<s>[INST] Finding the Derivative of |x| using the Limit Definition Please Help me derive the derivative of the absolute value of x using the following limit definition.
$$\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}
$$
I have no idea as to how to get started.Please Help.
Thank You
 [/INST] We can write $|x| = \sqrt{x^2}$. Using the chain rule we then get
$$|x|' = \frac{1}{2\sqrt{x^2}} \cdot 2x = \frac{x}{\sqrt{x^2}} = \frac{x}{|x|}$$

EDIT Using limit:
$$\begin{align}
\frac{\sqrt{(x+\Delta x)^2}-\sqrt{x^2}}{\Delta x} 
& = \frac{(\sqrt{(x+\Delta x)^2}-\sqrt{x^2})(\sqrt{(x+\Delta x)^2}+\sqrt{x^2})}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\
& = \frac{(x+\Delta x)^2-x^2}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\
& = \frac{2 x \Delta x + \Delta x^2}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\
& = \frac{2x + \Delta x}{\sqrt{(x+\Delta x)^2}+\sqrt{x^2}} \\
& \to \frac{2x}{\sqrt{x^2}+\sqrt{x^2}} = \frac{2x}{2|x|} = \frac{x}{|x|}
\end{align}$$
</s>