【微分積分】7-1-2 偏微分の計算｜問題集

\(f_x(x,y)=6x-y\)
\(f_y(x,y)=-x+1\)

\(f_x(x,y)=2xe^{-y}\)
\(f_y(x,y)=-x^2e^{-y}\)

\(\displaystyle f_x(x,y)=\frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{\sqrt{x^2+y^2}}\)
\(\displaystyle f_y(x,y)=\frac{2y}{2\sqrt{x^2+y^2}}=\frac{y}{\sqrt{x^2+y^2}}\)

\(f_x(x,y)=\sin y\)
\(f_y(x,y)=x\cos y\)

\(\displaystyle f_x(x,y)=\frac{x+y-(x-y)}{(x+y)^2}=\frac{2y}{(x+y)^2}\)
\(\displaystyle f_y(x,y)=\frac{-(x+y)-(x-y)}{(x+y)^2}=\frac{-2x}{(x+y)^2}\)

\(f_x(x,y)=3x^2+y^2\)
\(f_y(x,y)=2xy+3y^2\)

\(f_x(x,y)=e^x\sin y\)
\(f_y(x,y)=e^x\cos y\)

\(\displaystyle f_x(x,y)=\frac{2x}{x^2+y^2}\)
\(\displaystyle f_y(x,y)=\frac{2y}{x^2+y^2}\)

\(f_x(x,y)=4(x^2+2y)^3・2x=8x(x^2+2y)^3\)
\(f_y(x,y)=4(x^2+2y)^3・2=8(x^2+2y)^3\)

\(\displaystyle f_x(x,y)=\frac{1}{2}・\frac{1}{x^2+y^2}・2x=\frac{x}{x^2+y^2}\)
\(\displaystyle f_y(x,y)=\frac{1}{2}・\frac{1}{x^2+y^2}・2y=\frac{y}{x^2+y^2}\)

\(\displaystyle f_x(x,y)\)
\(\displaystyle =2x\tan^{-1}\frac{y}{x}+(x^2+y^2)・\frac{1}{1+(\frac{y}{x})^2}・\left(-\frac{y}{x^2}\right)\)
\(\displaystyle =2x\tan^{-1}\frac{y}{x}-\frac{(x^2+y^2)y}{x^2+y^2}\)
\(\displaystyle =2x\tan^{-1}\frac{y}{x}-y\)

\(\displaystyle f_y(x,y)\)
\(\displaystyle =2y\tan^{-1}\frac{y}{x}+(x^2+y^2)・\frac{1}{1+(\frac{y}{x})^2}・\frac{1}{x}\)
\(\displaystyle =2y\tan^{-1}\frac{y}{x}+\frac{(x^2+y^2)x}{x^2+y^2}\)
\(\displaystyle =2y\tan^{-1}\frac{y}{x}+x\)

\(x^y=e^{y\log x}\)なので
\(\displaystyle f_x(x,y)=e^{y\log x}・y・\frac{1}{x}=yx^{y-1}\)
\(\displaystyle f_y(x,y)=e^{y\log x}・\log x=x^y\log x\)

\(\displaystyle z_x=\frac{-ny^n}{x^{n+1}}e^{-\frac{y}{x}}+\frac{y^n}{x^n}e^{-\frac{y}{x}}・\frac{y}{x^2}\)
\(\displaystyle =\frac{(-nx+y)y^n}{x^{n+2}}e^{-\frac{y}{x}}\)
\(\displaystyle z_y=\frac{ny^{n-1}}{x^n}e^{-\frac{y}{x}}+\frac{y^n}{x^n}e^{-\frac{y}{x}}・\left(-\frac{1}{x}\right)\)
\(\displaystyle =\frac{(nx-y)y^{n-1}}{x^{n+1}}e^{-\frac{y}{x}}\)
よって、
\(xz_x+yz_y\)
\(\displaystyle =\frac{(-nx+y)y^n}{x^{n+2}}e^{-\frac{y}{x}}+\frac{(nx-y)y^{n-1}}{x^{n+1}}e^{-\frac{y}{x}}\)
\(=0\)