# Let $f:\mathbb R \rightarrow \mathbb R$ be a differentiable even function

I came across the above problem. I see that $G'(x)=f'(x)f(\sqrt{\tan(f(x))})$. Now since $f$ is a differentiable even function, $f(-x)=f(x)$ and so $-f'(-x)=f'(x)$ and thus $f'(0)=0$ and hence we can conclude $G'(0)=0.$ Am I going in the right direction?

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If $f(0)<0$, $G$ does not defiened as real function. – Hanul Jeon Jan 17 '13 at 4:10
Differentiation not quite right, is $f'(x)\sqrt{\tan(f(x))}$. – André Nicolas Jan 17 '13 at 4:11
Thanks a lot sir for pointing out the mistake.I have corrected my post. – learner Jan 17 '13 at 4:33
could you post the source of your questions like this? – Un Chien Andalou Jan 17 '13 at 4:45

The answer is not always $B$ either. Following Babak Sorouh's idea, let $f(x) = \pi/2 - x^2$. Then

$$G'(x) = -2 x \sqrt{\tan\!\left(\pi/2-x^2\right)}$$

where it exists and

$$\lim_{x \to 0} G'(x) = -2.$$

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The example can be modified to have $f(x)$ positive everywhere since all that matters here is the behavior in a neighborhood of $x=0$. – Antonio Vargas Jan 17 '13 at 7:02
Thanks a lot sir for the detailed clarification. I have got it. – learner Jan 17 '13 at 10:02
Great example Thanks for sharing me that. +1 – Babak S. Jan 18 '13 at 3:31

Firstly, note that according to Leibniz rule, $G'(x)=f'(x)\sqrt{\tan(f(x))}$. Secondly, buy taking $f(x)=x^{2n}$ as a family of even functions with real valued, we have: $$G'(0)=0$$ so A and C cannot be true for a general case under problem's assumptions.

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This doesn't say whether B or D is true. For that, learner had a good idea, but as tetori points out it may depend on whether we add an assumption of positivity of $f$. – Jonas Meyer Jan 17 '13 at 4:25
@JonasMeyer: Yes, exactly. Since we are facing a multiple choices here so I think we don't need to prove it theoretically. And, as tetori point and that is absolutely true, it seems D looks right. I am thinking of an even function in which $\sqrt{\tan(f(0))}$ is undefined. – Babak S. Jan 17 '13 at 5:35
Nice lead, and hits home, I think! +1 – amWhy Feb 15 '13 at 3:02
@amWhy: Thanks for considering this small hint. dear amWhy, my credit for accessing the web has been expired, and I will have the new one 2 days later I think. That's why I have not been here. – Babak S. Feb 15 '13 at 8:11

Letting $H(x) := \int_0^x \sqrt{\tan t} \, \mathrm{d}t$, $H'(x) = \sqrt{\tan t}$.

Since $G(x) = H(f(x)) - H(0)$, $$G'(x) = H'(f(x))f'(x) - H(0) = \sqrt{\tan f(x)} f'(x)$$

Since $f$ is an even function, $f'(0) = 0$, we have $G'(0) = 0$.

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Hello, welcome to Mathematics Stack Exchange. Please use MathJax in your post, there is a tutorial on Meta. – wythagoras Oct 27 '15 at 10:16