合成関数の導関数・偏導関数
合成関数の導関数・偏導関数
合成関数の導関数・偏導関数は次のようになる。
\[ \frac{df}{dt}=\sum_{k=1}^{n}\frac{\partial f}{\partial x_{k}}\frac{dx_{k}}{dt} \] が成り立つ。
\[ \frac{\partial f}{\partial u_{i}}=\sum_{k=1}^{n}\frac{\partial f}{\partial x_{k}}\frac{\partial x_{k}}{\partial u_{i}} \] が成り立つ。
合成関数の導関数・偏導関数は次のようになる。
(1)合成関数の導関数
\(n\)変数関数\(f\left(x_{1},x_{2},\cdots,x_{n}\right)\)が全微分可能であり、\(x_{i}=x_{i}\left(t\right)\)が微分可能であるとき、\[ \frac{df}{dt}=\sum_{k=1}^{n}\frac{\partial f}{\partial x_{k}}\frac{dx_{k}}{dt} \] が成り立つ。
(2)合成関数の偏導関数
\(n\)変数関数\(f\left(x_{1},x_{2},\cdots,x_{n}\right)\)が全微分可能であり、\(x_{i}=x_{i}\left(u_{1},u_{2},\cdots,u_{m}\right)\)が偏微分可能であるとき、\[ \frac{\partial f}{\partial u_{i}}=\sum_{k=1}^{n}\frac{\partial f}{\partial x_{k}}\frac{\partial x_{k}}{\partial u_{i}} \] が成り立つ。
(1)
\(f\left(x,y\right)=x^{2}+2xy,x=t,y=t^{2}\)とすると、\begin{align*} \frac{df}{dt} & =\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}\\ & =\left(2x+2y\right)\cdot1+2x\cdot2t\\ & =2x+2y+4xt\\ & =2t+2t^{2}+4t^{2}\\ & =2t+6t^{2} \end{align*} となる。
(2)
\(f\left(x,y\right)=x+2xy+y^{2},x=u+uv,y=u+v\)とすると、\begin{align*} \frac{\partial f}{\partial u} & =\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\\ & =\left(1+2y\right)\left(1+v\right)+\left(2x+2y\right)\cdot1\\ & =2y\left(2+v\right)+2x+1+v\\ & =2\left(u+v\right)\left(2+v\right)+2\left(u+uv\right)+1+v\\ & =4u+2uv+4v+2v^{2}+2u+2uv+1+v\\ & =6u+4uv+2v^{2}+5v+1 \end{align*} \begin{align*} \frac{\partial f}{\partial v} & =\frac{\partial f}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial v}\\ & =\left(1+2y\right)u+\left(2x+2y\right)\cdot1\\ & =2y\left(1+u\right)+2x+u\\ & =2\left(u+v\right)\left(1+u\right)+2\left(u+uv\right)+u\\ & =2u+2u^{2}+2v+2uv+2u+2uv+u\\ & =2u^{2}+5u+4uv+2v \end{align*} となる。
(1)
\(f\)は全微分可能であるので、\[ df=\sum_{k=1}^{n}\frac{\partial f}{\partial x_{k}}dx_{k} \] となり、両辺を\(dt\)で割ると、
\[ \frac{df}{dt}=\sum_{k=1}^{n}\frac{\partial f}{\partial x_{k}}\frac{dx_{k}}{dt} \] となるので与式が成り立つ。
(1)-2
\(f\)が2変数関数の場合で示す。\begin{align*} \frac{df}{dt} & =\lim_{h\rightarrow0}\frac{f\left(x\left(t+h\right),y\left(t+h\right)\right)-f\left(x\left(t\right),y\left(t\right)\right)}{h}\\ & =\lim_{h\rightarrow0}\frac{f\left(x\left(t+h\right),y\left(t+h\right)\right)-f\left(x\left(t\right),y\left(t+h\right)\right)+f\left(x\left(t\right),y\left(t+h\right)\right)-f\left(x\left(t\right),y\left(t\right)\right)}{h}\\ & =\lim_{h\rightarrow0}\frac{f\left(x\left(t+h\right),y\left(t+h\right)\right)-f\left(x\left(t\right),y\left(t+h\right)\right)}{h}+=\lim_{h\rightarrow0}\frac{f\left(x\left(t\right),y\left(t+h\right)\right)-f\left(x\left(t\right),y\left(t\right)\right)}{h}\\ & =\lim_{h\rightarrow0}\frac{f\left(x\left(t+h\right),y\left(t+h\right)\right)-f\left(x\left(t\right),y\left(t+h\right)\right)}{x\left(t+h\right)-x\left(t\right)}\cdot\frac{x\left(t+h\right)-x\left(t\right)}{h}+=\lim_{h\rightarrow0}\frac{f\left(x\left(t\right),y\left(t+h\right)\right)-f\left(x\left(t\right),y\left(t\right)\right)}{y\left(t+h\right)-y\left(t\right)}\cdot\frac{y\left(t+h\right)-y\left(t\right)}{h}\\ & =\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt} \end{align*} となるので与式が成り立つ。
(2)
\(f\)は全微分可能であるので、\[ df=\sum_{k=1}^{n}\frac{\partial f}{\partial x_{k}}dx_{k} \] となり、\(i\ne j\)となる\(u_{j}\)を一定にして両辺を\(du_{i}\)で割ると、
\[ \frac{\partial f}{\partial u_{i}}=\sum_{k=1}^{n}\frac{\partial f}{\partial x_{k}}\frac{\partial x_{k}}{\partial u_{i}} \] となるので与式が成り立つ。
(2)-2
\(f\)が2変数関数\(f\left(x_{1},x_{2}\right)\)で\(x_{i}=x_{i}\left(u_{1},u_{2}\right),i\in\left\{ 1,2\right\} \)の場合で示す。\begin{align*} \frac{\partial f}{\partial u_{1}} & =\lim_{h\rightarrow0}\frac{f\left(x_{1}\left(u_{1}+h,u_{2}\right),x_{2}\left(u_{1}+h,u_{2}\right)\right)-f\left(x_{1}\left(u_{1},u_{2}\right),x_{2}\left(u_{1},u_{2}\right)\right)}{h}\\ & =\lim_{h\rightarrow0}\frac{f\left(x_{1}\left(u_{1}+h,u_{2}\right),x_{2}\left(u_{1}+h,u_{2}\right)\right)-f\left(x_{1}\left(u_{1},u_{2}\right),x_{2}\left(u_{1}+h,u_{2}\right)\right)+f\left(x_{1}\left(u_{1},u_{2}\right),x_{2}\left(u_{1}+h,u_{2}\right)\right)-f\left(x_{1}\left(u_{1},u_{2}\right),x_{2}\left(u_{1},u_{2}\right)\right)}{h}\\ & =\lim_{h\rightarrow0}\frac{f\left(x_{1}\left(u_{1}+h,u_{2}\right),x_{2}\left(u_{1}+h,u_{2}\right)\right)-f\left(x_{1}\left(u_{1},u_{2}\right),x_{2}\left(u_{1}+h,u_{2}\right)\right)}{h}+\lim_{h\rightarrow0}\frac{f\left(x_{1}\left(u_{1},u_{2}\right),x_{2}\left(u_{1}+h,u_{2}\right)\right)-f\left(x_{1}\left(u_{1},u_{2}\right),x_{2}\left(u_{1},u_{2}\right)\right)}{h}\\ & =\lim_{h\rightarrow0}\frac{f\left(x_{1}\left(u_{1}+h,u_{2}\right),x_{2}\left(u_{1}+h,u_{2}\right)\right)-f\left(x_{1}\left(u_{1},u_{2}\right),x_{2}\left(u_{1}+h,u_{2}\right)\right)}{x_{1}\left(u_{1}+h,u_{2}\right)-x_{1}\left(u_{1},u_{2}\right)}\cdot\frac{x_{1}\left(u_{1}+h,u_{2}\right)-x_{1}\left(u_{1},u_{2}\right)}{h}+\lim_{h\rightarrow0}\frac{f\left(x_{1}\left(u_{1},u_{2}\right),x_{2}\left(u_{1}+h,u_{2}\right)\right)-f\left(x_{1}\left(u_{1},u_{2}\right),x_{2}\left(u_{1},u_{2}\right)\right)}{x_{2}\left(u_{1}+h,u_{2}\right)-x_{2}\left(u_{1},u_{2}\right)}\cdot\frac{x_{2}\left(u_{1}+h,u_{2}\right)-x_{2}\left(u_{1},u_{2}\right)}{h}\\ & =\frac{\partial f}{\partial x_{1}}\frac{\partial x_{1}}{\partial u_{1}}+\frac{\partial f}{\partial x_{2}}\frac{\partial x_{2}}{\partial u_{1}} \end{align*} となるので与式が成り立つ。
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