Find $f'(x)$ where $f(x) = x^\frac{3}{2}$ using $h \to 0$ method $$f(x) = x^{\frac{3}{2}}$$
$$f'(x) = ?$$
How would you solve this using the $h\to 0$ method?
Here's the initial setup.
$$f'(x) = \lim_{h \to 0}\frac{(x+h)^\frac{3}{2}-x^{3/2}}{h}$$
I tried difference of cubes but it didn't get me anywhere.

${(\sqrt {x + h} )^3} - {\left( {\sqrt x } \right)^3} = \left( {\sqrt {x + h}  - \sqrt x } \right)\left( {\sqrt {{{\left( {x + h} \right)}^2}}  + \sqrt {\left( {x + h} \right)x}  + \sqrt {{x^2}} } \right)$ and $h = x + h - x = \left( {\sqrt {x + h}  - \sqrt x } \right)\left( {\sqrt {x + h}  + \sqrt x } \right)$ so
$$\mathop {\lim }\limits_{h \to 0} \frac{{{{(\sqrt {x + h} )}^3} - {{\left( {\sqrt x } \right)}^3}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\sqrt {{{\left( {x + h} \right)}^2}}  + \sqrt {\left( {x + h} \right)x}  + \sqrt {{x^2}} } \right)}}{{\left( {\sqrt {x + h}  + \sqrt x } \right)}} = \frac{3}{2}\sqrt x $$
