<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] This is a little bit not fun, but will give your algebra a good workout. First note that 
$$(x+h)^{4/3}-x^{4/3}=\left((x+h)^{2/3}+x^{2/3}\right)\left((x+h)^{2/3}-x^{2/3}\right)
.\tag{1}$$
and the right-hand side of (1) can be written as
$$\left((x+h)^{2/3}+x^{2/3}\right)\left((x+h)^{1/3}+x^{1/3}\right)\left((x+h)^{1/3}-x^{1/3}\right)
.\tag{2}$$ 
We have used the identity $a^2-b^2=(a+b)(a-b)$ a couple of times.
So we want to find
$$\lim_{h\to 0}\frac{\left((x+h)^{2/3}+x^{2/3}\right)\left((x+h)^{1/3}+x^{1/3}\right)\left((x+h)^{1/3}-x^{1/3}\right)}{h}.\tag{3}$$
The first two terms in the numerator of (3) behave very nicely as $h\to 0$. We need to worry only about
$$\frac{(x+h)^{1/3}-x^1/3}{h}.\tag{4}$$
Now we use the identity $(a-b)(a^2+ab+b^2)=a^3-b^3$. Letting $a=(x+h)^{1/3}$ and $b=x^{1/3}$ we multiply top and bottom of (4) by $a^2+ab+b^2$. We get
$$\frac{(x+h)-x}{h\left((x+h)^{2/3}+(x+h)^{1/3}x^{1/3}+x^{2/3}\right)}.$$
Now everything is under control, you can put the pieces together.
Remark: There are many other ways to do the algebra. For example instead of the initial factoring that we did, you could immediately multiply top and bottom by 
$$(x+h)^{8/3}+(x+h)^{4/3}x^{4/3}+x^{8/3}.$$ 
Then on top we end up with $(x+h)^4-x^4$. We can expand this as usual and get a simpler-looking path to the answer. 
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