Find derivative of $f(x)=\frac{1}{\sqrt{x+2}}$ by definition Use the definition of a derivative to find the derivative of:
$$f(x)=\frac{1}{\sqrt{x+2}}+2x$$
my work:
$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
$$\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h+2}}+2(x+h)-\frac{1}{\sqrt{x+2}}-2x}{h}$$
$$\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h+2}}+2h-\frac{1}{\sqrt{x+2}}}{h}$$
$$\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}}{h}+2$$
$$\lim_{h\to 0}\frac{1}{h}\frac{\sqrt{x+2}-\sqrt{x+h+2}}{\sqrt{x+h+2}\sqrt{x+2}}+2$$
I don't know what to do from here.
 A: Hint: Note that 
$$\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h+2}}+2h-\frac{1}{\sqrt{x+2}}}{h}=\lim_{h\to 0}\frac{1}{h}\left[\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}\right]+\lim_{h\to0}\frac{2h}{h}=\lim_{h\to 0}\frac{1}{h}\left[\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}\right]+2$$
Now look that:
$$\lim_{h\to 0}\frac{1}{h}\left[\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}\right]=\lim_{h\to 0}\frac{1}{h}\left[\frac{\sqrt{x+2}-\sqrt{x+h+2}}{\sqrt{x+h+2}\sqrt{x+2}}\right]$$
then now multiply numerator  and denominator  by $$\sqrt{x+2}+\sqrt{x+h+2}$$
then
$$\lim_{h\to 0}\frac{1}{h}\left[\frac{\sqrt{x+2}-\sqrt{x+h+2}}{\sqrt{x+h+2}\sqrt{x+2}}\right]=\lim_{h\to 0}\frac{1}{h}\left[\frac{(\sqrt{x+2}-\sqrt{x+h+2})(\sqrt{x+2}+\sqrt{x+h+2})}{\sqrt{x+h+2}\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+h+2})}\right]=\lim_{h\to 0}\frac{1}{h}\left[\frac{(\sqrt{x+2})^2-(\sqrt{x+h+2})^2}{\sqrt{x+h+2}\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+h+2})}\right]=\lim_{h\to 0}\frac{1}{h}\left[\frac{x+2-(x+h+2)}{\sqrt{x+h+2}\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+h+2})}\right]=\lim_{h\to 0}\frac{1}{h}\left[\frac{-h}{\sqrt{x+h+2}\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+h+2})}\right]=\lim_{h\to 0}\left[\frac{-1}{\sqrt{x+h+2}\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+h+2})}\right]=\frac{-1}{2(x+2)\sqrt{x+2}}$$
then $$f'(x)=\frac{-1}{2(x+2)\sqrt{x+2}}+2$$
A: Hint:  Apply the formula
$$a-b={a^2-b^2\over a+b}$$
to the expression $\sqrt{x+2}-\sqrt{x+h+2}$ in the numerator of your last line.
