<s>[INST] 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 
$(x+h)^\frac{4}{3}$
 [/INST] Hint:
The expansion of $(x+h)^\frac{4}{3}$ is 
$$h\sqrt[3]{x+h} + x\sqrt[3]{x+h}$$
So then you will have
$$\lim_{x \rightarrow \infty}\frac{h\sqrt[3]{x+h} + x\sqrt[3]{x+h} - x^{\frac{4}{3}}}{h}$$
Can you take it from there?
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