【微分積分】7-1-2 偏微分の計算｜要点まとめ

\(z_x=8x+5y\)
\(z_y=5x-21y^2\)

\(z=2x^2y^2-xy^3\)
\(z_x=4xy^2-y^3\)
\(z_y=4x^2y-3xy^2\)

\(z_x=(2x+y)\cos(x^2+xy)\)
\(z_y=x\cos(x^2+xy)\)

\(\displaystyle z_x=\sqrt{x^2+y^2}+(x-y)\frac{2x}{2\sqrt{x^2+y^2}}\)
\(\displaystyle =\frac{2x^2-xy+y^2}{\sqrt{x^2+y^2}}\)

\(\displaystyle z_y=-\sqrt{x^2+y^2}+(x-y)\frac{2y}{2\sqrt{x^2+y^2}}\)
\(\displaystyle =\frac{-x^2+xy-2y^2}{\sqrt{x^2+y^2}}\)

\(\displaystyle z_x=\frac{3(4x-3y)-(3x+2y)・4}{(4x-3y)^2}\)
\(\displaystyle =\frac{-17y}{(4x-3y)^2}\)

\(\displaystyle z_y=\frac{2(4x-3y)-(3x+2y)・(-3)}{(4x-3y)^2}\)
\(\displaystyle =\frac{17x}{(4x-3y)^2}\)

\(\displaystyle z_x=\frac{y\sqrt{x^2+xy+y^2}-xy\frac{2x+y}{2\sqrt{x^2+xy+y^2}}}{x^2+xy+y^2}\)
\(\displaystyle =\frac{2y(x^2+xy+y^2)-xy(2x+y)}{2(x^2+xy+y^2)^{\frac{3}{2}}}\)
\(\displaystyle =\frac{xy^2+2y^3}{2(x^2+xy+y^2)^{\frac{3}{2}}}\)

\(x,y\)が対称なので、
\(\displaystyle z_y=\frac{x^2y+2x^3}{2(x^2+xy+y^2)^{\frac{3}{2}}}\)

\(z_x=y(e^{xy}-e^{-xy})\)
\(z_y=x(e^{xy}-e^{-xy})\)

\(\displaystyle z_x=y\frac{x^2-y^2}{x^2+y^2}+xy\frac{2y^2・2x}{(x^2+y^2)^2}\)
\(\displaystyle =\frac{y(x^4-y^4+4x^2y^2)}{(x^2+y^2)^2}\)

\(\displaystyle z_y=x\frac{x^2-y^2}{x^2+y^2}-xy\frac{2x^2・2y}{(x^2+y^2)^2}\)
\(\displaystyle =\frac{x(x^4-y^4-4x^2y^2)}{(x^2+y^2)^2}\)

\(\displaystyle z_x=y\sin^{-1}\frac{y}{x}+xy\frac{1}{\sqrt{1-(\frac{y}{x})^2}}・\left(-\frac{y}{x^2}\right)\)
\(\displaystyle =y\sin^{-1}\frac{y}{x}-\frac{|x|y^2}{x\sqrt{x^2-y^2}}\)

\(\displaystyle z_y=x\sin^{-1}\frac{y}{x}+xy\frac{1}{\sqrt{1-(\frac{y}{x})^2}}・\frac{1}{x}\)
\(\displaystyle =x\sin^{-1}\frac{y}{x}+\frac{|x|y}{\sqrt{x^2-y^2}}\)

\(u_x=2xy+yz+z^3\)
\(u_y=x^2+xz\)
\(u_z=xy+3xz^2\)

\(u_x=yz\cos(xyz)\)
\(u_y=xz\cos(xyz)\)
\(u_z=xy\cos(xyz)\)

\(u_x=2xye^{x^2y}\log(z^2+1)\)
\(u_y=x^2e^{x^2y}\log(z^2+1)\)
\(\displaystyle u_z=\frac{2ze^{x^2y}}{z^2+1}\)

