Question

How do you find f' if `f(x)=cos^2(3sqrt x )`?

Answer 1

`f'(x)=(-3 cos(3sqrtx)sin(3sqrtx))/sqrtx`

Explanation:

One can use the chain rule break the function into two parts `u=3sqrtx or 3x^0.5` and `f(u)=(cos(u))^2`

Then derive each part `x^n` becomes `nx^(n-1)` and `cos(x)` becomes `-sin(x)` so

`u'=1.5x^-0.5`
and `f'(u)=2cos(u)*-sin(u)` as we have a function inside a function `( )^2` and `cos( )` inside it so we have to derive it the outside function and multiply it by the inside function

Then all we have to do now is take replace the `u`'s with `3sqrtx` and multiply the very inside derivative `u'` by `f'(u)`

`f'(x)=(-3 cos(3sqrtx)sin(3sqrtx))/sqrtx`