Derivative of $\sqrt{ \left | x\right | }$

I'm not sure at all about my result:
$\ f'(x)=\frac{\left | x \right |}{x} . \frac{1}{2\sqrt{\left | x \right |}} = \frac{\left | x \right |^{0.5}}{2x}$

I don't know if we can simplify like this:
$\frac{\left | x \right |}{\sqrt(\left | x \right |)} = \left | x \right |^{0.5}$

• For $x\ne 0$, sure. – user251257 May 14 '17 at 15:34
• if we put $g(x)=sign(x)\sqrt{|x|}$ we would have a continuous function not differentiable in $x=0$. – Veridian Dynamics May 14 '17 at 15:39

Best way to take derivative of these kind of functions is :

$$f(x)=\begin{cases} \sqrt{x} & x\geq0\\ \sqrt{-x} & x<0 \end{cases} \implies f'(x)=\begin{cases} \dfrac{1}{2\sqrt x}& x\gt0\\ \dfrac{-1}{2\sqrt{-x}} & x<0 \end{cases}$$ (Note that the function isn't differentiable at $x=0$)

Which can also be written as :

$$f'(x)=\frac{\operatorname {sgn} x}{2\sqrt{|x|}}$$

• I know it, but later in the exercice this function will be in a sequence with a recurrence, so I can't predict if x is positive of negative – Hugues May 14 '17 at 15:39

An idea:

For $\;x<0\implies f(x)=\sqrt{-x}\;$ , and thus

$$f'(x):=-\frac1{2\sqrt{-x}}=-\frac1{2\sqrt{|x|}}$$

For $\;x>0\implies f(x)=\sqrt x\;$ , and thus

$$f'(x)=\frac1{2\sqrt x}=\frac1{2\sqrt{|x|}}$$

For $\;x=0\;$ the function isn't differentiable.

By using the chain rule.

$\frac{d}{dx}\sqrt{\left | x \right |}$

$=\frac{\frac{d}{dx}(\left |x \right |)}{2\sqrt{\left | x \right |}}$

$=\frac{x}{\left |x \right |}\frac{1}{2\sqrt{\left | x \right |}}$

$=\frac{x}{2\left |x \right |^\frac{3}{2}}$

Your function is even and that means that it's derivative is odd function. Thus, for $x>0$ we have that $$f'(x) = (\sqrt x)' = \frac 1{2\sqrt x},$$ while for $x<0$, we have $$-f'(x) = f'(-x) = \frac1{2\sqrt{-x}}.$$ Combining the two, we get that $$f'(x) = \frac{\operatorname{sgn}x}{2\sqrt{|x|}}.$$