = 2.$ Finally, the chain rule says to multiply thses derivatives: $\frac {dy}{dx} = \frac {dy}{du} \cdot \frac {du}{dx} = \cos (u) \cdot 2 = 2\cos (2x).$

Here is a video of Example 1

(problem 1)

Compute $\frac {d}{dx} \sin (5x) = \answer {5\cos (5x)}.$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The “outside” function is $f(x) = \sin (x)$ and the “inside” function is $g(x) = 5x$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

example 2

Find $h'(x)$ if $h(x) = e^{5x}$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = 5x \quad \text {and} \quad f(x) = e^x.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) = 5; \quad \text {and}$ $f'(x) = e^x, \quad \text {so}$ $f'(g(x)) =e^{g(x)} = e^{5x}.$ By the chain rule, $h'(x) = f'(g(x))g'(x) = e^{5x} \cdot 5 = 5e^{5x}.$

Here is a video of the example

(problem 2)

Compute $\frac {d}{dx} e^{3x} = \answer {3e^{3x}}.$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = e^x$ and the inside function is $g(x) = 3x$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

example 3 Find $h'(x)$ if $h(x) = \cos (x^2 + 1)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = x^2 + 1 \quad \text {and} \quad f(x) = \cos (x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) = 2x; \quad \text {next}$ $f'(x) = -\sin (x), \quad \text {so}$ $f'(g(x))= -\sin (g(x)) = -\sin (x^2 + 1).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = -\sin (x^2 + 1) \cdot 2x = -2x\sin (x^2 + 1).$

Here is a video of the example

(problem 3) Compute $\frac {d}{dx} \cos (2-x^3) = \answer {3x^2\sin (2-x^3)}.$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \cos (x)$ and the inside function is $g(x) = 2-x^3$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

example 4

Find $h'(x)$ if $h(x) = (x^3 - 4x + 2)^7$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = x^3 - 4x + 2 \quad \text {and} \quad f(x) = x^7.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) =3x^2 - 4; \quad \text {next}$ $f'(x) = 7x^6 , \quad \text {so}$ $f'(g(x)) = 7[g(x)]^6 = 7(x^3 - 4x + 2)^6 .$ By the chain rule, $h'(x) = f'(g(x))g'(x) = 7(x^3 + 4x + 2)^6(3x^2 - 4)=7(3x^2 - 4)(x^3 - 4x + 2)^6.$

Here is a video of the example

(problem 4) Compute $\frac {d}{dx} (x^3 - 4x + 5)^8 = \answer {8(3x^2 - 4)(x^3 - 4x + 5)^7 }.$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = x^8$ and the inside function is $x^3 - 4x + 5$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

example 5

Find $h'(x)$ if $h(x) = \sqrt {x^2 + 4}$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = x^2 + 4 \quad \text {and} \quad f(x) = \sqrt x.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) = 2x; \quad \text {next}$ $f'(x) = \frac {1}{2\sqrt x} , \quad \text {so}$ $f'(g(x)) = \frac {1}{2\sqrt {g(x)}}=\frac {1}{2\sqrt {x^2 + 4}}.$ By the chain rule,

$\begin{align*} h'(x) &= f'(g(x))g'(x)\\ &= \frac{1}{2\sqrt {x^2 + 4}}\cdot 2x\\ &= \frac{2x}{2\sqrt {x^2 + 4}}\\ =\frac{x}{\sqrt {x^2 + 4}}. \end{align*}$

Here is a video of the example

(problem 5) Compute $\frac {d}{dx} \sqrt {x^2 - 3} = \answer {\frac {x}{\sqrt {x^2 - 3}}}.$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \sqrt {x}$ and the inside function is $g(x) = x^2 - 3$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

example 6

Find $h'(x)$ if $h(x) = \tan (3x)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = 3x \quad \text {and} \quad f(x) =\tan (x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) =3; \quad \text {next}$ $f'(x) =\sec ^2(x) , \quad \text {so}$ $f'(g(x)) = \sec ^2(g(x)) = \sec ^2(3x).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \sec ^2(3x) \cdot 3= 3\sec ^2(3x).$

Here is a video of the example

(problem 6a) Compute $\frac {d}{dx} \tan (x^4) = \answer {4x^3 \sec ^2(x^4)}.$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \tan (x)$ and the inside function is $g(x) = x^4$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

(problem 6b) What is the derivative of $\sec (2x)$ ?

$\sec (2x)\tan (2x)$ $2\sec (x)\tan (x)$ $-2\sec (2x)\tan (2x)$ $2\sec (2x)\tan (2x)$

example 7 Find $h'(x)$ if $h(x) = \frac {3}{(2x + 5)^4}$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = 2x+ 5 \quad \text {and} \quad f(x) = \frac {3}{x^4} = 3x^{-4}.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) = 2; \quad \text {next}$ $f'(x) = -12x^{-5}, \quad \text {so}$ $f'(g(x)) = -12[g(x)]^{-5} = -12(2x+5)^{-5} = -\frac {12}{(2x+5)^5}.$ By the chain rule, $h'(x) = f'(g(x))g'(x) = -\frac {12}{(2x+5)^5}\cdot 2 = -\frac {24}{(2x+5)^5}.$

