Find derivative of $f(x)=\frac{1}{\sqrt{x+2}}$ by definition

Question

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.

Answer 1

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

Answer 2

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.