# What is the equation of the line that is normal to f(x)= e^(2x-2) sqrt( 2x-2)  at  x=1 ?

May 4, 2018

First, find the derivative of the function using the product rule.
$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{2 x - 2} \cdot \frac{1}{2} {\left(2 x - 2\right)}^{- \frac{1}{2}} \cdot 2 + \sqrt{2 x - 2} \cdot {e}^{2 x - 2} \cdot 2$

Then, simplify:
$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{2 x - 2} / {\left(2 x - 2\right)}^{\frac{1}{2}} + 2 \sqrt{2 x - 2} \cdot {e}^{2 x - 2}$

Substituting $x = 1$ into this will give your derivative which is the slope of the line tangent to the curve at $x = 1$

The slope of the normal to the curve at $x = 1$ is the negative reciprocal of the derivative. In order to find the equation of your line, you substitute $x = 1$ into your original function to find the $y$ value and use these two values in the slope-point formula for the equation of a line.

$y - {y}_{1} = - \frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}} \left(x - {x}_{1}\right)$