Here is a video of the example

(problem 7) Compute $\frac {d}{dx} \frac {2}{(3x - 8)^5} = \answer {\frac {-30}{(3x - 8)^6}}.$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = 2x^{-5}$ and the inside function is $g(x) = 3x - 8$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

example 8 Find $h'(x)$ if $h(x) = \sin (x^3)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = x^3 \quad \text {and} \quad f(x) = \sin (x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) = 3x^2; \quad \text {next}$ $f'(x) =\cos (x) , \quad \text {so}$ $f'(g(x)) = \cos (g(x)) = \cos (x^3).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \cos (x^3)\cdot 3x^2 = 3x^2\cos (x^3).$

Here is a video of the example

(problem 8) Compute $\frac {d}{dx} \sin (3x^5) = \answer {\cos (3x^5)\cdot 15x^4}.$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \sin (x)$ and the inside function is $g(x) = 3x^5$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

example 9 Find $h'(x)$ if $h(x) = \sin ^3(x)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) =\sin (x) \quad \text {and} \quad f(x) = x^3.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) =\cos (x); \quad \text {next}$ $f'(x) = 3x^2, \quad \text {so}$ $f'(g(x)) = 3g^2(x) = 3\sin ^2(x).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = 3\sin ^2(x)\cos (x).$

Here is a video of the example

(problem 9) Compute $\frac {d}{dx} \sin ^4(x) = \answer {4\sin ^3(x)\cos (x)}.$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = x^4$ and the inside function is $g(x) = \sin (x)$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

example 10 Find $h'(x)$ if $h(x) = e^{-x^2}$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = -x^2 \quad \text {and} \quad f(x) =e^x.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) = -2x; \quad \text {next}$ $f'(x) = e^x, \quad \text {so}$ $f'(g(x)) = e^{g(x)} = e^{-x^2}.$ By the chain rule, $h'(x) = f'(g(x))g'(x) = e^{-x^2} (-2x) \quad \text {or} \quad -2xe^{-x^2}.$

Here is a video of the example

(problem 10)

Compute $\frac {d}{dx} e^{\sqrt x} = \answer {\frac {e^{\sqrt x}}{2\sqrt x}}$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = e^x$ and the inside function is $g(x) = \sqrt x$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

example 11 Find $h'(x)$ if $h(x) = \ln (x^4 +1)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = x^4 + 1 \quad \text {and} \quad f(x) = \ln (x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) = 4x^3; \quad \text {next}$ $f'(x) = \frac {1}{x}, \quad \text {so}$ $f'(g(x)) =\frac {1}{g(x)} = \frac {1}{x^4 + 1} .$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \frac {1}{x^4 + 1} \cdot 4x^3 = \frac {4x^3}{x^4 + 1}.$

Here is a video of the example

(problem 11) Compute $\frac {d}{dx} \ln (x^3 -2) = \answer {\frac {3x^2}{x^3 - 2}}.$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \ln (x)$ and the inside function is $g(x) = x^3 - 2$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

example 12 Find $h'(x)$ if $h(x) = \ln ^5(x)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = \ln (x) \quad \text {and} \quad f(x) = x^5.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) = \frac {1}{x}; \quad \text {next}$ $f'(x) = 5x^4, \quad \text {so}$ $f'(g(x)) = 5g^4(x) = 5\ln ^4(x).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = 5\ln ^4(x)\cdot \frac {1}{x} = \frac {5\ln ^4(x)}{x}.$

Here is a video of the example

(problem 12)

Compute $\frac {d}{dx} \ln ^3(x) = \answer {\frac {3\ln ^2(x)}{x}}$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = x^3$ and the inside function is $g(x) = \ln (x)$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

example 13 Find $h'(x)$ if $h(x) = \ln (\sin (x))$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = \sin (x) \quad \text {and} \quad f(x) = \ln (x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) = \cos (x); \quad \text {next}$ $f'(x) = \frac {1}{x}, \quad \text {so}$ $f'(g(x)) = \frac {1}{g(x)} = \frac {1}{\sin (x)}.$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \frac {1}{\sin (x)} \cdot \cos (x) = \frac {\cos (x)}{\sin (x)} = \cot (x).$

Here is a video of the example

(problem 13a) Compute $\frac {d}{dx} \ln (\sec (x)) = \answer {\tan (x)}$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \ln (x)$ and the inside function is $g(x) = \sec (x)$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

(problem 13b) Compute $\frac {d}{dx} \cos (\ln (x)) = \answer {\frac {-\sin (\ln (x))}{x}}.$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \cos (x)$ and the inside function is $g(x) = \ln (x)$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

example 14 Find $\frac {d}{dx} \left [\ln (g(x))\right ]$ .
We use the chain rule with $g(x)$ as the inside function and we get $\frac {d}{dx} \left [\ln (g(x))\right ] = \frac {1}{g(x)}\cdot g'(x) = \frac {g'(x)}{g(x)}.$

The above formula is both useful and important. We will see it again in the Logarithmic Differentiation section.

Find $\frac {d}{dx} \left [\ln (x^3 + 2)\right ]$ .

We use the above formula with $g(x) = x^3 + 2$ to get $\frac {d}{dx} \left [\ln (x^3 + 2)\right ] = \frac {3x^2}{x^3 + 2}.$

(problem 14) Compute $\frac {d}{dx} \left [\ln (x^4 + 3x - 1)\right ] = \answer {\frac {4x^3 + 3}{x^4 + 3x -1}}.$

Here are some detailed, lecture style videos on the chain rule: