$\frac{\partial }{\partial x}\left(e^{x^2\/y} \ln \left(y^2\/x\right)\right)\={\mathit{\ⅇ}}^{\frac{x^2}{y}}\left(\frac{2 x \ln \left(\frac{y^2}{x}\right)}{y}-\frac{1}{x}\right)$

$\frac{\partial }{\partial y}\left(e^{x^2\/y} \ln \left(y^2\/x\right)\right)\=\frac{{\mathit{\ⅇ}}^{\frac{x^2}{y}}}{y}\left(2-\frac{x^2 \ln \left(\frac{y^2}{x}\right)}{y}\right)$

$\genfrac{}{}{0ex}{}{\frac{\partial f}{\partial x}}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{\left(x\,y\right)\=\left(2\,-2\right)}$ =  $-2{\ⅇ}^{-2}\ln \left(2\right)-\frac{{\ⅇ}^{-2}}{2}$

$\genfrac{}{}{0ex}{}{\frac{\partial f}{\partial y}}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{\left(x\,y\right)\=\left(2\,-2\right)}$=  $-{\ⅇ}^{-2}\ln \left(2\right)-{\ⅇ}^{-2}$

Table 4.1.4(a)   First partial derivatives of $f$

Context Panel

$f_x$ 

$f_y$ 

${\ⅇ}^{x^2\/y} \ln \left(y^2\/x\right)$

${\ⅇ}^{\frac{x^2}{y}}\ln \left(\frac{y^2}{x}\right)$

$\stackrel{\text{differentiate w.r.t. x}}{\to }$

$\frac{2x{\ⅇ}^{\frac{x^2}{y}}\ln \left(\frac{y^2}{x}\right)}{y}-\frac{{\ⅇ}^{\frac{x^2}{y}}}{x}$

$\stackrel{\text{evaluate at point}}{\to }$

$-2{\ⅇ}^{\mathrm{-2}}\ln \left(2\right)-\frac{{\ⅇ}^{\mathrm{-2}}}{2}$

${\ⅇ}^{x^2\/y} \ln \left(y^2\/x\right)$

${\ⅇ}^{\frac{x^2}{y}}\ln \left(\frac{y^2}{x}\right)$

$\stackrel{\text{differentiate w.r.t. y}}{\to }$

$-\frac{x^2{\ⅇ}^{\frac{x^2}{y}}\ln \left(\frac{y^2}{x}\right)}{y^2}\+\frac{2{\ⅇ}^{\frac{x^2}{y}}}{y}$

$\stackrel{\text{evaluate at point}}{\to }$

$-{\ⅇ}^{-2}\ln \left(2\right)-{\ⅇ}^{-2}$

Define $f$ as an expression

${\ⅇ}^{x^2\/y} \ln \left(y^2\/x\right)$$\stackrel{\text{assign to a name}}{\to }$$f$

Obtain $f_x\left(x\,y\right)$ and $f_y\left(x\,y\right)$ 

$\frac{\partial }{\partial x} f$ = $\frac{2x{\ⅇ}^{\frac{x^2}{y}}\ln \left(\frac{y^2}{x}\right)}{y}-\frac{{\ⅇ}^{\frac{x^2}{y}}}{x}$$$

$\frac{\partial }{\partial y} f$ = $-\frac{x^2{\ⅇ}^{\frac{x^2}{y}}\ln \left(\frac{y^2}{x}\right)}{y^2}\+\frac{2{\ⅇ}^{\frac{x^2}{y}}}{y}$$$

Obtain $f_x\left(a\,b\right)$ and $f_y\left(a\,b\right)$

$\genfrac{}{}{0ex}{}{\frac{\partial }{\partial x} f}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=2\,y\=-2}$ = $-2{\ⅇ}^{-2}\ln \left(2\right)-\frac{1}{2}{\ⅇ}^{-2}$$$

$\genfrac{}{}{0ex}{}{\frac{\partial }{\partial y} f}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=2\,y\=-2}$ = $-{\ⅇ}^{-2}\ln \left(2\right)-{\ⅇ}^{-2}$

Define the functions  $f_x$ and $f_y$ 

$f_x$ 

$f_y$ 

$\mathrm{f\_\_x}\left(x\,y\right)\=\frac{\partial }{\partial x} f$$\stackrel{\text{assign as function}}{\to }$$\mathrm{f\_\_x}$

