# What is the equation of the line that is normal to f(x)= xe^sqrtx  at  x= 2 ?

Feb 13, 2016

$y = {e}^{\sqrt{2}} \left(1 + \frac{1}{\sqrt{2}}\right) x$

#### Explanation:

To find the equation of the line, we need a point the line passes through and a gradient.

As we wish to find the line that is tangent to $x = 2$ then we can use $f \left(2\right)$ to find a point where the line passes through:

$f \left(x\right) = x {e}^{\sqrt{x}} \to f \left(2\right) = 2 {e}^{\sqrt{2}}$

So we know at least the line passes through $\left(2 , 2 {e}^{\sqrt{2}}\right)$

Now we need the gradient which we can obtain from $f ' \left(x\right)$ the derivative.

Differentiate $f \left(x\right)$ using the product rule and chain rule to get:

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

We have simplified the function by factoring out ${e}^{\sqrt{x}}$ and dividing the $\frac{x}{2 \sqrt{x}}$ term by $\sqrt{x}$. We can now obtain the gradient:

$f ' \left(2\right) = {e}^{\sqrt{2}} \left(1 + \frac{\sqrt{2}}{2}\right) = {e}^{\sqrt{2}} \left(1 + \frac{1}{\sqrt{2}}\right) = m$

We can now put this into the equation of a line:

$y - b = m \left(x - a\right)$ where $\left(a , b\right)$ is our know point from above. That gives us:

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

We can the ${e}^{\sqrt{2}}$ over to the right by adding then factoring to simplify and get:

$y = {e}^{\sqrt{2}} \left[\left(1 + \frac{1}{\sqrt{2}}\right) \left(x - 2\right) + 2\right]$