Derivative of function = indeterminate form How can I calculate the derivative in $x=0$ of $$f(x)=(x-\sin x)^{1/3}?$$ 
I found the derivative and replaced $x$ but it would be an indeterminate form and if I try using limit of $(f(x)-f(0))/x$, it doesn't lead me anywhere so how could I find it?
 A: The formula for the derivative doesn't make sense at $0$. Indeed, for $x\ne0$,
$$
f'(x)=\frac{1}{3}(x-\sin x)^{-2/3}(1-\cos x)
$$
but the expression is undefined for $x=0$. The reason is that $x\mapsto x^{1/3}$ is not differentiable at $0$; however, this doesn't mean the function you're given isn't differentiable at $0$.
You have two ways out of the dilemma. One is to compute
$$
f'(0)=\lim_{x\to0}f'(x)
$$
(this is possible by l'Hôpital), the other is to use the definition:
$$
f'(0)=\lim_{x\to0}\frac{(x-\sin x)^{1/3}}{x}=
\lim_{x\to0}\left(\frac{x-\sin x}{x^3}\right)^{\!1/3}
$$
Can you find
$$
\lim_{x\to0}\frac{x-\sin x}{x^3}
$$
with Taylor expansion or l'Hôpital?
A: $$\left(\frac{f(x)-f(0)}x\right)^3=\frac{x-\sin x}{x^3}.$$
By L'Hospital, twice,
$$\frac{1-\cos x}{3x^2}$$ then $$\frac{\sin x}{6x}.$$
Can you conclude ?

Without L'Hospital, write
$$L=\lim_{x\to0}\frac{(x-\sin x)^{1/3}}x=\lim_{x\to0}\frac{(3x-\sin 3x)^{1/3}}{3x}=\lim_{x\to0}\left(\frac{3x-3\sin x}{27x^3}+\frac{4\sin^3x}{27x^3}\right)^{1/3}=\left(\frac{L^3}9+\frac4{27}\right)^{1/3}.$$
You can draw $L$.
A: HINT: the first derivative is given by $$f'(x)=1/3\,{\frac {1-\cos \left( x \right) }{ \left( x-\sin \left( x
 \right)  \right) ^{2/3}}}
$$ and the limit for $x$ tends to zero exists and is equal to $$\frac{1}{6^{1/3}}$$
A: Use Taylor:
$$f'(0)=\lim_{x\to 0}\frac{(x-\sin x)^{1/3}-0}x.$$
In which 
\begin{align}
\sin x &=x-\frac16x^3+O(x^6)\\
(x-\sin x)^{1/3}&=(\frac16x^3+O(x^6))^{1/3}=(\frac16)^{1/3}x(1+O(x^3))^{1/3}
\end{align}
Can you take it from here?
