Answer
$$f'(x)=\frac{3}{2} x^{1 / 2}$$
Work Step by Step
Since
$$ f^{\prime}(x) =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$
Then
\begin{aligned}
f^{\prime}(x) &=\lim _{h \rightarrow 0} \frac{(x+h)^{3 / 2}-x^{3 / 2}}{h}\\
&=\lim _{h \rightarrow 0} \frac{\sqrt{(x+h)^{3}}-\sqrt{x^{3}}}{h} \frac{\sqrt{(x+h)^{3}}+\sqrt{x^{3}}}{\sqrt{(x+h)^{3}}+\sqrt{x^{3}}} \\
&=\lim _{h \rightarrow 0} \frac{(x+h)^{3}-x^{3}}{h}\left(\frac{1}{\sqrt{(x+h)^{3}}+\sqrt{x^{3}}}\right)\\
&= \lim _{h \rightarrow 0} \frac{(x+h)^{3}-x^{3}}{h} \lim _{h \rightarrow 0}\left(\frac{1}{\sqrt{(x+h)^{3}}+\sqrt{x^{3}}}\right)\\
&= \lim _{h \rightarrow 0} \frac{3x^2h+3xh^2+h^3}{h} \lim _{h \rightarrow 0}\left(\frac{1}{\sqrt{(x+h)^{3}}+\sqrt{x^{3}}}\right)\\
&= (3x^2)\frac{1}{2 \sqrt{x^{3}}}\\
&=\frac{3}{2} x^{1 / 2}
\end{aligned}
