<s>[INST] How to find the derivative of $\sqrt{x+2} -x$ using limit definition? So the function is $f(x) = \sqrt{x+2} -x$, and I keep hitting dead ends trying to solve it using the definition of derivative. If anyone can help, it would be greatly appreciated!
 [/INST] $\require{cancel}$Note that
$$f(x)=\sqrt{x+2}-x$$
$$f(x+h)=\sqrt{x+h+2}-x-h$$
$$f(x+h)-f(x)=\sqrt{x+h+2}-\sqrt{x+2}-h$$
So we have
$$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x+h+2}-\sqrt{x+2}-h}{h}$$
$$= \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}-1$$
$$= \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}\cdot \frac{\sqrt{x+h+2} + \sqrt{x+2}} {\sqrt{x+h+2}+\sqrt{x+2}} - 1$$
$$=\frac{(\cancel{x}+h+\cancel{2})-(\cancel{x}+\cancel{2})}{h(\sqrt{x+h+2} + \sqrt{x+2})} - 1$$
$$=\frac{\cancel{h}}{\cancel{h}(\sqrt{x+h+2} + \sqrt{x+2})} - 1$$
$$= \frac{1}{\sqrt{x+h+2} + \sqrt{x+2}} - 1$$
Let $h\to 0$ to get
$$\boxed{f’(x) = \frac{1}{2\sqrt{x+2}}-1}$$
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