Question

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

Answer 1

`-6x^2sin(x^3)cos(x^3)`

Explanation:

`cos^2(x^3)=cos(x^3)cos(x^3)`
there is a differential rule
`d/dxf(x)g(x)=f'(x)g(x)+f(x)g'(x)`
and use chain rule
so
`d/dxcos^2(x^3)`
=`d/dxcos(x^3)cos(x^3)`
=`(-sin(x^3)3x^2)cos(x^3)+cos(x^3)*(-sin(x^3)3x^2)`
=`-6x^2sin(x^3)cos(x^3)`

Answer 2

The derivative of `cos^2(x^3)` is `-6x^2sin(x^3)cos(x^3)=-3x^2sin(2x^3)`

Explanation:

It could be differentiated using chain rule only
`cos^2(x^3)=(cos(x^3))^2`
usually we would say
`(f(g(x)))'=f'(g(x))*g'(x)`
but this time we have
`(cos(x^3))^2=f(g(h(x)))`
where `f(x)=x^2`, `g(x)=cos(x)` and `h(x)=x^3`

It turns out that the formula above is kind of "recursive". It means that if there is more functions inside, we can plug g(h(x)) as second function ant then find derivative of it by chain rule:

`(f(g(h(x))))'=f'(g(h(x)))*(g(h(x)))'=`
`=f'(g(h(x)))*g'(h(x))*h'(x)`

(That's the essence of "chain" rule)

For our example
`(cos(x^3))^2=2cos(x^3)*-sin(x^3)*3x^2=`
`=-6x^2sin(x^3)cos(x^3)` (same result)

But we can simplify more!

Trigonometry gives us bunch of formulas including this one:
`sin(2x)=2sin(x)cos(x)`
so we could simplify our result
`-6x^2sin(x^3)cos(x^3)=-3x^2sin(2x^3)`