Let $f : \Bbb R \to \Bbb R$ be the function defined by $f(x)= \frac {\sin x}{|x|+ \cos x}$. Then

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

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

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

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

Using the standard definition of derivative, I get that $f$ is differentiable at $0$ as well as $\frac {\pi}2$. This is as follows,

$\lim_{h \to 0} \frac {f(0+h)-f(0)}{h}=1=\lim_{h \to 0} \frac {f(0-h)-f(0)}{-h}$


$\lim_{h\to 0} \frac {f(\frac {\pi}2 +h)-f(\frac {\pi}{2})}{h}=0=\lim_{h \to 0} \frac {f(\frac {\pi}2 -h)-f(\frac {\pi}2)}{-h}$ using L'Hospital rule at one stage.

Hence I eliminate options B and D.

Since $0$ was the only doubtful point to check about differentiability, I choose option A as the answer.

I regret a little about not knowing how to check validity of option C. Can you tell me how to prove it wrong?

  • $\begingroup$ Just differentiate and see what happens when $x=0$. $\endgroup$ – Kushal Bhuyan Dec 14 '15 at 13:31
  • $\begingroup$ While I differentiate by $\frac uv$-rule, I get encountered by the term $|x|$. In that case how should I approach it? $\endgroup$ – Error 404 Dec 14 '15 at 13:35
  • 2
    $\begingroup$ $\frac{d}{dx}|x|=\frac{x}{|x|}$, to see this start with $t=|x|$ and then $t^2=x^2$ $\endgroup$ – Kushal Bhuyan Dec 14 '15 at 13:38
  • 1
    $\begingroup$ @Quintic. Nice. That really worked. I verified $ \lim_{h \to 0} f'(0+h)=\lim_{h \to 0} f'(0-h)=1$. $\endgroup$ – Error 404 Dec 14 '15 at 14:13
  • 2
    $\begingroup$ Then you can post your solution as an answer. It is a nice practice to answer your own question after you find out everything right. :) $\endgroup$ – Kushal Bhuyan Dec 14 '15 at 15:18

A is true answer

By definition of the derivative $$f'(0)=\lim_{h\to0}\frac{\frac{\sin h}{|h|+\cos h} - \frac{\sin 0}{0+1} }h= \lim_{h\to0}\frac{\sin h}{h(|h|+\cos h)} $$ Note that (using L'Hospital and then substitution) $$ \lim_{h\to0^+}\frac{\sin h}{h(|h|+\cos h)}=\lim_{h\to0^+}\frac{\sin h}{h(h+\cos h)}=\lim_{h\to0^+}\frac{\cosh}{2h+\cos h - h\sinh)}=1 $$

Verify by yourself that also $$ \lim_{h\to0^-}\frac{\sin h}{h(|h|+\cos h)}= \lim_{h\to0^-}\frac{\sin h}{h(-h+\cos h)}=1 $$

Conclude that $f'(0)=1$

So you got that $f'(0)$ is defined, to show $f'$ is continuous at $x=0$ one need to show for example that $$\lim_{x\to0} f'(x)=f'(0)$$

$$\lim_{x\to0} f'(x) = \lim_{x\to0}\lim_{h\to0}\frac{\frac{\sin x+h}{|x+h|+\cos (x+h)} - \frac{\sin x}{|x|+\cos x} }h= \lim_{h\to0}\lim_{x\to0}\frac{\frac{\sin x+h}{|x+h|+\cos (x+h)} - \frac{\sin x}{|x|+\cos x} }h=f'(0) $$ The switching the order of limits is allowed when the function is uniformly continuous. A continuous function on a compact set is uniformly continuous, so everything is fine here, but I guess that you won't like this sort of answer:)

Using $x\ne0$: $|x|'=\mathrm{sgn}(x)$ where $$\mathrm{sgn}(x) = \begin{cases}1&x>0\\0&x=0\\-1&x<0\end{cases}$$ $$f'(x)=\frac{\cos (x)}{\left| x\right| +\cos (x)}-\frac{\sin (x) (\mathrm{sgn}(x)-\sin (x))}{(\left| x\right| +\cos (x))^2},\quad x\ne0$$

$$\lim_{x\to0^+} f'(x) = \frac{1}{0 +1}-\frac{0 (0-0)}{(0 +1)^2}=1$$

Similarly $$\lim_{x\to0^-} f'(x) = 1$$ Conclude $$\lim_{x\to0} f'(x)=f'(0)$$ and you done.

  • 1
    $\begingroup$ How did you conclude that $f'$ is continuous at $0$? $\endgroup$ – Kushal Bhuyan Dec 14 '15 at 13:45
  • $\begingroup$ I just show why $f$ is differentiable at $x=0$, will add a note about continuity $\endgroup$ – Michael Medvinsky Dec 14 '15 at 13:52
  • $\begingroup$ One more thing as $f'(0)=1$ so we can conclude that $f'$ is continuous at $0$. Is that right? $\endgroup$ – Kushal Bhuyan Dec 14 '15 at 13:53
  • 1
    $\begingroup$ @Quintic nope, math.stackexchange.com/questions/292275/… $\endgroup$ – Michael Medvinsky Dec 14 '15 at 14:19
  • 1
    $\begingroup$ @Alex Nope, using L'Hospital we differentiate different function, the nominator and denominator. $\endgroup$ – Michael Medvinsky Dec 14 '15 at 14:20

I feel you are looking for a descriptive answer, still leaving it here as an answer since it explains your doubt.

From graph:

enter image description here

Where green's $f$, blue $f'$, red $f''$. $f'$ is clearly continuous but not differentiable at $0$

  • 1
    $\begingroup$ The graph certainly makes the things clear. Thanks :) $\endgroup$ – Error 404 Dec 14 '15 at 14:07

As per the discussion with @Quintic in the above comments,

$f'(x)=\frac {1+|x|\cos x-\frac {x}{|x|} \sin x}{(|x|+\cos x)^2}$.

Now, $\lim_{h \to 0} f'(0+h)=\lim_{h \to 0} \frac {1+|h|\cos h-\frac {h}{|h|}\sin h}{(|h|+\cos h)^2}=\lim_{h \to 0}\frac {1+h\cos h-\sin h}{(h+\cos h)^2}$. (Where $h \gt 0$)

$\Rightarrow \lim_{h\to 0} f'(0+h)=\frac {1+0-0}{(0+1)^2}=1.$

Similarly, $\lim_{h\to 0} f'(0-h)=\lim_{h\to 0} \frac {1+h\cos h-\sin h}{(h+\cos h)^2}$ (Where $h \gt 0)$

$\Rightarrow \lim_{h\to 0} f'(0-h)=1$.

Hence $\lim_{h\to 0} f'(0+h)=\lim_{h\to 0} f'(0-h) \Rightarrow$ $f'$ is continuous at $x=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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