Question

What is the derivative of cos (3x)^2?

I got -18x sin (3x)^2. Is this correct?

Answer 1

`f'(x)=-6cos(3x)sin(3x)`

Explanation:

So we have:

`f(x)=(cos(3x))^2`

We need to remember a few things:

The power rule: `d/dx(x^n)=nx^(n-1)` where `n` is a constant.

The chain rule: If `f(x)=g(h(x))`, then `f'(x)=g'(h(x))*h'(x)`

Therefore,
`f'(x)=2*(cos(3x))^(2-1)*d/dx(cos(3x))`

Notice here that we have three functions together: `x^2`,`cos(x)`, and `3x`

Now remember that `d/dx(cosx)=-sinx`

=>`f'(x)=2(cos(3x))^(1)*(-sin(3x))*d/dx(3x)`

=>`f'(x)=2cos(3x)*(-sin(3x))*3`

=>`f'(x)=6cos(3x)*(-1*sin(3x))`

=>`f'(x)=-6cos(3x)sin(3x)`