Using the definition of the derivative prove that if $f(x)=x^\frac{4}{3}$ then $f'(x)=\frac{4x^\frac{1}{3}}{3}$ So I have that $f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ 
and know that applying  
$f(x)=x^\frac{4}{3}=f\frac{(x+h)^\frac{4}{3} -x^\frac{4}{3}}{h}$
but am at a loss when trying to expand 
$(x+h)^\frac{4}{3}$

Let
$$D = \frac{(x+h)^{4/3}-x^{4/3}}{h}$$
Then
$$\begin{align}D^3 &= \frac{(x+h)^4-3 (x+h)^{8/3} x^{4/3} + 3 (x+h)^{4/3} x^{8/3} - x^4}{h^3}\\ &= \frac{4 x^3 h + 6 x^2 h^2 + 4 x h^3 + h^4 -3 (x+h)^{4/3} x^{4/3} \left [ (x+h)^{4/3}-x^{4/3}\right ]}{h^3}\\ &= \frac{4 x^3 + 6 x^2 h + 4 x h^2 +h^3}{h^2} - \frac{3 (x+h)^{4/3} x^{4/3}}{h^2} D\end{align}$$
Thus,
$$h^2 D^3 + 3 (x+h)^{4/3} x^{4/3} D = 4 x^3 + 6 x^2 h + 4 x h^2 +h^3$$
Taking the limit as $h \to 0$, (and assuming that $\bar{D}=\lim_{h \to 0} D$ remains finite as we do so), we see that
$$3 x^{8/3} \bar{D} = 4 x^3$$
or $\bar{D} = (4/3) x^{1/3}$ as was to be demonstrated.
