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

Jan 29, 2016

$y - {e}^{8} = - \frac{1}{e} ^ 8 \left(x - 4\right)$

#### Explanation:

First, we can simplify the function to make differentiation simpler:

$f \left(x\right) = {e}^{x \cdot {x}^{1 \text{/"2))=e^(x^(3"/} 2}}$

We should find the point on the function the normal line will intercept:

$f \left(4\right) = {e}^{{4}^{3 \text{/} 2}} = {e}^{8}$

The normal line will intercept the point $\left(4 , {e}^{8}\right)$.

To find the slope of the normal line, first find the slope of the tangent line. Since the normal line and tangent line are perpendicular, their slopes will be opposite reciprocals.

The slope of the tangent line can be found through finding the value of the derivative of the function at $x = 4$.

To differentiate the function, use the chain rule.

The specific application of the chain rule that applies here is $\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \cdot u '$. Here $u = {x}^{3 \text{/} 2}$. Applying this yields

$f ' \left(x\right) = {e}^{{x}^{3 \text{/"2))*d/dx(x^(3"/} 2}}$

The internal derivative can be found through the power rule.

f'(x)=e^(x^(3"/"2))(3/2x^(1"/"2))=(3sqrtx(e^(x^(3"/"2))))/2

The slope of the tangent line is

$f ' \left(4\right) = \frac{3 \sqrt{4} \left({e}^{{4}^{3 \text{/} 2}}\right)}{2} = 3 {e}^{8}$

The slope of the normal line will be the opposite reciprocal of this, or $- \frac{1}{e} ^ 8$.

We now must relate the information we know about the normal line into a single equation: its slope is $- \frac{1}{e} ^ 8$ and it passes through the point $\left(4 , {e}^{8}\right)$. The described line in point-slope form is

$y - {e}^{8} = - \frac{1}{e} ^ 8 \left(x - 4\right)$