Finding derivative using definition Using definition of derivative, find derivative of f at 0 when f(x) = $x^3sin \mid x \mid$
Since definition is:
$f'(x)=\frac{f(x)-f(a)}{x-a} = \frac{x^3sin \mid x \mid - a^3sin \mid a \mid}{x-a}$
What should I do from here? Should I attempt to multiply by $\frac{x^3sin \mid x \mid + a^3sin \mid a \mid}{x^3sin \mid x \mid + a^3sin \mid a \mid}$? Or should I use another method about doing this?
Also. when tackling these sorts of questions, what will I know what I am supposed to do? When seeing online examples, most solutions will simply multiply by the conjugate just like above. Are most examples like this? Thanks
 A: Your method is unnecessarily complicated. Assuming I read the question correctly, you really just want the derivative at $x=0$, not a general expression for the derivative. 
So let $$f'(0)=\lim\limits_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim\limits_{x\to 0}\frac{f(x)}{x}=\lim\limits_{x\to 0}\frac{x^3\sin(|x|)}{x}=\lim\limits_{x\to 0}x^2\sin(|x|)=0$$
Thus, from the definition of the derivative, we have shown that $f'(0)=0$. 
Note that the final limit above was calculated through direct substitution, because the limit of a continuous function at a point is just the function's value at that point. Because $x^2$ and $\sin(|x|)$ are both continuous functions, their product is also continuous. This is because the product of any two continuous functions is itself a continuous function. 
Also, if you're wondering why I originally called your method "unnecessarily complicated": That's because I believe the derivative of your function is actually a somewhat unwieldy piece-wise function. Much easier in this case just to go straight to calculating $f'(0)$. 
