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So I get that if only $\sin|x^2+x|$ was given it is not differentiable at $x=0$, but why does it become differentiable at $0$ when a factor of $b|x|$ is introduced? And if it does, then is the statement "A non differentiable function times a factor which becomes zero (at those non differentiable points) makes the whole function differetiable" universal?

For example, $y= |x^2-1|\sin\pi x$ is differentiable at both $1$ and $-1$ while the first factor, taken alone, is not. Is there a better logic?

Matthew Conroy
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James
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3 Answers3

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If $f$ is any function that satisfies $|f(x)| \le C x^2$ for $x$ close to $0,$ then $f'(0)=0.$

zhw.
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I will update this answer with more details when I have time, but here's something to think about:

Plug in $\sin |x^2+x|$ into the definition of the derivative, $\lim_{h \to 0} \dfrac {f(x+h)-f(x)}{h}$. Look at the limits as $h \to 0^{+}$, and as $h \to 0^{-}$. You are interested in it only when $x=0$, so you can plug that in for $x$.

Now do the same with $b |x| \sin |x^2+x|$ and compare the results. (also note the trivial case when $b=0$, since you let $b \in \mathbb{R}$ )

Ovi
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    Check $x=-1$ as well. – Joey Zou Jun 25 '16 at 19:39
  • The answer is not the issue but there must be a more generalised way ( (say ,a theoram ) by which its hapenning , putting each Q in the very defination and getting LHL and RHL is one way . Is there an other ? – James Jun 25 '16 at 20:48
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If $b=0$, then the function is obviously differentiable. If $b\ne0$, differentiability of $f$ does not depend on $b$, so we can assume $b=1$.

The derivative can be computed by using that the derivative of $|x|$ is $\operatorname{sgn} x$, for $x\ne0$. Thus we see there are problems at $0$ and $-1$.

On the interval $(-1/2,1/2)$ we have $$ f(x)=x\sin(x^2+x) $$ so the function is differentiable at $0$.

Look at what happens in the interval $(-2,-1)$ and in the interval $(-1,0)$ and conclude.

egreg
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