Question

How do you differentiate `f(x)=cos(=3x^2+2)^2`?

Answer 1

`f'(x)=-12xcos(3x^2+2)*sin(3x^2+2)`

Explanation:

We need to remember two things:
the chain rule and the power rule.

The chain rule states that if `h(x)=f(g(x))`, then `h'(x)=f'(g(x))*g'(x)`
The power rule states that if `f(x)=x^n`, then `f'(x)=nx^(n-1)`
For this problem, note that `d/dxcosx=-sinx`
Also, derivative of a constant is always 0.

Let's apply this to `f(x)=cos(3x^2+2)^2`
=>`f'(x)=2cos(3x^2+2)*-sin(3x^2+2)*6x`

=>`f'(x)=-12xcos(3x^2+2)*sin(3x^2+2)`

That is our answer!

We have basically thought of `cos(3x^2+2)^2` as a triple function of `h(g(f(x)))` where `h(x)=x^2`, `g(x)=cos(x)` and`f(x)=3x^2+2`