91Ó°ÊÓ

Given function is,

$\mathrm{f}(\mathrm{x},\mathrm{y})=\frac{\mathrm{x}}{{\mathrm{y}}^2}$

${\mathrm{f}}_{\mathrm{x}}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)\left(\mathrm{x}-{\mathrm{x}}_0\right)+{\mathrm{f}}_{\mathrm{y}}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)\left(\mathrm{y}-{\mathrm{y}}_0\right)=\mathrm{z}-\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$

${\mathrm{f}}_{\mathrm{x}}(-2,1)(\mathrm{x}+2)+{\mathrm{f}}_{\mathrm{y}}(-2,1)(\mathrm{y}-1)=\mathrm{z}-\mathrm{f}(-2,1).....1$

Functions are,

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

${\mathrm{f}}_{\mathrm{x}}(-2,1)=1......2$

${\mathrm{f}}_{\mathrm{y}}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)={\left.\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{f}(\mathrm{x},\mathrm{y})\right|}_{\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)}={\left.\frac{\mathrm{d}}{\mathrm{dy}}\frac{\mathrm{x}}{{\mathrm{y}}^2}\right|}_{\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)}$

localid="1650569173089" ${\mathrm{f}}_{\mathrm{y}}(-2,1)={\left.\frac{-2\mathrm{x}}{{\mathrm{y}}^3}\right|}_{(-2,1)}$

localid="1650569436165" ${\mathrm{f}}_{\mathrm{y}}(-2,1)=4.........3$

$\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)=\mathrm{f}(-2,1)=\frac{-2}{1}$

localid="1650569447717" $\mathrm{f}(-2,1)=-2....4$

Substitute $2,3,4$equation in $1$

we get,

$1(\mathrm{x}+2)+4(\mathrm{y}-1)=\mathrm{z}+2$

$\mathrm{x}+4\mathrm{y}-\mathrm{z}=2-2+4$

$\mathrm{x}+4\mathrm{y}-\mathrm{z}=4$

Regarded the equation of the normal line is,

$\mathrm{x}={\mathrm{x}}_0+\mathrm{Î\pm ³Ù},\mathrm{y}={\mathrm{y}}_0+\mathrm{β³Ù},\mathrm{z}={\mathrm{z}}_0+\mathrm{γ³Ù}$

${\mathrm{z}}_0=\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$and $\mathrm{γ}={\mathrm{D}}_{\mathrm{w}}\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$

The directional derivative is,

role="math" localid="1650569949338" ${\mathrm{D}}_{\mathrm{u}}\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)={Lim}_{\mathrm{h}→0}\frac{\mathrm{f}\left({\mathrm{x}}_0+\mathrm{Î\pm ³ó},{\mathrm{y}}_0+\mathrm{β³ó}\right)-\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)}{\mathrm{h}}$

${\mathrm{D}}_{\mathrm{u}}\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)={Lim}_{\mathrm{h}→0}\frac{\mathrm{f}\left({\mathrm{x}}_0+\mathrm{Î\pm ³ó},{\mathrm{y}}_0+\mathrm{β³ó}\right)-\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)}{\mathrm{h}}.......5$

$\mathrm{f}\left(-2+\frac{\sqrt{10}}{10}\mathrm{h},1-\frac{3\sqrt{10}}{10}\mathrm{h}\right)=\frac{-2+\frac{\sqrt{10}}{10}\mathrm{h}}{{\left(1-\frac{3\sqrt{10}}{10}\mathrm{h}\right)}^2}$

$=\frac{-2+\frac{\mathrm{h}}{\sqrt{10}}}{{\left(1-\frac{3}{\sqrt{10}}\mathrm{h}\right)}^2}$

$=\frac{\frac{-2\sqrt{10}}{\sqrt{10}}+\mathrm{h}}{\left(\frac{10-6\sqrt{10}\mathrm{h}+9{\mathrm{h}}^2}{10}\right)}$

role="math" localid="1650570094444" $=\frac{-2^{}×10+\mathrm{h}\sqrt{10}}{10-6\sqrt{10}\mathrm{h}+9{\mathrm{h}}^2}.........6$

And$\mathrm{f}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)=\mathrm{f}(-2,1)=\frac{-2}{1^2}=-2......7$

Substitute $6,7$equation in $5$equation,

we get,

${\mathrm{D}}_{\mathrm{u}}\mathrm{f}(-2,1)={Lim}_{\mathrm{h}→0}\frac{\frac{-2×10+\mathrm{h}\sqrt{10}}{10-6\sqrt{10}\mathrm{h}+9{\mathrm{h}}^2}-(-2)}{\mathrm{h}}$

$={Lim}_{\mathrm{h}→0}\frac{-20+\mathrm{h}\sqrt{10}+20-12\sqrt{10}\mathrm{h}+18{\mathrm{h}}^2}{\mathrm{h}\left(10-6\sqrt{10}\mathrm{h}+9{\mathrm{h}}^2\right)}$

$={Lim}_{\mathrm{h}→0}\frac{-11\mathrm{h}\sqrt{10}+18{\mathrm{h}}^2}{\mathrm{h}\left(10-6\sqrt{10}\mathrm{h}+9{\mathrm{h}}^2\right)}$

$={Lim}_{\mathrm{h}→0}\frac{(-11\sqrt{10}+18\mathrm{h})}{\left(10-6\sqrt{10}\mathrm{h}+9{\mathrm{h}}^2\right)}$

localid="1650570637628" ${\mathrm{D}}_{\mathrm{u}}\mathrm{f}(-2,1)=\frac{-11\sqrt{10}}{10}$