Derivative and limit of $\sqrt[3]{x-\sin x}$ Given $f:R\rightarrow R, f(x) = \sqrt[3]{x-\sin x}$, compute $f'(0)$.Now, this can be done by using the definition of the derivative : 
$$f'(x_0) = \lim_{x\to{x_0}} \frac{f(x)-f(x_0)}{x-x_0}$$This yields the right answer.However, why can't we derive the function and then plug in 0? I mean:
$$f'(x) = \frac{1-\cos x}{3\sqrt[3]{(x-\sin x)^2}}$$
And this is undefined for $x = 0$. Why does this happen?
 A: You can use the differentiation rules so long as they make sense; thus
$$
f'(x)=\frac{1-\cos x}{3\sqrt[3]{(x-\sin x)^2}}
$$
for $x\ne0$, because $x\ne \sin x$ for $x\ne0$.
The derivative at $0$ may still exist even if the above expression is undefined at $0$. You need to use other methods, though.
However, if the function is continuous at $0$ and the limit of the derivative exists, then l’Hôpital’s theorem tells us that the function is differentiable at $0$ and the derivative is the limit.
Now
$$
\lim_{x\to0}f'(x)=
\lim_{x\to0}\frac{1-\cos x}{3\sqrt[3]{(x-\sin x)^2}}=
\lim_{x\to0}\frac{1}{3}\sqrt[3]{\frac{(1-\cos x)^3}{(x-\sin x)^2}}
$$
and you should be able to compute it.
Note this is a sufficient condition; the limit of the derivative may not exist and the function still be differentiable. In this case you have to resort to the definition of the derivative as limit.
A: Stuff happens in math for no particular reason. This is why we have alternatives. For example, division by zero happens for no reason, but for this we have many options to handle (factoring, rationalizing). Similarly you can't "plug in" cause it would result in a division of 0, but you have another option for a reason, to avoid cases like these.
