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합성함수의 미분법

먼저 혼자 풀어 보고, 풀이로 확인한 다음, 아래 「한 번 더」 문제로 이어서 연습하세요.

CLP-1 · 2.9절 · 34번우리말 번역

f(x)=ecsc⁡x2x3−9tan⁡x3f(x)=\sqrt[3]{\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }} 의 도함수를 구하세요. 답을 간단히 하지 않아도 됩니다.

영어 원문 보기

Evaluate the derivative of f(x)=ecsc⁡x2x3−9tan⁡x3f(x)=\sqrt[3]{\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }}. You do not have to simplify your answer.

정답·풀이 보기

f(x)=ecsc⁡x2x3−9tan⁡x3=(ecsc⁡x2x3−9tan⁡x)13To begin the differentiation, we can choose our “outside" function to be g(x)=x13, and our “inside" function to be h(x)=ecsc⁡x2x3−9tan⁡x. Then f(x)=g(h(x)), so f′(x)=g′(h(x))⋅h′(x)=13(h(x))−23h′(x):f′(x)=13(ecsc⁡x2x3−9tan⁡x)−23⋅ddx{ecsc⁡x2x3−9tan⁡x}=13(x3−9tan⁡xecsc⁡x2)23⋅ddx{ecsc⁡x2x3−9tan⁡x}This leads us to use the quotient rule:=13(x3−9tan⁡xecsc⁡x2)23(x3−9tan⁡xddx{ecsc⁡x2}−ecsc⁡x2ddx{x3−9tan⁡x}(tan⁡2x)(x3−9))Let’s figure out those two derivatives on their own, then plug them in. Using the chain rule twice:ddx{ecsc⁡x2}=ecsc⁡x2ddx{csc⁡x2}=ecsc⁡x2⋅(−csc⁡(x2)cot⁡(x2))⋅ddx{x2}=−2xecsc⁡x2csc⁡(x2)cot⁡(x2)For the other derivative, we start with the product rule, then chain:ddx{x3−9tan⁡x}=ddx{x3−9}⋅tan⁡x+x3−9sec⁡2x=12x3−9ddx{x3−9}⋅tan⁡x+x3−9sec⁡2x=3x2tan⁡x2x3−9+x3−9sec⁡2xNow, we plug these into our equation for f′(x):f′(x)=13(x3−9tan⁡xecsc⁡x2)23(x3−9tan⁡xddx{ecsc⁡x2}−ecsc⁡x2ddx{x3−9tan⁡x}(tan⁡2x)(x3−9))=13(x3−9tan⁡xecsc⁡x2)23⋅ (x3−9tan⁡x(−2x)ecsc⁡x2csc⁡(x2)cot⁡(x2)−ecsc⁡x2(3x2tan⁡x2x3−9+x3−9sec⁡2x)(tan⁡2x)(x3−9))\begin{aligned} f(x)&=\sqrt[3]{\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }}\\ &=\left({\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }}\right)^{\frac{1}{3}} \\ \text{To begin the differentiation, we can choose our ``outside" function to be $g(x)=x^{\frac{1}{3}}$, and our ``inside" function to be $h(x)=\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }$. Then $f(x)=g(h(x))$, so $f'(x)=g'(h(x))\cdot h'(x)=\frac{1}{3}(h(x))^{-\frac{2}{3}}h'(x)$:} \\ f'(x)&=\frac{1}{3}\left( \dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }\right)^{\frac{-2}{3}}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }\right\}\\ &=\frac{1}{3} \left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}} \cdot \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }\right\} \\ \text{This leads us to use the quotient rule:} \\ &=\frac{1}{3}\left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}} \left(\frac{ \sqrt{x^3-9}\tan x \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\csc x^2}\right\}-e^{\csc x^2}\frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^3-9}\tan x \right\}}{(\tan^2 x)(x^3-9) }\right) \\ \text{Let's figure out those two derivatives on their own, then plug them in. Using the chain rule twice:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\csc x^2}\right\}&=e^{ \csc x^2}\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \csc x^2\right\}=e^{\csc x^2}\cdot (-\csc( x^2)\cot( x^2))\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ x^2\}\\&=-2xe^{\csc x^2}\csc(x^2)\cot(x^2) \\ \text{For the other derivative, we start with the product rule, then chain:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \sqrt{x^3-9} \tan x \right\}&= \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^3-9}\right\}\cdot\tan x+\sqrt{x^3-9}\sec^2 x\\ &=\frac{1}{2\sqrt{ x^3-9}}\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^3-9\right\}\cdot \tan x+\sqrt{x^3-9}\sec^2 x\\ &=\frac{3x^2 \tan x}{2\sqrt{{x^3-9}}}+\sqrt{x^3-9}\sec^2 x \\ \text{Now, we plug these into our equation for $f'(x)$:} \\ f'(x)&=\frac{1}{3}\left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}} \left(\frac{ \sqrt{x^3-9}\tan x \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\csc x^2}\right\}-e^{\csc x^2} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^3-9}\tan x \right\}}{(\tan^2 x)(x^3-9) }\right)\\ &=\frac{1}{3}\left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}}\cdot\\ &~\left(\frac{ \sqrt{x^3-9}\tan x (-2x)e^{\csc x^2}\csc(x^2)\cot(x^2)-e^{\csc x^2} \left(\frac{3x^2\tan x}{2\sqrt{{x^3-9}}}+\sqrt{x^3-9}\sec^2 x\right)}{(\tan^2 x)(x^3-9) }\right) \end{aligned}

