# What is the equation of the line that is normal to f(x)= e^(2x-2) /sqrt( 2x-2)  at  x=3 ?

Jul 4, 2018

$y = - \frac{8}{7 \cdot {e}^{4}} x + {e}^{4} / 2 + \frac{24}{7 \cdot {e}^{4}}$

#### Explanation:

Writing your function in the form

$f \left(x\right) = {e}^{2 x - 2} {\left(2 x - 2\right)}^{- \frac{1}{2}}$
then we can use the product and the chain rule, doing this we get

$f ' \left(x\right) = 2 {e}^{2 x - 2} {\left(2 x - 2\right)}^{- \frac{1}{2}} + {e}^{2 x - 2} {\left(2 x - 2\right)}^{- \frac{3}{2}} \cdot \left(- \frac{1}{2}\right) \cdot 2$
so we get

$f \left(3\right) = {e}^{4} - {e}^{4} / 8 = \frac{7}{8} {e}^{4}$

now we can find the slope of the normal line

${m}_{N} = - \frac{8}{7 {e}^{4}}$

now we get

$f \left(3\right) = {e}^{4} / 2$
so we get

$y = - \frac{8}{7 {e}^{4}} x + {e}^{4} / 2 + \frac{24}{7 {e}^{4}}$