Find the derivative of ${\mathbf{ze}}^{\mathbf{x}}\mathbf{cosy}$ at $\left(1,0,\frac{\pi }{3}\right)$ in the direction of the vector $\mathbf{i}\mathbf{+}\mathbf{2}\mathbf{j}$ .

The given function is $z{e}^x\cos y$ with respect to reference point $\left(1,0,\frac{\mathrm{\pi }}{3}\right)$ and vector $\mathrm{i}+2\mathrm{j}$.

Directional derivative represents the rate at which value of the function changes at fixed point and direction.

Use the formula $\frac{d\phi }{du}=\nabla \phi .u$

Here $\frac{\mathrm{d\phi }}{\mathrm{du}}$: directional derivative of function $\phi$, $\nabla \mathrm{\phi }$: the gradient of $\phi$ and u : a unit vector.

Given that $z{e}^x\cos y$ and point$\left(1,0,\frac{\mathrm{\pi }}{3}\right)$, let $\mathrm{A}=\mathrm{i}+2\mathrm{j}$

Then, unit vector $\mathrm{u}=\frac{\mathrm{A}}{\left|\mathrm{A}\right|}$

$\mathrm{u}=\frac{\mathrm{i}+2\mathrm{j}}{\sqrt{{\left(1\right)}^2+{\left(2\right)}^2}}$

$\mathrm{u}=\frac{\mathrm{i}+2\mathrm{j}}{\sqrt{5}}$

Use the formula $\nabla \mathrm{\phi }=\mathrm{i}\frac{\partial }{\partial \mathrm{x}}+\mathrm{j}\frac{\partial }{\partial \mathrm{y}}+\mathrm{k}\frac{\partial }{\partial \mathrm{z}}$

$$\begin{gathered} \nabla \mathrm{\phi }=\mathrm{i}\frac{\partial \mathrm{\phi }}{\partial \mathrm{x}}+\mathrm{j}\frac{\partial \mathrm{\phi }}{\partial \mathrm{y}}+\mathrm{k}\frac{\partial \mathrm{\phi }}{\partial \mathrm{z}} \\ \nabla \mathrm{\phi }=\mathrm{i}\frac{\partial }{\partial \mathrm{x}}\left({\mathrm{ze}}^{\mathrm{x}}\mathrm{cosy}\right)+\mathrm{j}\frac{\partial }{\partial \mathrm{y}}\left({\mathrm{ze}}^{\mathrm{x}}\mathrm{cosy}\right)+\mathrm{k}\frac{\partial }{\partial \mathrm{z}}\left({\mathrm{ze}}^{\mathrm{x}}\mathrm{cosy}\right) \\ \nabla \mathrm{\phi }=\mathrm{i}\left({\mathrm{ze}}^{\mathrm{x}}\mathrm{cosy}\right)+\mathrm{j}\left(‐{\mathrm{ze}}^{\mathrm{x}}\mathrm{siny}\right)+\mathrm{k}\left({\mathrm{e}}^{\mathrm{x}}\mathrm{cosy}\right) \\ \nabla {\mathrm{\phi }}_{\left(1,0\frac{\mathrm{\pi }}{3}\right)}=\mathrm{i}\left(\frac{\mathrm{e\pi }}{3}\right)+\mathrm{j}\left(0\right)+\mathrm{k}\left(\mathrm{e}\right) \end{gathered}$$

The formula for directional derivative given below:

$\frac{\mathrm{d\phi }}{\mathrm{du}}=\nabla \mathrm{\phi }.\mathrm{u}$

Put the values given below.

$$\begin{gathered} \nabla \mathrm{\phi }=\frac{\mathrm{e\phi }}{3}\mathrm{i}+\mathrm{ek} \\ \mathrm{u}=\frac{\mathrm{i}+2\mathrm{j}}{\sqrt{5}} \\ \mathrm{u}=\frac{\mathrm{i}+2\mathrm{j}}{\sqrt{5}} \end{gathered}$$

Then, it can be solved further.

$$\begin{gathered} \frac{\mathrm{d\phi }}{\mathrm{du}}=\left(\frac{\mathrm{e\pi }}{3}\mathrm{i}+\mathrm{ek}\right)\frac{1}{\sqrt{5}}\left(\mathrm{i}+2\mathrm{j}\right) \\ \frac{\mathrm{d\phi }}{\mathrm{du}}=\frac{1}{\sqrt{5}}\left(\frac{\mathrm{e\pi }}{3}0+0\right) \\ \frac{\mathrm{d\phi }}{\mathrm{du}}=\frac{\mathrm{e\pi }}{3\sqrt{5}} \end{gathered}$$

The directional derivative of the function ${\mathrm{ze}}^{\mathrm{x}}\mathrm{cosy}$ is $\frac{\mathrm{e\pi }}{3\sqrt{5}}$.