Finding the Derivative of $\sqrt{x}$ How can I find the derivative of $\sqrt{x}$ using first principle. Specifically I'm having difficulty expanding $\sqrt{x + h}$ or rather $(x + h)^.5$. 

Is there any generalized formula for the expansion of non integer exponents less than 1?
You can use the following
$$\begin{aligned}
\frac{\sqrt{x+h}-\sqrt{x}}{h}&amp; =\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\&amp; =\frac{1}{\sqrt{x+h}+\sqrt x}.
\end{aligned}$$
Notice, $$\frac{d\sqrt x}{dx}=\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$$
$$=\lim_{h\to 0}\frac{\sqrt x\left(1+\frac{h}{x}\right)^{1/2}-\sqrt{x}}{h}$$
Since, $h\to 0 \iff \frac{h}{x}\to 0$ hence using binomial expansion &amp; neglecting higher power terms, we get 
$$=\lim_{h\to 0}\frac{\sqrt x\left(1+\frac{1}{2}\frac{h}{x}\right)-\sqrt{x}}{h}$$
$$=\lim_{h\to 0}\frac{\sqrt x\left(1+\frac{h}{2x}-1\right)}{h}$$
$$=\lim_{h\to 0}\frac{\sqrt x\left(\frac{h}{2x}\right)}{h}$$
$$=\lim_{h\to 0}\frac{\left(\frac{h}{2\sqrt x}\right)}{h}$$
$$=\lim_{h\to 0}\frac{1}{2\sqrt x}=\frac{1}{2\sqrt x}$$