Finding derivative of $\frac{1}{\sqrt{x+2}}$ using only the definition of the derivative $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Question

I need to find $\frac{d}{dx}(\frac{1}{\sqrt{x+2}})$ only using the basic definition of the derivative $f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$, but I'm not having any luck with the algebra. Any hints?

Answer 1

$$\frac{d}{dx}\ \frac{1}{\sqrt{x+2}}=\lim_{h\to0}\frac{\left(\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}\right)}{h}$$

$$=\lim_{h\to0}\frac{\left(\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}\right)}{h}\times\frac{\frac{1}{\sqrt{x+h+2}}+\frac{1}{\sqrt{x+2}}}{\frac{1}{\sqrt{x+h+2}}+\frac{1}{\sqrt{x+2}}}$$

$$=\lim_{h\to0}\frac{\left(\frac{1}{x+h+2}-\frac{1}{x+2}\right)}{h\left(\frac{1}{\sqrt{x+h+2}}+\frac{1}{\sqrt{x+2}}\right)}$$

$$=\lim_{h\to0}\frac{\left(\frac{-h}{(x+h+2)(x+2)}\right)}{h\left(\frac{1}{\sqrt{x+h+2}}+\frac{1}{\sqrt{x+2}}\right)}$$

$$=\lim_{h\to0}\frac{\left(\frac{-1}{(x+h+2)(x+2)}\right)}{\left(\frac{1}{\sqrt{x+h+2}}+\frac{1}{\sqrt{x+2}}\right)}$$

$$=\frac{\frac{-1}{(x+2)^2}}{\frac{2}{\sqrt{x+2}}}$$

$$=\frac{-1}{2(x+2)^{\frac{3}{2}}}$$

Answer 2

By elementary algebra (try it!) you can show that $$\frac{f(x+h)-f(x)}h=-\frac1{\sqrt{x+h+2}\,\sqrt{x+2}\,\,(\sqrt{x+h+2}+\sqrt{x+2})}$$ and it is then easy to find the limit as $h\to0$.