$\mathrm{f\_\_y}\left(x\,y\right)\=\frac{\partial }{\partial y} f$$\stackrel{\text{assign as function}}{\to }$$\mathrm{f\_\_y}$

Obtain $f_x\left(x\,y\right)$ and $f_y\left(x\,y\right)$ 

$\mathrm{f\_\_x}\left(x\,y\right)$ = $\frac{2x{\ⅇ}^{\frac{x^2}{y}}\ln \left(\frac{y^2}{x}\right)}{y}-\frac{{\ⅇ}^{\frac{x^2}{y}}}{x}$$$

$\mathrm{f\_\_y}\left(x\,y\right)$ = $-\frac{x^2{\ⅇ}^{\frac{x^2}{y}}\ln \left(\frac{y^2}{x}\right)}{y^2}\+\frac{2{\ⅇ}^{\frac{x^2}{y}}}{y}$$$

Obtain $f_x\left(a\,b\right)$ and $f_y\left(a\,b\right)$ 

$\mathrm{f\_\_x}\left(2\,-2\right)$ = $-2{\ⅇ}^{-2}\ln \left(2\right)-\frac{1}{2}{\ⅇ}^{-2}$$$

$\mathrm{f\_\_y}\left(2\,-2\right)$ = $-{\ⅇ}^{-2}\ln \left(2\right)-{\ⅇ}^{-2}$

Define $f$ as an expression

$f≔{\ⅇ}^{x^2\/y} \ln \left(y^2\/x\right)\:$

Obtain $f_x\left(x\,y\right)$ and $f_y\left(x\,y\right)$ 

$\mathrm{fx}≔\mathrm{diff}\left(f\,x\right)$

$\frac{2x{\ⅇ}^{\frac{x^2}{y}}\ln \left(\frac{y^2}{x}\right)}{y}-\frac{{\ⅇ}^{\frac{x^2}{y}}}{x}$

$\mathrm{fy}≔\mathrm{diff}\left(f\,y\right)$

$-\frac{x^2{\ⅇ}^{\frac{x^2}{y}}\ln \left(\frac{y^2}{x}\right)}{y^2}\+\frac{2{\ⅇ}^{\frac{x^2}{y}}}{y}$

Obtain $f_x\left(a\,b\right)$ and $f_y\left(a\,b\right)$ 

$\mathrm{eval}\left(\mathrm{fx}\,\left[x\=2\,y\=-2\right]\right)$

$-2{\ⅇ}^{-2}\ln \left(2\right)-\frac{1}{2}{\ⅇ}^{-2}$

$\mathrm{simplify}\left(\mathrm{eval}\left(\mathrm{fy}\,\left[x\=2\,y\=-2\right]\right)\right)$

$-{\ⅇ}^{-2}\left(\ln \left(2\right)\+1\right)$

Define $f$ as a function

$f≔\left(x\,y\right)\to {\ⅇ}^{x^2\/y} \ln \left(y^2\/x\right)\:$

Obtain $f_x\left(x\,y\right)$ and $f_y\left(x\,y\right)$ 

${\mathrm{D}}_1\left(f\right)\left(x\,y\right)$ = $\frac{2x{\ⅇ}^{\frac{x^2}{y}}\ln \left(\frac{y^2}{x}\right)}{y}-\frac{{\ⅇ}^{\frac{x^2}{y}}}{x}$$$

${\mathrm{D}}_2\left(f\right)\left(x\,y\right)$ = $-\frac{x^2{\ⅇ}^{\frac{x^2}{y}}\ln \left(\frac{y^2}{x}\right)}{y^2}\+\frac{2{\ⅇ}^{\frac{x^2}{y}}}{y}$$$

Obtain $f_x\left(a\,b\right)$ and $f_y\left(a\,b\right)$ 

${\mathrm{D}}_1\left(f\right)\left(2\,-2\right)$ = $-2{\ⅇ}^{-2}\ln \left(2\right)-\frac{1}{2}{\ⅇ}^{-2}$$$

$\left({\mathrm{D}}_2\left(f\right)\left(2\,-2\right)\right)$ = $-{\ⅇ}^{-2}\ln \left(2\right)-{\ⅇ}^{-2}$$$