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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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    Please show your effort..to provide extra context. – 2017-02-10
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    Taking the derivative of the equation, what do you get? – 2017-02-10
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    Differentiating both sides, we get that $$2xf'(x^2)-f'(x)\cos (f(x))=0$$ – 2017-02-10
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    @ThomasAndrews,Yes I just corrected it. – 2017-02-10
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    @S.C.B. I would show it if I had any idea. – 2017-02-10

1 Answers 1

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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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    but you can factor out f(x) and f(x^2) if x = 1, as f(1) – 2017-02-10
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    Nicely done. (+1). Yeah, I meant it became untrue in general. – 2017-02-10
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    So I was getting there then... thank you very much – 2017-02-10
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    +1 liked that you also considered $2-\cos(f(1))$, sometimes is is easy to miss such details :). – 2017-02-10
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    Are we assuming that we cannot venture into the complex domain? Because if so $\cos^{-1} 2=-i\ln (2-\sqrt3)$ – 2017-02-10