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

Jun 14, 2016

It is $y = - \frac{4}{e} ^ 2 x + 9.55$.

#### Explanation:

The line passing from $x = 4$ and tangent to ${e}^{\sqrt{x}}$ has as slope

$\frac{d}{\mathrm{dx}} {e}^{\sqrt{x}}$ evaluated when $x = 4$. The derivative can be calculated with the chain's rule

$\frac{d}{\mathrm{dx}} {e}^{\sqrt{x}} = {e}^{\sqrt{x}} \cdot \frac{d}{\mathrm{dx}} \sqrt{x} = {e}^{\sqrt{x}} / \left(2 \sqrt{x}\right)$.

The slope is

$m = {e}^{\sqrt{4}} / \left(2 \sqrt{4}\right) = {e}^{2} / 4$.

So the tangent has equation

$y = {e}^{2} / 4 x + q$

The $q$ can be found imposing the passage of the line from the point $\left(4 , {e}^{\sqrt{4}}\right) = \left(4 , {e}^{2}\right)$

${e}^{2} = {e}^{2} / 4 \cdot 4 + q$ then $q = 0$ and the equation of the tangent is

$y = {e}^{2} / 4 x$

The orthogonal has the negative inverse slope, so the equation is

$y = - \frac{4}{e} ^ 2 x + q$

Again, to find $q$ we have to impose the passage from $\left(4 , {e}^{2}\right)$

${e}^{2} = - \frac{4}{e} ^ 2 \cdot 4 + q$

$q = {e}^{2} + \frac{16}{e} ^ 2 = \frac{{e}^{4} + 16}{e} ^ 2 \setminus \approx 9.55$ and finally the equation of the orthogonal is

$y = - \frac{4}{e} ^ 2 x + 9.55$ or

$y = - 0.54 x + 9.55$.