Find the right derivative of $\sin^2\left(x \sin \frac1x\right)$ at $x = 0$. $$f(x)=\sin^2\left(x \sin \frac1x\right)$$
I was trying to find the right derivative of $f$ at $x = 0$.
( In this question, it says to take the f(0) as f is right continuous at x = 0. Therefore, considering the right continuous at x = 0, I obtained f(0) = 0 )
before that, I know,
I should find whether $f$ is the right differentiable at $x = 0$.
Is $f$ right differentiable at $x = 0?$ If so, what is the right derivative of $f$ at $0?$
I think $f$ is not right differentiable at $x = 0$. It would be great if someone could clarify this.
Any help would be appreciated.
 A: Hint
$$\vert f(x) \vert = \left\vert \sin^2(x \sin \frac1x) \right\vert \le x^2 \sin^2\left(\frac{1}{x} \right) \le x^2$$
Hence
$$ \frac{\vert f(x) \vert}{\vert x \vert} \le \vert x \vert$$
A: it is not right differentiable at x=0 let us solve it: $ f(x) = \sin^2(x\cdot  \sin(\frac{1}{x})) \therefore $ a little bit of basic algebra and calculus $ \therefore f\prime (x) = (\sin \frac{1}{x} - \frac{1}{x}\cdot \cos\frac{1}{x})\sin(2x\cdot \sin\frac{1}{x}) $
as you can see this function is not right differentiable at $x=0
$
A: You want to find out the limit $\lim_{x\to 0+} \frac{f(x)-f(0)}{x-0}$.
$f(x)=\sin^2(x\sin \frac 1x), x\ne 0$
Since $f(0)$ is not defined (in OP), the aforementioned limit does not make sense.
If however, in addition, you had $f(0)=0$, then it can be shown (using the fact that $|\sin y|\le |y|$ for all $y\in \mathbb R$) that: $$|\frac{f(x)-f(0)}{x-0}|=|\frac {f(x)}{x}|\leq |x|$$
So by Squeeze principle, it follows that the required limit is $0$.
Note that if $f(0)=k\ne 0$, then the limit can't exist. For, existence of the above limit would then imply:
$$\lim_{x\to 0+}(f(x)-f(0)=\lim_{x\to 0+}(\frac{f(x)-f(0)}{x-0})\lim_{x\to 0+} x=0$$
Note that $\lim_{x\to 0+} f(x)=0$ and hence by limit rules $\lim_{x\to 0+} (f(x)-f(0))=0=0-k$, which is a contradiction as $k\ne 0$.
