Derivative of {f(x)}^2

Question

$h(x)= f^2(x)$

Find $h'(x)$.

I initially thought that $h'(x)=2f(x)$ but it is wrong. When I googled it, I found it to be $h'(x) = 0$ but I'm not sure if that's true.

Answer 1

If $u(x) = x^2$, then $h(x) = u(f(x))$, and by the chain rule $h'(x) = u'(f(x)) f'(x) = 2 f(x) f'(x)$.

Answer 2

Well, if $$h(x)=\left(f(x)\right)^2$$ then using the chain rule we get $$h'(x)=2f(x)f'(x)$$

So, I'm not sure how you're getting $h'(x)$ to be $0$, the derivative is $0$ only when the function is a constant so $h'(x)$ being $0$ means that $h(x)=c$ where c is some constant.

Now if that's the case $f(x)$ would be the square root of $c$ so $f(x)=\sqrt{c}$ and this would make $h'(x)=0$