Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

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?

share|cite|improve this question
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
up vote 4 down vote accepted

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

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

share|cite|improve this answer
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.

share|cite|improve this answer
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 $G'(0) = 0$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.