Question

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

Answer 1

`-(2sqrt3)/(12sqrt3-1)`

Explanation:

`f'(x)=2x-1/(2sqrtx)`

`f'(3)=6-1/(2sqrt3)`
This is the slope of the tangent at x = 3.

Slope of the normal is the negative reciprocal -1/f'(3)=`-(2sqrt3)/(12sqrt3-1)`