This question aims to find the **directional derivative** of the function f at the given point in the direction indicated by the angle $\theta$.

A directional Derivative is a type of derivative that tells us the **change of the function** at a **point** with **time** in the **vector direction**.

We also find partial derivatives according to the directional derivative formula. The **partial derivatives** can be found by keeping one of the variables constant while applying derivation of the other.

## Expert Answer

The given function is:

\[f(x, y) = e^x cos y\]

\[(x, y) = ( 0, 0 )\]

The angle is given by:

\[\theta = \frac{\pi}{4}\]

The formula to find the directional derivative of the given function is:

\[D_u f(x, y) = f_x (x, y) a + f_y (x, y) b\]

To find the partial derivatives:

$f_x = e ^ x cos y$ and $f_y = – e ^ x sin y$

Here, a and b represent the angle. In this case, the angle is $\theta$.

By putting values in the above-mentioned formula of directional derivative:

\[D_u f(x, y ) = ( e ^ x cos y ) cos ( \frac { \pi } { 4 } ) + ( – e ^ x sin y ) sin ( \frac { \pi } { 4 } ) \]

\[D_u f(x, y) = ( e ^ x cos y ) ( \frac { 1 } { \sqrt { 2 } } ) + ( – e ^ x sin y ) ( \frac { 1 } { \sqrt { 2 } } ) \]

\[ D _ u f ( x, y ) = \frac { \sqrt { 2 }} { 2 } [ ( e ^ x cos y ) + ( – e ^ x sin y ) \]

By putting values of x and y:

\[ D _ u f ( x, y ) = \frac { \sqrt { 2 }} { 2 } [ ( e ^ 0 cos 0 ) + ( – e ^ 0 sin 0 ) \]

\[ D _ u f ( 0, 0 ) = \frac { \sqrt { 2 }} { 2 } \]

## Numerical Solution

**The directional derivative of the function f at the given point in the direction indicated by the angle $\theta$ is $ \frac {\sqrt {2}} {2} $.**

## Example

Find the directional derivative at $ \theta = \frac{\pi}{3} $

\[D_u f(x, y) = (e^x cos y) cos(\frac{\pi}{3}) + (-e^x sin y) sin(\frac{\pi}{3})\]

\[= (e ^ x cos y ) (\frac{1}{2}) + (-e^x sin y)(\frac {\sqrt{3}}{2})\]

\[= \frac { \sqrt { 3 } +1}{2} [(e^x cos y) + (- e^x sin y ) \]

\[= \frac { \sqrt {3} + 1}{2} [(e^0 cos 0 ) + ( – e ^ 0 sin 0 )\]

\[D _ u f ( 0, 0 ) = \frac { \sqrt {3} + 1} { 2 } \]

*Image/Mathematical drawings are created in Geogebra*