Finding the Fourth Derivative Let $f(x)$ be a four-times differential function such that 
$f(2x^2-1)=2xf(x)$
What is the value of $f''''(0)$?
A brute force approach is differentiating the given condition 4 times and finding values of $f(x)$ and its derivatives at required numbers or at a value of $x$ where the function can be simplified. Considering the LHS has a composite function and the RHS has the product of 2 functions, differentiating both sides leads to a lot of terms. I tried shifting values of $x$ but  did not find a convenient substitution to simplify the problem. How should I proceed?
 A: I managed to figure this out a little late.
In the given functional equation, replace $x$ with $-x$.
The equation is now written as:
$f(2x^2-1)=-2xf(-x)$
But it is given that $f(2x^2-1)=2xf(x)$
Therefore, $-2xf(-x)=2xf(x)$
$=>$ $-f(-x)=f(x)$
A rather simple functional equation which can be easily differentiated 4 times.
The answer is $0$.
A: Let $x=\cos\theta$. This tells us that $f(\cos 2\theta)=2\cos\theta f(\cos\theta)$. So, if we set $f(\cos\theta)=g(\theta)$, we now have that:
$$g(2\theta)=2\cos\theta g(\theta)$$
This is reminiscent of the double angle formula for the sine function, and we can see that:
$$\frac{g(2\theta)}{\sin2\theta}=\frac{g(\theta)}{\sin\theta}$$
Given that $f$, hence $g$, is continuous [four times differentiable $\implies$ continuous], we're thus able to say that $g(\theta)\propto\sin\theta$. This technically requires a little more work [see appendix], but is intuitively a fairly reasonable claim. Thus:
$$f(\cos\theta)=K\sin\theta\implies f(x)=K\sqrt{1-x^2}$$
We can then either differentiate or use a power series expansion to find $f''''(0)$ - if we write $\sqrt{1-x^2}=(1-x^2)^{1/2}=1-\frac{x^2}{2}-\frac{x^4}{8}+O(x^6)$, which gives $f''''(0)=K\frac{-24}{8}=-3K$.
[Appendix]
Given $\frac{g(2\theta)}{\sin2\theta}=\frac{g(\theta)}{\sin\theta}$, we define $h(\theta)=\frac{g(\theta)}{\sin\theta}$, noting that $h(\theta)=h(2\theta)$, and aim to prove that $h$ is constant. Indeed, let $\theta\in\mathbb{R}$. Then $h(\theta)=h(\theta/2)=h(\theta/4)...=h(\theta/2^n)$ for all $n\in\mathbb{Z}$. Continuity of all the functions involved means that we can take $n\to\infty$ and see that $h(\theta)=h(0)$, and thus that $h$ is constant. The desired result follows.
