Finding the Derivative of $\sqrt{x}$

Question

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?

Answer 1

You can use the following $$\begin{aligned} \frac{\sqrt{x+h}-\sqrt{x}}{h}& =\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\& =\frac{1}{\sqrt{x+h}+\sqrt x}. \end{aligned}$$

Answer 2

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 & neglecting higher power terms, we get $$=\lim_{h\to 0}\frac{\sqrt x\left(1+\frac{1}{2}\frac{h}{x}\right)-\sqrt{x}}{h}$$ $$=\lim_{h\to 0}\frac{\sqrt x\left(1+\frac{h}{2x}-1\right)}{h}$$ $$=\lim_{h\to 0}\frac{\sqrt x\left(\frac{h}{2x}\right)}{h}$$ $$=\lim_{h\to 0}\frac{\left(\frac{h}{2\sqrt x}\right)}{h}$$ $$=\lim_{h\to 0}\frac{1}{2\sqrt x}=\frac{1}{2\sqrt x}$$