Derivative and integral of the abs function I would like to ask about how to find the derivative of the absolute value function for example :
$\dfrac{d}{dx}|x-3|$
My try:$$
f(x)=|x-3|\\
f(x) =
\begin{cases}
x-3, & \text{if }x \geq3 \\
3-x, & \text{if }x \leq3
\end{cases}
$$
So:
$$f'(x) =
\begin{cases}
1, & \text{if }x \geq3 \\
-1, & \text{if }x \leq3
\end{cases}
$$
What is wrong with this approcah?.Please clarify.
Also I want also like to find out how to integrate the absolute value function.
Thanks
 A: It is almost right..But you have to consider that, at $x=3$, $f(3)=0$.
So:
$$f(x)=\left\{\begin{matrix}
x-3, x>3\\ 
0, x=3\\ 
3-x, x<3
\end{matrix}\right.$$
Therefore,
$$f'(x)=\left\{\begin{matrix}
1, x>3\\  
-1, x<3
\end{matrix}\right.$$
EDIT: The derivative does not exist at $x=3$,because:
$$\lim_{h \to 0} \frac{f(3+h)-f(3)}{h}=\lim_{h \to 0} \frac{|h|-|0|}{h}=\lim_{h \to 0} \frac{|h|}{h}$$
The last limit does not exist,because:
$$\lim_{h \to 0^-}\frac{|h|}{h}=-1 \text{ BUT } \lim_{h \to 0^+}\frac{|h|}{h}=+1 $$
The limits are different on either side,so the limit does not exist.
Therefore,the function $f(x)=|x-3|$ is not differentiable at $x=3$.
A: In your answer, the derivative isn't even not well defined since $f(3) = -1$ and $f(3) = 1$. The same note goes for the original function. The function is 
$$
f(x) =
\begin{cases}
x-3, & \text{if }x \geq3 \\
3-x, & \text{if }x \color{red}{<}3
\end{cases}
$$
There is nothing wrong with your approach. The only thing is that the function is not differentiable at $3$. So you should get
$$
f'(x) =
\begin{cases}
1, & \text{if }x  >3 \\
-1, & \text{if }x <3
\end{cases}
$$

So show that the function isn't differentiable at $3$, you consider the limit
$$
\lim_{h\to 0}\frac{f(3+h) - f(3)}{h}.
$$
Here the left hand limit is
$$
\lim_{h\to 0^-}\frac{f(3+h) - f(3)}{h} = -1
$$
while the right hand limit
$$
\lim_{h\to 0^+}\frac{f(3+h) - f(3)}{h} = 1.
$$
(You can probably work out the details.) Since these two limit do not equal
$$
\lim_{h\to 0}\frac{f(3+h) - f(3)}{h}
$$
does not exist, and so by definition the function is not differentiable at $3$.
A: 1) Differentiation:
Define the signum function 
$$\mathop{sgn}{(x)}=
\begin{cases}
-1 \quad \text{if } x<0 \\
+1 \quad \text{if } x>0 \\
0 \quad   \text{if } x=0 \\
\end{cases}$$
Claim: $$ \frac{d |x|}{dx} = \mathop{sgn}(x), x\neq 0$$
Proof: Use the definition of the absolute value function and observe the left and right limits at $x=0$.
Hence, $$
\frac{d |x-3|}{dx} = 
\begin{cases}
-1 \quad \text{if } x-3<0 \quad(x<3)\\
+1 \quad \text{if } x-3>0 \quad(x>3)
\end{cases}$$
2) Indefinite integration:
$$\int |x| \, \mathrm{d}x = \frac{x|x|}{2} + C$$
Proof:
$$
\frac{d}{dx}\left(\frac{x|x|}{2}\right)=\frac{1}{2}[ |x|+x \mathop{sgn}(x)] = \frac{1}{2}(2|x|)=|x|
$$
3)Definite integration:
Look at the interval over which you need to integrate, and if needed break the integral in two pieces - one over a negative interval, the other over the positive. For example, if $a<0, b>0$,
$$
\int_a^b |x| \, \mathrm{d}x = \int_a^0 (-x)\, \mathrm{d}x + \int_0^b x \, \mathrm{d}x = \frac{b^2+a^2}{2}.
$$
