Derivative of $|x|$ using definition of the derivative.

Question

Using the definition: $f’(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$ i try to find the derivative of the absolute value function. My first two steps are, $f’(x)=\lim_{h\to 0} \frac{|x+h|-|x|}{h}$ rewritten as $\lim_{h\to 0} \frac{\sqrt{(x+h)^2}-\sqrt{x^2}}{h}$ It’s at this point that I’m not sure about my last step, and if it’s correct then I’m lost on how to proceed in order to get the h cancellation to evaluate the limit. I apologize if this is a duplicate. Thank you.

Answer 1

Although it is a duplicate, here is an alternate answer to the one given there:

\begin{eqnarray} \frac{d}{dx}|x|&=&\lim_{h\to0}\frac{|x+h|-|x|}{h}\\ &=&\lim_{h\to0}\frac{|x+h|-|x|}{h}\cdot\frac{|x+h|+|x|}{|x+h|+|x|}\\ &=&\lim_{h\to0}\frac{|x+h|^2-|x|^2}{h(|x+h|+|x|)}\\ &=&\lim_{h\to0}\frac{(x+h)^2-x^2}{h(|x+h|+|x|)}\\ &=&\lim_{h\to0}\frac{h(2x+h)}{h(|x+h|+|x|)}\\ &=&\lim_{h\to0}\frac{2x+h}{|x+h|+|x|}\\ &=&\frac{x}{|x|}=\frac{|x|}{x} \end{eqnarray}