Find first derivative of a function $f(x) = x\sqrt[3]{x}$ using definition I'm having trouble finding first derivative of a function:
$f(x) = x\sqrt[3]{x}$
using definition:
$\displaystyle \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
I should get:
$\displaystyle \lim_{h \to 0} \frac{(x+h)(\sqrt[3]{x+h})-x\sqrt[3]{x}}{h} $
But I don't know what my next step should be.
Could somebody give me a hint please?
 A: Hint: $\dfrac{(x+h)\sqrt[3]{x+h} - x\sqrt[3]{x}}{h}= x\cdot\dfrac{\sqrt[3]{x+h} - \sqrt[3]{x}}{h}+\sqrt[3]{x+h}$. From this you can simplify the factor next to $x$ of the first term using the identity: $a - b = \dfrac{a^3 - b^3}{a^2+ab+b^2}$
A: If you multiply both the numerator and the denominator of that fraction by$$(x+h)^2\sqrt[3]{x+h}^2+(x+h)\sqrt{x+h}\,x\sqrt[3]x+x^2\sqrt[3]x^2,$$then the limit becomes\begin{multline}\lim_{h\to0}\frac{(x+h)^3(x+h)-x^4}{h\bigl((x+h)^2\sqrt[3]{x+h}^2+(x+h)\sqrt{x+h}\,x\sqrt[3]x+x^2\sqrt[3]x^2\bigr)}=\\=\lim_{h\to0}\frac{6hx^2+4h^2x+h^3+4x^3}{(x+h)^2\sqrt[3]{x+h}^2+(x+h)\sqrt{x+h}\,x\sqrt[3]x+x^2\sqrt[3]x^2}.\end{multline}Can you take it from here?
A: You seemed to have trapped yourself by writing $f(x)=x^{4/3}$ in the form of a product that makes the task less transparent. Doing so leads more naturally into the need for the basic technique of the product rule of "plus then minus the cross term".
Namely, take $F(x) = x$ and $G(x) = \sqrt[3]{x}$ such that $f(x)=F(x)G(x)$.
$$\begin{align}
&F(x+h)G(x+h)-F(x)G(x) \\
&= F(x+h)G(x+h) ~\mathbf{{}- F(x+h)G(x) + F(x+h)G(x) }~- F(x)G(x) \\
&= F(x+h)\bigl( G(x+h)-G(x) \bigr) - \bigl(F(x+h) - F(x) \bigr)G(x)
\end{align}$$
Then you have two separate simpler derivatives that becomes the product rule $(FG)'=FG'+F'G$.
