Q: Does $f'(0)$ exist?
$ f(x) = \begin{cases} 0 & x= 0 \\ x\sin(\frac{9}{x}) & x\ne 0 \end{cases} $
Why or why not?
How about if the $x$ in front of sine changes to $x^3$ or $x^2$ ?
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1Welcome to MSE ! What've you tried ? Did you try to compute $f'(0)$ by definition? – Arpit Kansal Sep 11 '15 at 16:45
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possible duplicate of Is there a tangent of $x\sin(1/x)$ at $x = 0$? – Sep 11 '15 at 23:11
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Hint: If $f(x)=x^a\sin\frac{9}{x}$ for $x\neq 0$, then $$\frac{f(x)-f(0)}{x-0}=x^{a-1}\sin\frac{9}{x},$$ from which you can deduce that the limit as $x\longrightarrow 0$ exists if and only if $a>1$.
Damian Reding
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You can compute the derivative of each possibility in the piecewise function and check for consistency at $x=0$
$$ f'=\left\{ \begin{array}{ll} 0 & \quad x=0 \\ \sin\left(\frac{9}{x}\right)-\frac{9}{x}\cos\left(\frac{9}{x}\right) & \quad x \neq 0 \end{array} \right. $$ If both approach the same value as $x \to 0$ then the derivative exists.
For different $x^a$ you have:
$$ f'=\left\{ \begin{array}{ll} 0 & \quad x=0 \\ ax^{a-1}\sin\left(\frac{9}{x}\right)-9x^{a-2}\cos\left(\frac{9}{x}\right) & \quad x \neq 0 \end{array} \right. $$
Again, check if the limits are equal.
crb233
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Firstly, you cannot first compute the derivative (which presupposes that it exists) and then check if it exists. And even if $f'(0)$ exists, we do not necessarily have that $\lim\limits_{0\neq x\longrightarrow 0}f'(x)=f'(0)$, because that's the statement that $f'$ is continuous at $0$. – Damian Reding Sep 11 '15 at 17:11
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@Mathgemini The converse is true though, which is what my answer says. If $f'$ is continuous at zero then $f'(0)$ exists – crb233 Sep 11 '15 at 17:17
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You cannot go about showing that $f'$ is continuous at $0$ without already knowing that $f'(0)$ exists. – Damian Reding Sep 11 '15 at 17:25