Let $\alpha, \beta \ge 1 \in \mathbb{N}$ and:

$$f(x) = \left\{ {\matrix{ {{x^\beta }\sin \left( {{1 \over {{x^\alpha }}}} \right)} \cr {0,x = 0} \cr } } \right.$$

I checked for continuity and found that the function is continuous for every $\alpha, \beta \in \mathbb{N}$.

Now, the derivative $f'(x_0)$ is:
$$f'(x) = \mathop {\lim }\limits_{x \to {x_0}} {{f(x) - f({x_0})} \over {x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} {{{x^\beta }\sin \left( {{1 \over {{x^\alpha }}}} \right) - {x_0}^\beta \sin \left( {{1 \over {{x_0}^\alpha }}} \right)} \over {x - {x_0}}}$$

For, $x_0 = 0$ we have:

$$f'(0) = \mathop {\lim }\limits_{x \to 0} {{{x^\beta }\sin \left( {{1 \over {{x^\alpha }}}} \right) - 0} \over {x - 0}} = \mathop {\lim }\limits_{x \to 0} {x^{\beta - 1}}\sin \left( {{1 \over {{x^\alpha }}}} \right)$$

If $\beta = 1$ then the limit doesn't exist. Hence, $f$ is not differentiable at $x_0=0$.

If $\beta > 1$ then the limit is $0$ (Hence, $f'(0) = 0$).

I'm not sure if I covered all cases. Is there another point which is non-differentiable?

  • $\begingroup$ Why did you disregard the case $\beta \neq 1$? It needs care. $\endgroup$ – Git Gud May 17 '14 at 10:36
  • $\begingroup$ Because it's clear that if $\beta > 1$ then the limit is $0$. $\endgroup$ – AnnieOK May 17 '14 at 11:17
  • $\begingroup$ Fair enough. The function is differentiable at all points. You're done. $\endgroup$ – Git Gud May 17 '14 at 11:30

When you are finding $f'(x_{0})$, where $x_{0}=0$, the term $x_{0}^\beta\sin\left(\frac{1}{x^\alpha}\right)$ is not $0$ because of the term $\frac{1}{x^\alpha}$.

Given that your function is continuous, you can use the following result (due to the Mean Value Theorem).

Let $f$ be continuous at $x_{0}$ and suppose that $\displaystyle\lim_{x\rightarrow x_{0}}f'(x)$ exits. Then $f$ is differentiable at $x_{0}$ and $f'(x_{0})=\displaystyle\lim_{x\rightarrow x_{0}}f'(x)$.

Now, for all $x\neq 0$, the given function is differentiable.

Next, consider $x=0$. Then, $\displaystyle\lim_{x\rightarrow 0}f'(x)=\displaystyle\lim_{x\rightarrow 0}\left[\beta x^{\beta-1}\sin\left(\frac{1}{x^\alpha}\right)-\alpha x^{\beta-(\alpha+1)}\cos\left(\frac{1}{x^\alpha}\right)\right]$.

For $f$ to be differentiable at $x=0$:

The above limit exists (and equates to $0$) if $\beta-1\neq 0$ and $\beta-(\alpha+1)\neq 0$. Given that $\alpha,\beta\geq 1$, we get $\beta-(\alpha+1)\neq 0$, that is, $\beta\neq \alpha+1$.


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