# What is the slope of the line normal to the tangent line of f(x) = x^2-sqrtx  at  x= 3 ?

Apr 15, 2016

$- \frac{2 \sqrt{3}}{12 \sqrt{3} - 1}$

#### Explanation:

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

$f ' \left(3\right) = 6 - \frac{1}{2 \sqrt{3}}$
This is the slope of the tangent at x = 3.

Slope of the normal is the negative reciprocal -1/f'(3)=$- \frac{2 \sqrt{3}}{12 \sqrt{3} - 1}$