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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.

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  • $\begingroup$ Please show your effort..to provide extra context. $\endgroup$
    – S.C.B.
    Feb 10 '17 at 15:54
  • $\begingroup$ Taking the derivative of the equation, what do you get? $\endgroup$ Feb 10 '17 at 15:56
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    $\begingroup$ Differentiating both sides, we get that $$2xf'(x^2)-f'(x)\cos (f(x))=0$$ $\endgroup$
    – S.C.B.
    Feb 10 '17 at 15:56
  • $\begingroup$ @ThomasAndrews,Yes I just corrected it. $\endgroup$
    – Lorenzo B.
    Feb 10 '17 at 15:57
  • $\begingroup$ @S.C.B. I would show it if I had any idea. $\endgroup$
    – Lorenzo B.
    Feb 10 '17 at 15:57
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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 $f'(1)=0$.

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  • $\begingroup$ but you can factor out f(x) and f(x^2) if x = 1, as f(1) $\endgroup$
    – Cato
    Feb 10 '17 at 16:15
  • $\begingroup$ Nicely done. (+1). Yeah, I meant it became untrue in general. $\endgroup$
    – S.C.B.
    Feb 10 '17 at 16:16
  • $\begingroup$ So I was getting there then... thank you very much $\endgroup$
    – Lorenzo B.
    Feb 10 '17 at 16:17
  • $\begingroup$ +1 liked that you also considered $2-\cos(f(1))$, sometimes is is easy to miss such details :). $\endgroup$
    – MrYouMath
    Feb 10 '17 at 16:24
  • $\begingroup$ Are we assuming that we cannot venture into the complex domain? Because if so $\cos^{-1} 2=-i\ln (2-\sqrt3)$ $\endgroup$
    – Teh Rod
    Feb 10 '17 at 16:44

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