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$$f:\mathbb R\rightarrow \mathbb R$$ be the function defined by $$f(x)={{\sin x}\over {|x|+\cos x}}.$$ Then ,

$A.$ $f$ is differentiable at all $x\in \mathbb R.$

$B.$ $f$ is not differentiable at $0.$

$C.$ $f$ is differentiable at $x=0$ but $f'$ is not continuous at $x=0.$

$D.$ $f$ is not differentiable at $x={{\pi}\over 2}.$

I think option $B$ is correct as we know $|x|$ is not differentiable at $x=0.$ So that immediately makes options $A$ and $C$ incorrect.

And the derivative can actually be worked out for option $D$ .

So , the correct answer is option $B.$ Am I right $?$

Thanks.

user118494
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    Just because $|x|$ is not differentiable at $x=0$ doesn't mean that any function where $|x|$ is used it not differentiable at $x=0$, e.g. $f:\mathbb R\rightarrow\mathbb R,~f(x)=x|x|$. You have to check using the differential quotient. – Hirshy Dec 18 '15 at 10:20
  • Similarly $sinx$ is differentiable at all x but $\frac 1{sinx}$ is not – Jasser Dec 18 '15 at 10:30

1 Answers1

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Hint:
Do differentiation of $f(x)$ for both $x>0$ and $x<0$. and on observing the solutions you'll get
$f'(x) = \left(\frac{1-(\sin|x|-|x|\cos x)}{(|x|+\cos x)^2}\right)$. Now,$f(x)$ to be differentiable at $x=a$, $f'(a^+) = f'(a^-)$ and $f(x)$ to be continuous at $x=a$, $f(a^+) = f(a^-) = f(a) $.

Element118
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