Calculus: Early Transcendentals 8th Edition

$$f'(x)=\frac{3}{2}x^{1/2}$$ The domain of $f(x)$ and $f'(x)$ is $[0,\infty)$
$$f(x)=x^{3/2}$$ According to definition, $$f'(x)=\lim\limits_{h\to0}\frac{(x+h)^{3/2}-x^{3/2}}{h}$$ $$f'(x)=\lim\limits_{h\to0}\frac{\sqrt{(x+h)^3}-\sqrt {x^3}}{h}$$ Multiply both numerator and denominator by $\sqrt{(x+h)^3}+\sqrt {x^3}$, the numerator would become $$(\sqrt{(x+h)^3}-\sqrt {x^3})(\sqrt{(x+h)^3}+\sqrt {x^3})$$$$=(x+h)^3-x^3$$$$=(x+h-x)[(x+h)^2+x^2+x(x+h)]$$$$=h[x^2+h^2+2xh+x^2+x^2+xh]$$$$=h(3x^2+h^2+3xh)$$ Therefore, $$f'(x)=\lim\limits_{h\to0}\frac{h(3x^2+h^2+3xh)}{h(\sqrt{(x+h)^3}+\sqrt {x^3})}$$ $$f'(x)=\lim\limits_{h\to0}\frac{3x^2+h^2+3xh}{\sqrt{(x+h)^3}+\sqrt {x^3}}$$ $$f'(x)=\frac{3x^2+0^2+3x\times0}{\sqrt{(x+0)^3}+\sqrt{x^3}}$$ $$f'(x)=\frac{3x^2}{2\sqrt{x^3}}=\frac{3x^2}{2x^{3/2}}$$ $$f'(x)=\frac{3}{2}x^{1/2}$$ *Domain of $f(x)$ $f(x)=x^{3/2}=\sqrt{x^3}$. Therefore, $f(x)$ is defined for $R$ except where $x^3\lt0$, or $x\lt0$ So, the domain of $f(x)$ is $[0,\infty)$ *Domain of $f'(x)$ $f'(x)=\frac{3}{2}x^{1/2}=\frac{3}{2}\sqrt{x}$. Therefore, $f'(x)$ is defined for $R$ except where $x\lt0$ So, the domain of $f'(x)$ is $[0,\infty)$