Find the coefficient of $x^2$ in Taylor's expansion

Question

Let $f$ be thrice differentiable function on $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ such that $f'(x)= 1+(f(x))^2$.

If $f(0)=1$, then the coefficient of $x^2$ in Taylor's expansion of $f$ about $0$ is?

Answer 1

You know that $$ f''(x)=2f'(x)f(x) $$ by the chain rule. Also, $f'(0)=1+(f(0))^2=2$, so $$ f''(0)=2\cdot2\cdot1=4 $$