# Prove that $f'(1)=0$ if $f(x^2)-\sin f(x)=1$ [closed]

Let $f$ be a function continuous and differentiable on $\mathbb R$ such that:

$$f(x^2)-\sin f(x)=1 \quad \forall x\in\mathbb R$$

Prove that $f'(1)=0.$

Attempt:

I tried to differentiate and I got $$2xf'(x^2)-f'(x)\cos f(x)=0$$ then I put $x=1$ and I got $0=0$

I assume it is wrong.

• Please show your effort..to provide extra context. Feb 10 '17 at 15:54
• Taking the derivative of the equation, what do you get? Feb 10 '17 at 15:56
• Differentiating both sides, we get that $$2xf'(x^2)-f'(x)\cos (f(x))=0$$ Feb 10 '17 at 15:56
• @ThomasAndrews,Yes I just corrected it. Feb 10 '17 at 15:57
• @S.C.B. I would show it if I had any idea. Feb 10 '17 at 15:57

Differentiating $$2xf'(x^2)-f'(x)\cos(f(x))=0\ ,$$ for all $x$. Compute it for $x=1$ $$2 f'(1)-f'(1)\cos(f(1))=0\Rightarrow f'(1)\left[2-\cos(f(1))\right]=0$$ Now, $AB=0$ iff $A=0$ or $B=0$ [zero-product property]. So either $f'(1)=0$, or $\cos(f(1))=2$. But the cosine is bounded between $-1$ and $1$, so the only possibility is that $f'(1)=0$.
• +1 liked that you also considered $2-\cos(f(1))$, sometimes is is easy to miss such details :). Feb 10 '17 at 16:24
• Are we assuming that we cannot venture into the complex domain? Because if so $\cos^{-1} 2=-i\ln (2-\sqrt3)$ Feb 10 '17 at 16:44