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

Jan 2, 2016

The slope of tangent can be found by taking the first derivative. The slope of the normal is the negative reciprocal of the slope of the tangent.

#### Explanation:

$f \left(x\right) = {x}^{3} - \sqrt{x}$
Slope of tangent $= f ' \left(x\right) = 3 {x}^{2} - \frac{1}{2 \sqrt{x}}$
Slope of tangent at $x = 3$ is $f ' \left(3\right) = 3 {\left(3\right)}^{2} - \frac{1}{2 \sqrt{3}}$

Slope of tangent = 27 - 1/(2sqrt(3)

Simplifying we get $\frac{54 \sqrt{3} - 1}{2 \sqrt{3}}$

The slope of normal would be the negative reciprocal of this value.

Slope of normal $= \frac{- 2 \sqrt{3}}{54 \sqrt{3} - 1}$