# What is the equation of the line normal to  f(x)=e^(sqrtx/x) at  x=4?

Aug 16, 2017

$y - \sqrt{e} = 4 \sqrt{\frac{2}{e}} \left(x - 4\right)$

#### Explanation:

The normal line is perpendicular to the tangent line. So, if we use the derivative of the function to find the slope of the tangent line, we can then find the slope of the normal line and write its equation.

To find the derivative, first simplify the function:

$f \left(x\right) = {e}^{\frac{\sqrt{x}}{x}} = {e}^{{x}^{1 / 2} / {x}^{1}} = {e}^{{x}^{- 1 / 2}}$

Remember that $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$, so according to the chain rule, $\frac{d}{\mathrm{dx}} {e}^{u} = {e}^{u} \frac{\mathrm{du}}{\mathrm{dx}}$.

$f ' \left(x\right) = {e}^{{x}^{- 1 / 2}} \frac{d}{\mathrm{dx}} {x}^{- 1 / 2}$

And from here use the power rule:

$f ' \left(x\right) = {e}^{{x}^{- 1 / 2}} \left(- \frac{1}{2} {x}^{- 3 / 2}\right)$

So the slope of the tangent line at $x = 4$ is:

$f ' \left(4\right) = {e}^{{4}^{- 1 / 2}} \left(- \frac{1}{2} \cdot {2}^{- \frac{3}{2}}\right) = - {e}^{1 / 2} / {2}^{5 / 2} = - \frac{\sqrt{e}}{4 \sqrt{2}}$

A line of slope $m$ has a perpendicular line with slope $- \frac{1}{m}$, so the slope of the normal line is $- \frac{1}{f ' \left(4\right)}$, or:

$\text{slope normal} = 4 \sqrt{\frac{2}{e}}$

The only thing left is to find the any point on the normal line so we can write its equation, and we know that the normal line will intersect $f$ at $x = 4$:

$f \left(4\right) = {e}^{{4}^{- 1 / 2}} = {e}^{1 / 2} = \sqrt{e}$

The equation of a line that passes through $\left(4 , \sqrt{e}\right)$ with slope $4 \sqrt{\frac{2}{e}}$ is:

$y - \sqrt{e} = 4 \sqrt{\frac{2}{e}} \left(x - 4\right)$