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!
 A: $\require{cancel}$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
$$\boxed{f’(x) = \frac{1}{2\sqrt{x+2}}-1}$$
A: Let’s look at
$$\begin{align}
\Delta(x,h)&={\sqrt{x+h+2}-\sqrt{x+2}-h\over h}\\
&={\sqrt{x+h+2}-\sqrt{x+2}\over h}-{h\over h}\\
&={h\over h\left(\sqrt{x+h+2}+\sqrt{x+2}\right)}-1
\end{align}$$
Where the last equality is the result of multiplying numerator and denominator by the conjugate of $\sqrt{x+h+2}-\sqrt{x+2}$.
And we can now get the limit
$$\lim_{h\to 0}\Delta(x,h)={1\over 2\sqrt{x+2}}-1$$
