Question

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

Answer 1

Read below.

Explanation:

We use the chain rule:

`d/dx[f(g(x))]=f'(g(x))*g'(x)`

Power rule:

`d/dx[x^n]=nx^(n-1)`

`d/dx[sinx]=cosx`

Therefore:

`f'(x)=cos(x^3)*d/dx[x^3]`

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

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

Answer 2

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

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"Given "f(x)=g(h(x))" then"`

`f'(x)=g'(h(x))xxg'(x)larrcolor(blue)"chain rule"`

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

`rArrf'(x)=cos(x^3)xxd/dx(x^3)=3x^2cos(x^3)`