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 $x=0$ ?
Best Answer
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.