91Ó°ÊÓ

$$\begin{gathered} \text{The directional derivative of a function}f(x,y,z)\text{at a point}P=\left(x_0,y_0,z_0\right)\text{in the direction of a} \\ \text{unit vector}\mathbf{u}=⟨a,b,câŸ\copyright \text{is given by} \\ {D}_{u}f\left(x_0,y_0,z_0\right)=\underset{h→0}{lim} \frac{f\left(x_0+ah,y_0+bh,z_0+ch\right)−f\left(x_0,y_0,z_0\right)}{h} \\ \text{Here}f(x,y,z)=\sqrt{\frac{xy}{z}}\text{, and}P=(2,3,1). \\ Alsothevectorisv=(2,1,−2) \end{gathered}$$

$$\begin{gathered} \text{The unit vector "}\mathit{u}\text{" in the direction of "}v\text{" is given by} \\ \begin{array}{l}\mathbf{u}=\frac{1}{âˆ\yen \mathbf{v}âˆ\yen }\mathbf{v}\\ =\frac{1}{\sqrt{2^2+1^2+(−2{)}^2}}⟨2,1,−2âŸ\copyright \\ =\frac{1}{3}⟨2,1,−2âŸ\copyright \\ Nowbydefinition:\\ \begin{array}{l}{D}_{u}f(2,3,1)=\underset{h→0}{lim} \frac{f\left(2+\frac{2h}{3},3+\frac{h}{3},1−\frac{2h}{3}\right)−f(2,3,1)}{h}\\ =\underset{h→0}{lim} \frac{\sqrt{\left(2+\frac{2h}{3}\right)\left(3+\frac{h}{3}\right)÷\left(1−\frac{2h}{3}\right)}−\sqrt{\frac{6}{1}}}{h}\\ =\underset{h→0}{lim} \frac{\sqrt{6+\frac{8h}{3}+\frac{2h}{9}−\sqrt{6−4h}}}{h\sqrt{1−\frac{2h}{3}}}\\ Rationalizethenumeratorgives:\\ \begin{array}{l}{D}_{u}f(2,3,1)=\underset{h→0}{lim} \frac{\sqrt{6+\frac{8h}{3}+\frac{2h^2}{9}}−\sqrt{6−4h}}{h\sqrt{1−\frac{2h}{3}}}×\frac{\sqrt{6+\frac{8h}{3}+\frac{2h^2}{9}}+\sqrt{6−4h}}{\sqrt{6+\frac{8h}{3}+\frac{2h^2}{9}}+\sqrt{6−4h}}\\ =\underset{h→0}{lim} \frac{6+\frac{8h}{3}−\frac{2h^2}{9}−6+4h}{h\sqrt{1−\frac{2h}{3}}\left(\sqrt{6+\frac{8h}{3}+\frac{2h^2}{9}}+\sqrt{6−4h}\right)}\\ =\underset{h→0}{lim} \frac{\sqrt{\frac{20}{3}−\frac{2}{9}h}}{\sqrt{1−\frac{2h}{3}}\left(\sqrt{6+\frac{8h}{3}+\frac{2h^2}{9}}+\sqrt{6−4h}\right)}\\ Takinglimitash→0gives:D_{\mathrm{u}}f(2,3,1)=\frac{10}{\sqrt{6}}\end{array}\end{array}\end{array} \end{gathered}$$