빌드라인 스쿨

문제 은행 · 미적분 · 대학 미적분 · 다변수 미분

편도함수

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

CLP-3 · 2.3절 · 3번대학 수준우리말 번역

주어진 편도함수를 구하세요.

(a) f(x,y)=x2y3f(x,y) = x^2y^3; fxx(x,y)f_{xx}(x,y), fxyy(x,y)f_{xyy}(x,y), fyxy(x,y)f_{yxy}(x,y) (b) f(x,y)=exy2f(x,y) = e^{xy^2}; fxx(x,y)f_{xx}(x,y), fxy(x,y)f_{xy}(x,y), fxxy(x,y)f_{xxy}(x,y), fxyy(x,y)f_{xyy}(x,y) (c) f(u,v,w)=1u+2v+3w \displaystyle f(u,v,w) = \frac{1}{u+2v+3w}\ , ∂3f∂u∂v∂w(u,v,w) \displaystyle \frac{\partial^3 f}{\partial u\partial v\partial w}(u,v,w)\ , ∂3f∂u∂v∂w(3,2,1)\displaystyle \frac{\partial^3 f}{\partial u\partial v\partial w}(3,2,1)

영어 원문 보기

Find the specified partial derivatives.

(a) f(x,y)=x2y3f(x,y) = x^2y^3; fxx(x,y)f_{xx}(x,y), fxyy(x,y)f_{xyy}(x,y), fyxy(x,y)f_{yxy}(x,y) (b) f(x,y)=exy2f(x,y) = e^{xy^2}; fxx(x,y)f_{xx}(x,y), fxy(x,y)f_{xy}(x,y), fxxy(x,y)f_{xxy}(x,y), fxyy(x,y)f_{xyy}(x,y) (c) f(u,v,w)=1u+2v+3w \displaystyle f(u,v,w) = \frac{1}{u+2v+3w}\ , ∂3f∂u∂v∂w(u,v,w) \displaystyle \frac{\partial^3 f}{\partial u\partial v\partial w}(u,v,w)\ , ∂3f∂u∂v∂w(3,2,1)\displaystyle \frac{\partial^3 f}{\partial u\partial v\partial w}(3,2,1)

정답·풀이 보기

(a) 다음과 같다.

fx(x,y)=2xy3fxx(x,y)=2y3fxy(x,y)=6xy2fyxy(x,y)=fxyy(x,y)=12xy\begin{aligned} f_x(x,y) &= 2xy^3 & f_{xx}(x,y) &= 2y^3 \\ & & f_{xy}(x,y) &= 6xy^2 & f_{yxy}(x,y) = f_{xyy}(x,y) &= 12xy \end{aligned}

(b) 다음과 같다.

fx(x,y)=y2exy2fxx(x,y)=y4exy2fxxy(x,y)=4y3exy2+2xy5exy2fxy(x,y)=2yexy2+2xy3exy2fxyy(x,y)=(2+4xy2+6xy2+4x2y4)exy2=(2+10xy2+4x2y4)exy2\begin{aligned} f_x(x,y) &= y^2e^{xy^2} & f_{xx}(x,y) &= y^4e^{xy^2} & f_{xxy}(x,y) &= 4y^3e^{xy^2} + 2xy^5e^{xy^2} \\ & & f_{xy}(x,y) &= 2ye^{xy^2}+2xy^3e^{xy^2} & f_{xyy}(x,y) &= \big(2+4xy^2+6xy^2+4x^2y^4\big)e^{xy^2}\\ & & & & &= \big(2+10xy^2+4x^2y^4\big)e^{xy^2} \end{aligned}

(c) 다음과 같다.

∂f∂u(u,v,w)=−1(u+2v+3w)2∂2f∂u ∂v(u,v,w)=4(u+2v+3w)3∂3f∂u ∂v ∂w(u,v,w)=−36(u+2v+3w)4\begin{aligned} \frac{\partial f}{\partial u}(u,v,w) &= -\frac{1}{(u+2v+3w)^2} \\ \frac{\partial^2 f}{\partial u\,\partial v}(u,v,w) &= \frac{4}{(u+2v+3w)^3} \\ \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(u,v,w) &= -\frac{36}{(u+2v+3w)^4} \end{aligned}

특히

∂3f∂u ∂v ∂w(3,2,1)=−36(3+2×2+3×1)4=−36104=−92500\begin{aligned} \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(3,2,1) &= -\frac{36}{(3+2\times 2+3\times 1)^4} = -\frac{36}{10^4} = -\frac{9}{2500} \end{aligned}

원본 교재 풀이(영어)

(a) We have

fx(x,y)=2xy3fxx(x,y)=2y3fxy(x,y)=6xy2fyxy(x,y)=fxyy(x,y)=12xy\begin{aligned} f_x(x,y) &= 2xy^3 & f_{xx}(x,y) &= 2y^3 \\ & & f_{xy}(x,y) &= 6xy^2 & f_{yxy}(x,y) = f_{xyy}(x,y) &= 12xy \end{aligned}

(b) We have

fx(x,y)=y2exy2fxx(x,y)=y4exy2fxxy(x,y)=4y3exy2+2xy5exy2fxy(x,y)=2yexy2+2xy3exy2fxyy(x,y)=(2+4xy2+6xy2+4x2y4)exy2=(2+10xy2+4x2y4)exy2\begin{aligned} f_x(x,y) &= y^2e^{xy^2} & f_{xx}(x,y) &= y^4e^{xy^2} & f_{xxy}(x,y) &= 4y^3e^{xy^2} + 2xy^5e^{xy^2} \\ & & f_{xy}(x,y) &= 2ye^{xy^2}+2xy^3e^{xy^2} & f_{xyy}(x,y) &= \big(2+4xy^2+6xy^2+4x^2y^4\big)e^{xy^2}\\ & & & & &= \big(2+10xy^2+4x^2y^4\big)e^{xy^2} \end{aligned}

(c) We have

∂f∂u(u,v,w)=−1(u+2v+3w)2∂2f∂u ∂v(u,v,w)=4(u+2v+3w)3∂3f∂u ∂v ∂w(u,v,w)=−36(u+2v+3w)4\begin{aligned} \frac{\partial f}{\partial u}(u,v,w) &= -\frac{1}{(u+2v+3w)^2} \\ \frac{\partial^2 f}{\partial u\,\partial v}(u,v,w) &= \frac{4}{(u+2v+3w)^3} \\ \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(u,v,w) &= -\frac{36}{(u+2v+3w)^4} \end{aligned}

In particular

∂3f∂u ∂v ∂w(3,2,1)=−36(3+2×2+3×1)4=−36104=−92500\begin{aligned} \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(3,2,1) &= -\frac{36}{(3+2\times 2+3\times 1)^4} = -\frac{36}{10^4} = -\frac{9}{2500} \end{aligned}

한 번 더 — 비슷한 문제OpenStax 3권 · 4.3절 · 18번우리말 번역

z=e−xcos⁡(y)z={e}^{-x}\cos (y) 에 대하여 (0,1)(0,1) 에서 ∂z∂x\frac{\partial z}{\partial x} 를 구하세요. 점 P(0,1)P(0,1) 에서 편도함수의 값을 구하세요.

다 풀었으면 정답 확인

∂z∂x=−cos⁡(1)\frac{\partial z}{\partial x}=-\cos (1)

이 문제로 이어서 풀기
# 편도함수계산보통자동 분류

「편도함수」 문제 목록으로