# What is the equation of the normal line of f(x)= x^e*e^x  at x=3?

Feb 4, 2016

$y - {3}^{e} {e}^{3} = - \frac{{3}^{1 - e}}{{e}^{3} \left(e + 3\right)} \left(x - 3\right)$

#### Explanation:

First, find the point the normal line will intercept.

$f \left(3\right) = {3}^{e} {e}^{3}$

The normal line will pass through the point $\left(3 , {3}^{e} {e}^{3}\right)$.

To find the slope of the normal line, we should first find the derivative of the function. This can be found through the product rule:

$f ' \left(x\right) = {e}^{x} \frac{d}{\mathrm{dx}} \left[{x}^{e}\right] + {x}^{e} \frac{d}{\mathrm{dx}} \left[{e}^{x}\right]$

Note that $\frac{d}{\mathrm{dx}} \left[{e}^{x}\right] = {e}^{x}$ and that, through the power rule, $\frac{d}{\mathrm{dx}} \left[{x}^{e}\right] = e {x}^{e - 1}$.

Thus,

$f ' \left(x\right) = {e}^{x} \left(e {x}^{e - 1}\right) + {x}^{e} {e}^{x}$

$f ' \left(x\right) = {e}^{x} \left(e {x}^{e - 1} + {x}^{e}\right)$

The slope of the tangent line at $x = 3$ is

$f ' \left(3\right) = {e}^{3} \left(e \left({3}^{e - 1}\right) + {3}^{e}\right)$

$f ' \left(3\right) = {e}^{3} {3}^{e - 1} \left(e + 3\right)$

However, we want to find the normal line of the function. The slopes of the tangent line and normal line are perependicular, to they are opposite reciprocals. The opposite reciprocal of $f ' \left(3\right)$ is

$- {\left({e}^{3} {3}^{e - 1} \left(e + 3\right)\right)}^{-} 1 = - \frac{{3}^{1 - e}}{{e}^{3} \left(e + 3\right)}$

The slope of $- \frac{{3}^{1 - e}}{{e}^{3} \left(e + 3\right)}$ and point of $\left(3 , {3}^{e} {e}^{3}\right)$ can be related as an equation in point slope form:

$y - {3}^{e} {e}^{3} = - \frac{{3}^{1 - e}}{{e}^{3} \left(e + 3\right)} \left(x - 3\right)$