i have a few questions regarding differentiation using the first principle

Question

Differentiate the following using the first principle

$$f(x)=x^\frac{3}{2}$$

$$f(x)=\frac{1}{\sqrt x}$$

i have tried the questions but i am stuck in the second step

Answer 1

To apply first principles to fractional powers of $x$ the process involves rationalizing the numerator to deal with the roots.

For $f(x)=x^\frac{3}{2}$ do:

$$f'(x)=\lim_{h\to0}\frac{(x+h)^\frac{3}{2}-x^\frac{3}{2}}{h}$$ $$=\lim_{h\to0}\frac{(x+h)^\frac{3}{2}-x^\frac{3}{2}}{h}\times\frac{(x+h)^\frac{3}{2}+x^\frac{3}{2}}{(x+h)^\frac{3}{2}+x^\frac{3}{2}}$$ $$=\lim_{h\to0}\frac{(x+h)^3-x^3}{h\times\left((x+h)^\frac{3}{2}+x^\frac{3}{2}\right)}$$ $$=\lim_{h\to0}\frac{x^3+3x^2h+3xh^2+h^3-x^3}{h\times\left((x+h)^\frac{3}{2}+x^\frac{3}{2}\right)}$$ $$=\lim_{h\to0}\frac{3x^2h+3xh^2+h^3}{h\times\left((x+h)^\frac{3}{2}+x^\frac{3}{2}\right)}$$ $$=\lim_{h\to0}\frac{3x^2+3xh+h^2}{(x+h)^\frac{3}{2}+x^\frac{3}{2}}$$ $$=\frac{3x^2}{2x^\frac{3}{2}}$$ $$=\frac{3}{2}x^\frac{1}{2}$$

For $f(x)=x^{-\frac{1}{2}}$ its basically the same:

$$f'(x)=\lim_{h\to0}\frac{(x+h)^{-\frac{1}{2}}-x^{-\frac{1}{2}}}{h}$$ $$=\lim_{h\to0}\frac{(x+h)^{-\frac{1}{2}}-x^{-\frac{1}{2}}}{h}\times\frac{(x+h)^{-\frac{1}{2}}+x^{-\frac{1}{2}}}{(x+h)^{-\frac{1}{2}}+x^{-\frac{1}{2}}}$$ $$=\lim_{h\to0}\frac{(x+h)^{-1}-x^{-1}}{h\times\left((x+h)^{-\frac{1}{2}}+x^{-\frac{1}{2}}\right)}$$ $$=\lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h\times\left((x+h)^{-\frac{1}{2}}+x^{-\frac{1}{2}}\right)}$$ $$=\lim_{h\to0}\frac{\frac{-h}{x(x+h)}}{h\times\left((x+h)^{-\frac{1}{2}}+x^{-\frac{1}{2}}\right)}$$ $$=\lim_{h\to0}\frac{\frac{-1}{x(x+h)}}{(x+h)^{-\frac{1}{2}}+x^{-\frac{1}{2}}}$$ $$=\frac{\frac{-1}{x^2}}{2x^{-\frac{1}{2}}}$$ $$=-\frac{1}{2}x^{-\frac{3}{2}}$$