Why is the derivative of $y = f(x^3)$ equal to $f'(x^3) * 3x^2$

Question

Why is the derivative of this function:
$ y = f(x^3)$
equal to
$y'= f'(x^3)* 3x^2$

instead of just:
$y' = f'(x^3)$

I know it has something to do with one of the rules (chain?) but I am not sure how it would apply in this situation.

Answer 1

This is because of a particular rule that we call the chain rule. It can be derived as follows: let $f(g(x))$ be a composition of two differentiable functions. Then, $$ f(g(x))' = \lim_{x \to x_0} \frac{f(g(x))-f(g(x_0))}{x-x_0},\quad x=x_0 $$ Which is precisely $$\lim_{x\to x_0} \frac {f(g(x))-f(g(x_0))}{g(x)-g(x_0)}\cdot\frac{g(x)-g(x_0)}{x-x_0}$$ since $$\lim_{y\to y_0} \frac {f(y)-f(y_0)}{y-y_0} = f'(y_0)$$ and $$\lim_{x\to x_0} \frac {g(x)-g(x_0)}{x-x_0}= g'(x_0)$$ we have, setting $g(x)=y,\quad g(x_0)=y_0$ $$ f(g(x_0))' = f'(g(x_0)) g'(x_0)$$ which is precisley the chain rule. Plugging in $g(x)=x^3$, we get $$f(g(x_0))'=f'(x_0^3)\cdot 3x_0^2$$ which is just the derivative of $f(x^3)$ at $x=x_0$.