Partial

at $\left(x\,y\right)$ 

at $\left(a\,b\right)$ 

$f_{\mathrm{xx}}$

$-\frac{1}{x^2}$

$-\frac{1}{4}$

$f_{\mathrm{xy}}$

$0$

$0$

$f_{\mathrm{yx}}$

$0$

$0$

$f_{\mathrm{yy}}$

$\frac{1}{y^2}$

$\frac{1}{9}$

Table 4.2.1(a)   Second partials of $f$ at $\left(x\,y\right)$ and $\left(2\,-3\right)$ 

$f\=\ln \left(x\/y\right)$$\stackrel{\text{assign}}{\to }$

Second partial derivatives at $\left(x\,y\right)$ 

$\frac{{\partial }^2}{{\partial }^{}{x}^2} f$ = $-\frac{1}{x^2}$

$\frac{{\partial }^2}{\partial y\partial x} f$ = $0$

$\frac{{\partial }^2}{\partial x\partial y} f$ = $0$

$\frac{{\partial }^2}{{\partial }^{}{y}^2} f$ = $\frac{1}{y^2}$

Second partial derivatives at $\left(a\,b\right)$ 

$\mathrm{f\_\_xx}\left(x\,y\right)\=\frac{{\partial }^2}{{\partial }^{}{x}^2} f$$\stackrel{\text{assign as function}}{\to }$$\mathrm{f\_\_xx}$

$\mathrm{f\_\_xy}\left(x\,y\right)\=\frac{{\partial }^2}{\partial y\partial x} f$$\stackrel{\text{assign as function}}{\to }$$\mathrm{f\_\_xy}$

$\mathrm{f\_\_yy}\left(x\,y\right)\=\frac{{\partial }^2}{{\partial }^{}{y}^2} f$$\stackrel{\text{assign as function}}{\to }$$\mathrm{f\_\_yy}$

$\mathrm{f\_\_yx}\left(x\,y\right)\=\frac{{\partial }^2}{\partial x\partial y} f$$\stackrel{\text{assign as function}}{\to }$$\mathrm{f\_\_yx}$

Obtain the second partial derivatives at $\left(x\,y\right)$ 

$\mathrm{f\_\_xx}\left(x\,y\right)$ = $-\frac{1}{x^2}$

$\mathrm{f\_\_xy}\left(x\,y\right)$ = $0$

$\mathrm{f\_\_yx}\left(x\,y\right)$ = $0$

$\mathrm{f\_\_yy}\left(x\,y\right)$ = $\frac{1}{y^2}$

Obtain the second partial derivatives at $\left(a\,b\right)$ 

$f≔\ln \left(x\/y\right)\:$

Second partials at $\left(x\,y\right)$ 

Second partials at $\left(a\,b\right)$ 

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

$-\frac{1}{x^2}$

$\mathrm{eval}\left(\mathrm{f\_\_xx}\,\left[x\=2\,y\=-3\right]\right)$

$-\frac{1}{4}$

$\mathrm{f\_\_xy}≔\mathrm{diff}\left(f\,x\,y\right)$

$0$

$\mathrm{eval}\left(\mathrm{f\_\_xy}\,\left[x\=2\,y\=-3\right]\right)$

$0$

$\mathrm{f\_\_yx}≔\mathrm{diff}\left(f\,y\,x\right)$

$0$

$\mathrm{eval}\left(\mathrm{f\_\_yx}\,\left[x\=2\,y\=-3\right]\right)$

$0$

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

$\frac{1}{y^2}$

$\mathrm{eval}\left(\mathrm{f\_\_yy}\,\left[x\=2\,y\=-3\right]\right)$

$\frac{1}{9}$

$f≔\left(x\,y\right)\to \ln \left(x\/y\right)\:$

Second partials at $\left(x\,y\right)$ 

Second partials at $\left(a\,b\right)$ 

${\mathrm{D}}_{1\,1}\left(f\right)\left(x\,y\right)$

$-\frac{1}{x^2}$

${\mathrm{D}}_{1\,1}\left(f\right)\left(2\,-3\right)$

$-\frac{1}{4}$

${\mathrm{D}}_{1\,2}\left(f\right)\left(x\,y\right)$

$0$

${\mathrm{D}}_{1\,2}\left(f\right)\left(2\,-3\right)$

$0$

${\mathrm{D}}_{2\,1}\left(f\right)\left(x\,y\right)$

$0$

${\mathrm{D}}_{2\,1}\left(f\right)\left(2\,-3\right)$

$0$

${\mathrm{D}}_{2\,2}\left(f\right)\left(x\,y\right)$

$\frac{1}{y^2}$

${\mathrm{D}}_{2\,2}\left(f\right)\left(2\,-3\right)$

$\frac{1}{9}$