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For $x\in\Bbb{R}$, $f(x) = |\log2 – \sin x|$ and $g(x) = f(f(x))$, then:

  1. $g$ is differentiable at $x = 0$, and $g'(0) = –\sin(\log 2)$
  2. $g$ is not differentiable at $x = 0$
  3. $g'(0) = \cos(\log2)$
  4. $g'(0) = –\cos(\log2)$

Please someone help me with this question.

Sorry to say but I tried hard but was not able to find any thing. So I tried different things which were useless and I am not sharing those very very useless solution with you. Pls help me

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    $\begingroup$ Do not write plain text between dollar signs...as you can see things go terribly wrong. If you REALLY need to write some plain text between dollar signs, do write \text{text} ...or better: close dollar signs and write down. $\endgroup$ – DonAntonio Jan 24 at 14:35
  • $\begingroup$ Are you asked to prove (1) to (4)? But this doesn't make sense since (1) and (2) contradict each other. Something seems wrong in your question. $\endgroup$ – Bernard Massé Jan 24 at 14:38
  • $\begingroup$ @BernardMassé It is multiple choice. $\endgroup$ – Ben W Jan 24 at 14:39
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    $\begingroup$ @Ben W. Where does it say it is multiple choice? The question might be "which of the following is true?" or "prove all of the following" or "only one of the following is true, identify which one?" or "only one of the following is false, identify which one". $\endgroup$ – Bernard Massé Jan 24 at 14:42
  • $\begingroup$ @BernardMassé sorry that was too bad from me $\endgroup$ – Ram Sale Jan 24 at 15:05
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Since the absolute value function isn't everywhere-differentiable, we will get rid of them. Luckily everything is continuous, so we can plug in $x=0$ to find out the signs we need.

In particular, note that $\log(2)-\sin(0)$ is positive, and so $f(x)=\log(2)-\sin(x)$ (without the absolute value) in a neighborhood of zero. Hence, $g(x)=|\log(2)-\sin(\log(2)-\sin(x))|$ in the same neighborhood.

But we still need to get rid of the other absolute value. As before, we note that $\log(2)-\sin(\log(2)-\sin(0))$ is also positive, and so $$g(x)=\log(2)-\sin(\log(2)-\sin(x))$$ in a (possibly smaller) neighborhood of zero. Now you can compute the derivative in the usual way, using the chain rule.

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  • $\begingroup$ thanks for answering $\endgroup$ – Ram Sale Jan 24 at 15:05

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