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

Dec 4, 2016

The equation of the line normal to $f \left(x\right) = \sqrt{{e}^{\sqrt{x}}}$ at $x = 4$ is:

$y - e = - \frac{8}{e} \left(x - 4\right)$

#### Explanation:

The equation of the line normal to $f \left(x\right)$ in $x = \overline{x}$ is given by:

$y - f \left(\overline{x}\right) = - \frac{1}{f ' \left(\overline{x}\right)} \left(x - \overline{x}\right)$

So we calculate the first derivative of $f \left(x\right)$:

f(x) = sqrt (e ^ sqrtx) = e^(1/2x^(1/2)

using the chain rule:

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

For $\overline{x} = 4$:

$f \left(\overline{x}\right) = e$
$f ' \left(\overline{x}\right) = \frac{e}{8}$

So the equation of the normal line is:

$y - e = - \frac{8}{e} \left(x - 4\right)$