Whats the derivative of $\sqrt{4+|x|}$ using first principle Here is my attempt:
$$f(x)=\sqrt{4+|x|}$$
$$f`(x) = \lim_{h\to0} \frac{\sqrt{4+|x-h|}-\sqrt{4+|x|}}{h}$$
multiplying by the conjugate:
$$\lim_{h\to0} \frac{4+|x-h|-4-|x|}{h[\sqrt{4+|x-h|}+\sqrt{4+|x|}]}$$
$$\lim_{h\to0} \frac{|x-h|-|x|}{h[\sqrt{4+|x-h|}+\sqrt{4+|x|}]}$$
What is the next step? how can I deal with absloute value of |x|?
 A: The graph of the function looks like this: it isn't differentiable at x = 0.
As noted in the comments, Split the domain of the function into 
x>0 $f(x)=\sqrt{4+x}$ and 
x≤0 f(x)= $\sqrt{4-x}$. 
Both halves are easily differentiable, but have different values at x = 0 (or to be more precise, the limiting value for x > 0 differs from the value for x = 0).

(P.S - this is quite an interesting web site: http://fooplot.com/
A: First note that if $f(x)=\sqrt{4+|x|}$, then
$$ \frac{d}{dx}f(0) = \lim_{h\to0} \frac{\sqrt{4+|0+h|}-\sqrt{4+|0|}}{h}= \lim_{h\to0} \frac{\sqrt{4+|h|}-\sqrt{4}}{h}
$$
$$= \lim_{h\to0} \frac{\left(\sqrt{4+|h|}-\sqrt{4}\right)\left(\sqrt{4+|h|}+\sqrt{4}\right)}{h\left(\sqrt{4+|h|}+\sqrt{4}\right)}
$$
$$= \lim_{h\to0} \frac{4+|h|-4}{h\left(\sqrt{4+|h|}+2\right)}= \lim_{h\to0} \frac{|h|}{h\left(\sqrt{4+|h|}+2\right)} $$
From the left of zero, we have
$$
\lim_{h\to0^-} \frac{|h|}{h\left(\sqrt{4+|h|}+2\right)}= \lim_{h\to0^-} \frac{-h}{h\left(\sqrt{4-h}+2\right)}
$$
$$= \lim_{h\to0^-} \frac{-1}{\left(\sqrt{4-h}+2\right)}=-\frac{1}{\left(\sqrt{4}+2\right)}=-\frac14
$$
From the right of zero, we have
$$
\lim_{h\to0^+} \frac{|h|}{h\left(\sqrt{4+|h|}+2\right)}= \lim_{h\to0^+} \frac{h}{h\left(\sqrt{4+h}+2\right)}
$$
$$= \lim_{h\to0^+} \frac{1}{\left(\sqrt{4+h}+2\right)}=\frac{1}{\left(\sqrt{4}+2\right)}=\frac14
$$
Therefore, $f(x)$ is not differentiable at $x=0$.
