Let $f(x^4+8x)=24x$ where $x < 0$, find $f^\prime(0)$

Question

Let $f(x^4+8x)=24x$ where $x < 0$, which of the following equals to $f^\prime(0)$ ?

$\text{a)}~1~~~~\text{b)}~-1~~~~\text{c)}~2~~~~\text{d)}~ -2$

This question was asked in high school science competition. The official solution is $\text{a)}$ but I don't know how to solve it.

My solution

From my observation, $f(x)$ takes a function and output $3^{rd}$ drivative of it. So,
$$f(x) = \frac{d^3x}{dt^3}$$

Therefore, for a constant function $c$. $f(c) = 0$. which imply

$$ f^\prime(0) = 0 $$

which is not correct. I couldn't find where the error is. so a help would be greatly appreciated.

Answer 1

$$f'(x^4+8x)(4x^3+8)=24,$$ which for $x=-2$ gives $$(-24)f'(0)=24$$ or $$f'(0)=-1.$$

Answer 2

Differentiating it with respect to $x$, putting $0$ or $-2$ can make the argument of $f'(x^4+8x)$ equal to $0$, but since the equation is valid only for $x<0$, put $x=-2$ and answer comes out to be $-1$.

Answer 3

$f′(x^4+8x)(4x^3+8)=24 $,
now u want $f'(0)$ for $x<0$.
So put $x^4+8x=0$,
which gives $x=0,-2$,
$x=0$ is rejected since $x<0$.
So, $x=-2$.
Put $x=-2$ in $$f′(x^4+8x)(4x^3+8)=24$$
You will get $f'(0)\times(-24)=24$,
or, $f'(0)=-1$.
You get option $\text{b}$ instead of $\text{a}$.