Question

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

Answer 1

`d/(dx)cos(x^3)=-3x^2sin(x^3)`

Explanation:

Use chain rule: `(dy)/(dx)=(dy)/(du)*(du)/(dx)`

`y=cos(x^3)`, let `u=x^3`
Then `(du)/(dx)=3x^2` and `(dy)/(du)=-sinu=-sin(x^3)`
So `(dy)/(dx)=3x^2*-sin(x^3)=-3x^2sin(x^3)`

Answer 2

The answer is `-3x^2 sin(x^3)`

Explanation:

I mainly use formulas because some of them are easy to memorize and they help you see the answer right away, but you can also use the "u substitution." I think that's what is officially known as the "Chain Rule"

`color (red) (d/dx cos x = (cosx)'=-(x)'sinx=-sinx)` and when it's not `x` but any other variable, like `5x` for example, the formula is `color (red) (d/(du) cos u =(cos u)'= -(u)'sinu=-u'sinu)`

Note that `color (red) (u')` is the derivative of `color (red) u`

Our problem `f(x)=cos(x^3)`

Since it's not simply `x` but `x^3`, the first formula will not work but the second will.

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

Another method: "u substitution"

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

Let's say `u=x^3 => f(u)=cosu`

`f'(u)=-u'sinu`

And the derivative of `u=(u)'=(x^3)'=3x^2`

`=>f'(u)=-3x^2(sin(u))`

Substitute back `u=x^3`

`f'(x)=-3x^2(sin(x^3))=-3x^2sin(x^3)`

Hope this helps :)