# 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

• Your definition is incorrect. There should be a limit as $x \to a$ involved, and the derivative should be $f'(a)$. – Tob Ernack Nov 2 at 2:43
• Oh yes, I forgot to add the limits – Jisbon Nov 2 at 3:11

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 $$f'(0)$$.