원본 교재 풀이(영어)

f(x)=ecsc⁡x2x3−9tan⁡x3=(ecsc⁡x2x3−9tan⁡x)13To begin the differentiation, we can choose our “outside" function to be g(x)=x13, and our “inside" function to be h(x)=ecsc⁡x2x3−9tan⁡x. Then f(x)=g(h(x)), so f′(x)=g′(h(x))⋅h′(x)=13(h(x))−23h′(x):f′(x)=13(ecsc⁡x2x3−9tan⁡x)−23⋅ddx{ecsc⁡x2x3−9tan⁡x}=13(x3−9tan⁡xecsc⁡x2)23⋅ddx{ecsc⁡x2x3−9tan⁡x}This leads us to use the quotient rule:=13(x3−9tan⁡xecsc⁡x2)23(x3−9tan⁡xddx{ecsc⁡x2}−ecsc⁡x2ddx{x3−9tan⁡x}(tan⁡2x)(x3−9))Let’s figure out those two derivatives on their own, then plug them in. Using the chain rule twice:ddx{ecsc⁡x2}=ecsc⁡x2ddx{csc⁡x2}=ecsc⁡x2⋅(−csc⁡(x2)cot⁡(x2))⋅ddx{x2}=−2xecsc⁡x2csc⁡(x2)cot⁡(x2)For the other derivative, we start with the product rule, then chain:ddx{x3−9tan⁡x}=ddx{x3−9}⋅tan⁡x+x3−9sec⁡2x=12x3−9ddx{x3−9}⋅tan⁡x+x3−9sec⁡2x=3x2tan⁡x2x3−9+x3−9sec⁡2xNow, we plug these into our equation for f′(x):f′(x)=13(x3−9tan⁡xecsc⁡x2)23(x3−9tan⁡xddx{ecsc⁡x2}−ecsc⁡x2ddx{x3−9tan⁡x}(tan⁡2x)(x3−9))=13(x3−9tan⁡xecsc⁡x2)23⋅ (x3−9tan⁡x(−2x)ecsc⁡x2csc⁡(x2)cot⁡(x2)−ecsc⁡x2(3x2tan⁡x2x3−9+x3−9sec⁡2x)(tan⁡2x)(x3−9))\begin{aligned} f(x)&=\sqrt[3]{\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }}\\ &=\left({\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }}\right)^{\frac{1}{3}} \\ \text{To begin the differentiation, we can choose our ``outside" function to be $g(x)=x^{\frac{1}{3}}$, and our ``inside" function to be $h(x)=\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }$. Then $f(x)=g(h(x))$, so $f'(x)=g'(h(x))\cdot h'(x)=\frac{1}{3}(h(x))^{-\frac{2}{3}}h'(x)$:} \\ f'(x)&=\frac{1}{3}\left( \dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }\right)^{\frac{-2}{3}}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }\right\}\\ &=\frac{1}{3} \left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}} \cdot \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }\right\} \\ \text{This leads us to use the quotient rule:} \\ &=\frac{1}{3}\left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}} \left(\frac{ \sqrt{x^3-9}\tan x \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\csc x^2}\right\}-e^{\csc x^2}\frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^3-9}\tan x \right\}}{(\tan^2 x)(x^3-9) }\right) \\ \text{Let's figure out those two derivatives on their own, then plug them in. Using the chain rule twice:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\csc x^2}\right\}&=e^{ \csc x^2}\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \csc x^2\right\}=e^{\csc x^2}\cdot (-\csc( x^2)\cot( x^2))\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ x^2\}\\&=-2xe^{\csc x^2}\csc(x^2)\cot(x^2) \\ \text{For the other derivative, we start with the product rule, then chain:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \sqrt{x^3-9} \tan x \right\}&= \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^3-9}\right\}\cdot\tan x+\sqrt{x^3-9}\sec^2 x\\ &=\frac{1}{2\sqrt{ x^3-9}}\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^3-9\right\}\cdot \tan x+\sqrt{x^3-9}\sec^2 x\\ &=\frac{3x^2 \tan x}{2\sqrt{{x^3-9}}}+\sqrt{x^3-9}\sec^2 x \\ \text{Now, we plug these into our equation for $f'(x)$:} \\ f'(x)&=\frac{1}{3}\left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}} \left(\frac{ \sqrt{x^3-9}\tan x \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\csc x^2}\right\}-e^{\csc x^2} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^3-9}\tan x \right\}}{(\tan^2 x)(x^3-9) }\right)\\ &=\frac{1}{3}\left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}}\cdot\\ &~\left(\frac{ \sqrt{x^3-9}\tan x (-2x)e^{\csc x^2}\csc(x^2)\cot(x^2)-e^{\csc x^2} \left(\frac{3x^2\tan x}{2\sqrt{{x^3-9}}}+\sqrt{x^3-9}\sec^2 x\right)}{(\tan^2 x)(x^3-9) }\right) \end{aligned}

