How to finish the derivative calculation? I know that this can be calculated as:
$x^n=n \cdot x^{n-1}$ but I need to find a solution as a limit:
$f(x)=\sqrt[3]{x} ;f^{'}(x)=\lim _{\Delta x\to 
0}\frac{\sqrt[3]{x_0+\Delta x} -\sqrt[3]{x_0}}{\Delta x}=\lim 
_{\Delta x\to 0}\frac{(\sqrt[3]{x_0+\Delta x} -\sqrt[3]{x_0} 
)(\sqrt[3]{x_0+\Delta x} +\sqrt[3]{x_0} )}{\Delta 
x(\sqrt[3]{x_0+\Delta x} +\sqrt[3]{x_0} )}=\lim _{\Delta x\to 
0}\frac{(\sqrt[3]{x_0+\Delta x})^{2}-(\sqrt[3]{x_0})^{2}}{\Delta 
x(\sqrt[3]{x_0+\Delta x} +\sqrt[3]{x_0} )}=\lim _{\Delta x\to 
0}\frac{(\sqrt[3]{x_0+\Delta x})^{2}-(\sqrt[3]{x_0})^{2}}{\Delta 
x(\sqrt[3]{x_0+\Delta x} +\sqrt[3]{x_0} )}=...$
I don't know how to calculate further due to: $(\sqrt[3]{x_{0}+\Delta x})^{2}-(\sqrt[3]{x_{0}})^{2}$.
If instead it would be: $(x_{0}+\Delta x)^{2}-(x_{0})^{2}$, then I would write $(x_{0}+\Delta x)^{2}$ as: $(x_{0})^{2}+2 \cdot x_{0} \cdot \Delta x +(\Delta x)^{2}$, but i don't know how to be with cube root ?
 A: Use that $A^3-B^3=(A-B)(A^2+AB+B^2)$. So
$$\begin{align}
&\lim_{\Delta x\to 0}\frac{\sqrt[3]{x_0+\Delta x}-\sqrt[3]{x_0}}{\Delta x} \\
&= \lim_{\Delta x\to 0}\frac{\left(\sqrt[3]{x_0+\Delta x}-\sqrt[3]{x_0}\right)\left(\sqrt[3]{\left(x_0+\Delta x\right)^2}+\sqrt[3]{\left(x_0+\Delta x\right)x_0}+\sqrt[3]{x_0^2}\right)}{\Delta x\left(\sqrt[3]{\left(x_0+\Delta x\right)^2}+\sqrt[3]{\left(x_0+\Delta x\right)x_0}+\sqrt[3]{x_0^2}\right)} \\
&=\lim_{\Delta x\to 0}\frac{x_0+\Delta x-x_0}{\Delta x\left(\sqrt[3]{\left(x_0+\Delta x\right)^2}+\sqrt[3]{\left(x_0+\Delta x\right)x_0}+\sqrt[3]{x_0^2}\right)} \\
&=\lim_{\Delta x\to 0}\frac{1}{\sqrt[3]{\left(x_0+\Delta x\right)^2}+\sqrt[3]{\left(x_0+\Delta x\right)x_0}+\sqrt[3]{x_0^2}}
\end{align}$$
Note that the limit can be computed only for $x_0\ne 0$.
A: You are considering $$y=(\sqrt[3]{x_{0}+\Delta x})^{2}-(\sqrt[3]{x_{0}})^{2}=(x_0+\Delta x)^{\frac 23}-x_0^{\frac 23}$$ Using the binomial theorem or Taylor expansion around $\Delta x=0$, you should have
$$(\sqrt[3]{x_{0}+\Delta x})^{2}=x_0^{2/3}+\frac{2 \Delta x }{3 x_0^{1/3}}-\frac{\Delta x ^2}{9 x_0^{4/3}}+O\left(\Delta x^3\right)$$ making
$$y=\frac{2 \Delta x }{3 x_0^{1/3}}-\frac{\Delta x ^2}{9 x_0^{4/3}}+O\left(\Delta x^3\right)=\frac{2 \Delta x }{3 x_0^{1/3}}+O\left(\Delta x^2\right)$$
