I had learnt about the shortcut method of checking differentiability from a certain book few months back.

I am illustrating the shorcut method I had learnt here using an example:

Is $|x-1/9|^3$ differentiable at $x=1/9$

For $x>1/9$ $f(x)=(x-1/9)^3$ Differentiate w.r.t x and put $x=1/9$.Say the value of the derivative is $a$.

For $x<1/9$ $f(x)=-(x-1/9)^3$ Differentiate w.r.t x and put $x=1/9$.Say the value of the derivative is $b$.

If $a=b$ then it is differentiable at $x=1/9$

This method works because $f(x)$ is continuous at $x=1/9$

But recently I came across a function like:

$f(x)= \begin{cases} 0 & x= 0 \\ 2x+x^2\sin(\frac{1}{x}) &x \neq0 \end{cases}$

Even though $f(x)$ is continuous at $x=0$ the shortcut method does not seem to work here.

For $x>0$ $f'(x)=2+2x\sin(\frac{1}{x})+x^2\cos(\frac{1}{x})(\frac{-1}{x^2})$.

But here when I put $x=0$, $f'(x)$ becomes undefined.

However using the limit definition of derivative I am getting the value of derivative at $x=0$ as $2$.

Why isn't the shortcut method not valid here ? On the other hand why is the limit definition of derivative valid and working ? Does the derivative of $f(x)$ really exist at $x=0$ ?

  • 1
    $\begingroup$ You need that the limit of $f'$ at $x=0$ exists and $f$ is continuous in $0$. You misunderstood the shortcut. $\endgroup$
    – user251257
    Dec 3, 2016 at 15:07
  • $\begingroup$ Why should the limit of f' exist at x=0 ? Can you point me to any online resource which says so ? Or could you explain the reason behind it ? @user251257 $\endgroup$
    – user220382
    Dec 3, 2016 at 15:13
  • $\begingroup$ look in the proof of your shortcut $\endgroup$
    – user251257
    Dec 3, 2016 at 15:14
  • $\begingroup$ @user251257 I don't have the proof of my shortcut. I can't remember where I learnt it from. $\endgroup$
    – user220382
    Dec 3, 2016 at 15:15
  • $\begingroup$ then ask a new question for the proof or use the search function. $\endgroup$
    – user251257
    Dec 3, 2016 at 15:23

1 Answer 1


There is no contradiction:

This is a great example of the derivative existing pointwise, although not being continuous.

Hence the given function is continuous and differentiable, but not continuously differentiable.


Your Answer

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