Answer
$f'(x)=-\dfrac{2}{x\sqrt{x}}$
Work Step by Step
$f(x)=\dfrac{4}{\sqrt{x}}$
The derivative of $f$ at $x$ is $f'(x)=\lim_{\Delta x\to0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}$
Find $f(x+\Delta x)$ by substituting $x$ by $x+\Delta x$ in $f(x)$:
$f(x+\Delta x)=\dfrac{4}{\sqrt{x+\Delta x}}$
Substitute $f(x+\Delta x)$ and $f(x)$ into the formula and evaluate:
$f'(x)=\lim_{\Delta x\to0}\dfrac{\dfrac{4}{\sqrt{x+\Delta x}}-\dfrac{4}{\sqrt{x}}}{\Delta x}=...$
$...=\lim_{\Delta x\to0}\dfrac{\dfrac{4\sqrt{x}-4\sqrt{x+\Delta x}}{\sqrt{x^{2}+x\Delta x}}}{\Delta x}=...$
$...=\lim_{\Delta x\to0}\dfrac{4\sqrt{x}-4\sqrt{x+\Delta x}}{\Delta x\sqrt{x^{2}+x\Delta x}}=...$
Rationalize the numerator:
$...=\lim_{\Delta x\to0}\dfrac{4(\sqrt{x}-\sqrt{x+\Delta x})}{\Delta x\sqrt{x^{2}+x\Delta x}}\cdot\dfrac{\sqrt{x}+\sqrt{x+\Delta x}}{\sqrt{x}+\sqrt{x+\Delta x}}=...$
$...=\lim_{\Delta x\to0}\dfrac{4[(\sqrt{x})^{2}-(\sqrt{x+\Delta x})^{2}]}{\Delta x(\sqrt{x^{2}+x\Delta x})(\sqrt{x}+\sqrt{x+\Delta x})}=...$
$...=\lim_{\Delta x\to0}\dfrac{4(x-x-\Delta x)}{\Delta x(\sqrt{x^{2}+x\Delta x})(\sqrt{x}+\sqrt{x+\Delta x})}=...$
$...=\lim_{\Delta x\to0}\dfrac{-4\Delta x}{\Delta x(\sqrt{x^{2}+x\Delta x})(\sqrt{x}+\sqrt{x+\Delta x})}=...$
$...=\lim_{\Delta x\to0}\dfrac{-4}{(\sqrt{x^{2}+x\Delta x})(\sqrt{x}+\sqrt{x+\Delta x})}=...$
$...=\dfrac{-4}{[\sqrt{x^{2}+x(0)}][\sqrt{x}+\sqrt{x+0}]}=\dfrac{-4}{(\sqrt{x^{2}})(\sqrt{x}+\sqrt{x})}=...$
$...=\dfrac{-4}{(x)(2\sqrt{x})}=-\dfrac{2}{x\sqrt{x}}$
$f'(x)=-\dfrac{2}{x\sqrt{x}}$
