# How to Find the directional derivative of?:

## $f \left(x , y\right) = {e}^{- x y} , P \left(1 , - 1\right) , v = - i + \sqrt{3}$

May 1, 2018

$= \frac{e}{\sqrt{2}} \left(\sqrt{3} + 1\right)$

#### Explanation:

If the function $f \left(m a t h b f x\right)$ is differentiable at $m a t h b f {x}_{o}$, then the directional derivative exists along any vector $m a t h b f v$, and is:

${\nabla}_{\setminus m a t h b f \left\{v\right\}} f \left(m a t h b f {x}_{o}\right) = \nabla f \left(m a t h b f {x}_{o}\right) \setminus \cdot m a t h b f v$

You're looking at:

$\nabla f {\left(x , y\right)}_{P} \cdot \frac{\left\langle \sqrt{3} , - 1\right\rangle}{2}$

$= {\left\langle - y {e}^{- x y} , - x {e}^{- x y}\right\rangle}_{P} \cdot \frac{\left\langle \sqrt{3} , - 1\right\rangle}{2}$

$= \left\langle e , - e\right\rangle \cdot \frac{\left\langle \sqrt{3} , - 1\right\rangle}{2}$

$= \frac{e}{\sqrt{2}} \left(\sqrt{3} + 1\right)$