Question

Question #39ad9

Answer 1

`f'(4)=-1/16`

Explanation:

Using the power rule

• `d/dx x^n = nx^(n-1)`

we have

`f'(x) = d/dx2/x^2`

`=d/dx 2x^(-2)`

`=2(-2x^(-3))`

`=-4x^(-3)`

`=-4/x^3`

Evaluated at `x = 4`, we get

`f'(4) = -4/4^3`

`=-1/4^2`

`=-1/16`