How to differentiate $(x+1)^{3/2}$ using the limit definition of the derivative? I want to use the definition of the derivative to find $f'$ of $f(x)=(x+1)^{3/2}$. 

I solved it using the chain rule. Would like to try to solve it using the definition of a derivative: 

$$
\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}.
$$

Don't really know how I should start it. Should I make it $\sqrt{(x+1)^3}$? Or keep the exponent as $3/2$? Other guidance would be appreciated.
$$\lim_{h\rightarrow0}\frac{\sqrt{(x+h+1)^3}-\sqrt{(x+1)^3}}{h}=\lim_{h\rightarrow0}\frac{(x+h+1)^3-(x+1)^3}{h\left(\sqrt{(x+h+1)^3}+\sqrt{(x+1)^3}\right)}=$$
$$=\lim_{h\rightarrow0}\frac{3h(x+1)^2+3h^2(x+1)+h^3}{h\left(\sqrt{(x+h+1)^3}+\sqrt{(x+1)^3}\right)}=\frac{3(x+1)^2}{\sqrt{(x+1)^3}+\sqrt{(x+1)^3}}=\frac{3}{2}\sqrt{x+1}.$$
You can also use the product rule:
$$\left[(x+1)^{3/2}\right]'=\left[(x+1)\sqrt{x+1}\right]'=$$
$$\sqrt{x+1}\cdot\lim_{h\rightarrow 0} \frac{(x+h+1)-(x+1)}{h}+(x+1)\cdot\lim_{h\rightarrow 0} \frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}=$$
$$\sqrt{x+1}+(x+1)\cdot \lim_{h\rightarrow 0}\frac{h}{h(\sqrt{x+h+1}+\sqrt{x+1})}=$$
$$\sqrt{x+1}+\frac{\sqrt{x+1}}{2}=\frac{3}{2}\sqrt{x+1}.$$