How do I differentiate $f^3(x^2)$? I am learning about the chain rule and I'm not sure about kind of having to do it twice?
So say I have the function $f^{3}(g(x))$, how would I differentiate this to find $f'(x)$ using the chain rule?
Ok so what I have tried to do is say:
$$\frac{d}{dx}(f^3(g(x)) = 3f^2(g(x))f'(g(x))g'(x).$$
However, something tells me this is incorrect.
HINT
Let's differentiate $f^2(x^2) = f(f(x^2))$. You get
$$
\frac{d}{dx} f(f(x^2))
 = f'(f(x^2)) \frac{d}{dx} f(x^2)
 = f'(f(x^2))f'(x^2)\frac{d}{dx} x^2
 = f'(f(x^2))f'(x^2) \cdot 2x
$$
Another possible interpretation is that you mean $f^2(x^2) = (f(x^2))^2$ (but this is unlikely since we usually write that as $f(x^2)^2$, not as $f^2(x^2)$. If this is what you mean though, you would get
$$
\frac{d}{dx} (f(x^2))^2
 = 2f(x^2)\frac{d}{dx}f(x^2)
 = 2f(x^2) f'(x^2) \frac{d}{dx} x^2
 = 4x f(x^2) f'(x^2)
$$
HINT
Let's differentiate $f^2(x^2) = f(f(x^2))$. You get
$$
\frac{d}{dx} f(f(x^2))
 = f'(f(x^2)) \frac{d}{dx} f(x^2)
 = f'(f(x^2))f'(x^2)\frac{d}{dx} x^2
 = f'(f(x^2))f'(x^2) \cdot 2x
$$
Another possible interpretation is that you mean $f^2(x^2) = (f(x^2))^2$ (but this is unlikely since we usually write that as $f(x^2)^2$, not as $f^2(x^2)$. If this is what you mean though, you would get
$$
\frac{d}{dx} (f(x^2))^2
 = 2f(x^2)\frac{d}{dx}f(x^2)
 = 2f(x^2) f'(x^2) \frac{d}{dx} x^2
 = 4x f(x^2) f'(x^2)
$$