Finding derivative of squrtx

Question

Why does $\frac{1}{\sqrt{z} +\sqrt x }=\frac{1}{2\sqrt{x}}$?

Can anyone explain all steps in layman's terms for limit as $z$ approaches $x$ of $\frac{f(z)-f(x)}{z-x}$ when $z= (x+h)$ and $h= (z-x)$.

Answer 1

\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ &=\lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \\ &= \lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \times \frac{\sqrt{x+h}+x}{\sqrt{x+h}+x} \\ &= \lim_{h \to 0} \frac{(\sqrt{x+h})^{2}-(\sqrt{x})^{2}}{h} \times \frac{1}{\sqrt{x+h}+x} \\ &= \frac{1}{2\sqrt{x}} \end{align*}