한 번 더 — 비슷한 문제CLP-1 · 2.9절 · 29번우리말 번역

f(x)=excos⁡3(5x−7)f(x)=\dfrac{e^{x}}{\cos^3 (5x-7)} 의 도함수를 구하세요.

답 쓰고 채점
다 풀었으면 정답 확인

풀이를 두 가지 제시한다. 하나는 바로 몫의 미분법을 쓰는 풀이이고, 다른 하나는 먼저 간단히 한 뒤 곱의 미분법을 쓰는 풀이이다.

  • 풀이 1: 몫의 미분법으로 시작한다:

f′(x)=cos⁡3(5x−7)ddx{ex}−exddx{cos⁡3(5x−7)}cos⁡6(5x−7)=cos⁡3(5x−7)ex−exddx{cos⁡3(5x−7)}cos⁡6(5x−7)이제 합성함수의 미분법을 쓴다. cos⁡3(5x−7)=[cos⁡(5x−7)]3 이므로 “바깥" 함수는 g(x)=x3 이고, “안쪽" 함수는 h(x)=cos⁡(5x−1) 이다.=cos⁡3(5x−7)ex−ex⋅3cos⁡2(5x−7)⋅ddx{cos⁡(5x−7)}cos⁡6(5x−7)합성함수의 미분법이 다시 필요하다!=cos⁡3(5x−7)ex−ex⋅3cos⁡2(5x−7)⋅[−sin⁡(5x−7)⋅ddx{5x−7}]cos⁡6(5x−7)=cos⁡3(5x−7)ex−ex⋅3cos⁡2(5x−7)⋅[−sin⁡(5x−7)⋅5]cos⁡6(5x−7)간단히 하여 마무리한다:=excos⁡2(5x−7)(cos⁡(5x−7)+15sin⁡(5x−7))cos⁡6(5x−7)=excos⁡(5x−7)+15sin⁡(5x−7)cos⁡4(5x−7)=ex(sec⁡3(5x−7)+15tan⁡(5x−7)sec⁡3(5x−7))=exsec⁡3(5x−7)(1+15tan⁡(5x−7))\begin{aligned} f'(x) &= \frac{\cos^3(5x-7)\frac{\mathrm{d}}{\mathrm{d}x}\{e^x\}-e^x\frac{\mathrm{d}}{\mathrm{d}x}\{\cos^3(5x-7)\}}{\cos^6(5x-7)}\\ &= \frac{\cos^3(5x-7)e^x-e^x\frac{\mathrm{d}}{\mathrm{d}x}\{\cos^3(5x-7)\}}{\cos^6(5x-7)} \\ \text{이제 합성함수의 미분법을 쓴다. $\cos^3(5x-7)=[\cos(5x-7)]^3$ 이므로 ``바깥" 함수는 $g(x)=x^3$ 이고, ``안쪽" 함수는 $h(x)=\cos(5x-1)$ 이다.} \\ &= \frac{\cos^3(5x-7)e^x-e^x\cdot3 \cos^2 (5x-7) \cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ \cos(5x-7)\}}{\cos^6(5x-7)} \\ \text{합성함수의 미분법이 다시 필요하다!} \\ &= \frac{\cos^3(5x-7)e^x-e^x\cdot3{\cos}^2{(5x-7)} \cdot[{-\sin( 5x-7)\cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ 5x-7\}}]}{\cos^6(5x-7)}\\ &= \frac{\cos^3(5x-7)e^x-e^x\cdot3{\cos}^2{(5x-7)} \cdot[{-\sin({5x-7})\cdot5}]}{\cos^6(5x-7)} \\ \text{간단히 하여 마무리한다:} \\ &= \frac{e^x\cos^2(5x-7)\left(\cos(5x-7)+15\sin(5x-7)\right)}{\cos^6(5x-7)}\\ &=e^x \frac{\cos(5x-7)+15\sin(5x-7)}{\cos^4(5x-7)}\\ &=e^x(\sec^3(5x-7)+15\tan(5x-7)\sec^3(5x-7))\\ &=e^x\sec^3(5x-7)(1+15\tan(5x-7)) \end{aligned}