\((x,y)\neq(0,0)\)のとき、
\(\displaystyle f_x(x,y)=\frac{3x^2(x^2+y^2)-2x(x^3-y^3)}{(x^2+y^2)^2}\)
\(\displaystyle =\frac{x^4+3x^2y^2+2xy^3}{(x^2+y^2)^2}\)
\(\displaystyle f_y(x,y)=\frac{-3y^2(x^2+y^2)-2y(x^3-y^3)}{(x^2+y^2)^2}\)
\(\displaystyle =-\frac{3x^2y^2+2x^3y+y^4}{(x^2+y^2)^2}\)

\((x,y)=(0,0)\)のとき、
\(\displaystyle f_x(0,0)=\lim_{h\to0}\frac{f(0+h,0)-f(0,0)}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{h-0}{h}\)
\(\displaystyle =1\)
\(\displaystyle f_y(0,0)=\lim_{k\to0}\frac{f(0,0+k)-f(0,0)}{k}\)
\(\displaystyle =\lim_{k\to0}\frac{-k-0}{k}\)
\(\displaystyle =-1\)

\(\displaystyle f_x(0,0)=\lim_{h\to0}\frac{f(0+h,0)-f(0,0)}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{h^3+0}{\sqrt{h^2+0}}}{h}\)
\(\displaystyle =\lim_{h\to0}|h|\)
\(\displaystyle =0\)
よって、\(x\)は原点で偏微分可能。

\(\displaystyle f_y(0,0)=\lim_{k\to0}\frac{f(0,0+k)-f(0,0)}{k}\)
\(\displaystyle =\lim_{k\to0}\frac{\frac{0+k^2}{\sqrt{0+k^2}}}{k}\)
\(\displaystyle =\lim_{k\to0}\frac{|k|}{k}\)
\(\displaystyle \lim_{k\to+0}\frac{|k|}{k}=1,\lim_{k\to-0}\frac{|k|}{k}=-1\)なので、
\(y\)は原点で偏微分可能ではない。

\(\displaystyle f_x(0,0)=\lim_{h\to0}\frac{f(0+h,0)-f(0,0)}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{\frac{h^2・0}{h^4+0}}{h}\)
\(\displaystyle =0\)
よって、\(x\)は原点で偏微分可能。

\(\displaystyle f_y(0,0)=\lim_{k\to0}\frac{f(0,0+k)-f(0,0)}{k}\)
\(\displaystyle =\lim_{k\to0}\frac{\frac{0・k}{0+k^2}}{k}\)
\(\displaystyle =0\)
よって、\(y\)は原点で偏微分可能。

\(\displaystyle z_x=\frac{2x}{2\sqrt{x^2-y^2}}\sin^{-1}\frac{y}{x}\)
\(\displaystyle \ \ \ +\sqrt{x^2-y^2}\frac{1}{\sqrt{1-(\frac{y}{x})^2}}・\left(-\frac{y}{x^2}\right)\)
\(\displaystyle =\frac{x}{\sqrt{x^2-y^2}}\sin^{-1}\frac{y}{x}-\frac{|x|y}{x^2}\)
\(\displaystyle z_y=\frac{-2y}{2\sqrt{x^2-y^2}}\sin^{-1}\frac{y}{x}\)
\(\displaystyle \ \ \ +\sqrt{x^2-y^2}\frac{1}{\sqrt{1-(\frac{y}{x})^2}}・\frac{1}{x}\)
\(\displaystyle =\frac{-y}{\sqrt{x^2-y^2}}\sin^{-1}\frac{y}{x}+\frac{|x|}{x}\)
よって、
\(xz_x+yz_y\)
\(\displaystyle =\frac{x^2-y^2}{\sqrt{x^2-y^2}}\sin^{-1}\frac{y}{x}-\frac{|x|y}{x}+\frac{|x|y}{x}\)
\(\displaystyle =\sqrt{x^2-y^2}\sin^{-1}\frac{y}{x}\)
\(=z\)