  • 풀이 2: 몫의 미분법을 피하기 위해 먼저 간단히 한다:

f(x)=excos⁡3(5x−7)=exsec⁡3(5x−7)이제 곱의 미분법으로 미분한다:f′(x)=exsec⁡3(5x−7)+exddx{sec⁡3(5x−7)}여기서는 합성함수의 미분법이 필요하다. sec⁡3(5x−7)=[sec⁡(5x−7)]3 이므로 “바깥" 함수는 g(x)=x3 이고 “안쪽" 함수는 h(x)=sec⁡(5x−7) 이며, 따라서 g(h(x))=[sec⁡(5x−7)]3=sec⁡3(5x−7) 이다.=exsec⁡3(5x−7)+ex⋅3 sec⁡2(5x−7)⋅ddx{sec⁡(5x−7)}합성함수의 미분법이 다시 필요하다! ddx{sec⁡x}=sec⁡xtan⁡x 임을 떠올리자.=exsec⁡3(5x−7)+ex⋅3 sec⁡2(5x−7)⋅sec⁡(5x−7)tan⁡(5x−7)⋅ddx{5x−7}=exsec⁡3(5x−7)+ex⋅3 sec⁡2(5x−7)⋅sec⁡(5x−7)tan⁡(5x−7)⋅5간단히 하여 마무리한다:=exsec⁡3(5x−7)(1+15tan⁡(5x−7))\begin{aligned} f(x)&=\dfrac{e^{x}}{\cos^3 (5x-7)}\\ &=e^x\sec^3(5x-7) \\ \text{이제 곱의 미분법으로 미분한다:} \\ f'(x)&=e^x\sec^3(5x-7)+e^x\frac{\mathrm{d}}{\mathrm{d}x}\{\sec^3(5x-7)\} \\ \text{여기서는 합성함수의 미분법이 필요하다. $\sec^3(5x-7)=[\sec (5x-7)]^3$ 이므로 ``바깥" 함수는 $g(x)=x^3$ 이고 ``안쪽" 함수는 $h(x)=\sec(5x-7)$ 이며, 따라서 $g(h(x))=[\sec(5x-7)]^3=\sec^3(5x-7)$ 이다.} \\ &=e^x\sec^3(5x-7)+e^x\cdot3~ \sec^2 (5x-7) \cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ \sec(5x-7)\} \\ \text{합성함수의 미분법이 다시 필요하다! $\frac{\mathrm{d}}{\mathrm{d}x}\{\sec x\}=\sec x \tan x$ 임을 떠올리자.} \\ &=e^x\sec^3(5x-7)+e^x\cdot3~{\sec}^2{(5x-7)} \cdot {\sec( 5x-7)\tan( 5x-7)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ 5x-7\}}\\ &=e^x\sec^3(5x-7)+e^x\cdot3~{\sec}^2{(5x-7)} \cdot {\sec({5x-7})\tan({5x-7})\cdot 5} \\ \text{간단히 하여 마무리한다:} \\ &=e^x\sec^3(5x-7)(1+15\tan({5x-7})) \end{aligned}

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