Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I noticed that if the absolute value definition $\lvert{x}\rvert=\sqrt{x^2}$ is used then we can get derivatives of functions with absolute value, without having to redefine them as piece-wise.

For example, to get the derivative of $f(x)=x\lvert{x}\rvert$ we write $f(x)=x(x^2)^\frac{1}{2}$ and then $f'(x)=\sqrt{x^2}+x\frac{1}{2}(x^2)^{-\frac{1}{2}}(2x)=\sqrt{x^2}+\frac{x^2}{\sqrt{x^2}}=\frac{2x^2}{\sqrt{x^2}} =\frac{2x^2}{\lvert{x}\rvert}=2\lvert{x}\rvert $ which is correct.

You just have to avoid using the law of exponents to simplify $\lvert{x}\rvert = (x^2)^\frac{1}{2}=x^{2(\frac{1}{2})}=x$. My question is why does using $\lvert{x}\rvert=\sqrt{x^2}$ to get derivatives work, and why does the law of exponents seem to show that $\lvert{x}\rvert=x$ ?

share|improve this question
    
Well, why shouldn't differentiating $\sqrt{x^2}$ using the ordinary rules work, as long as you stay away from $x=0$ of course? After all, you are composing perfectly differentiable functions. Though it is somewhat easier to differentiate $\lvert x\rvert$ as $x/\lvert x\rvert$, or $\operatorname{sgn}x$ if you wish. –  Harald Hanche-Olsen Dec 4 '12 at 20:28
1  
The problem with the law of exponents there is that the square root function isn't totally well-defined. Every nonzero number has two square roots, a positive one and a negative one; when you say $|x| = \sqrt{x^2}$, you're always taking the positive one; when you say $x = \sqrt{x^2}$, you're taking the one with the same sign as $x$. These issues become a lot more apparent over the complex numbers, where every fractional power involves a choice of 'branch'. –  Paul VanKoughnett Dec 4 '12 at 20:57

1 Answer 1

$\lvert x\rvert$ is partially-differentiable function, since piece-wise notation is more clear and natural for it. By expressing and differentiating it as $\sqrt{x^2}$ you introduce $\sqrt x$ function that is still non-differentiable in point $x=0$. So, writing smooth analytical formulas that work towards disguising the fact of non-differentiable points doesn't introduce any great value.

If you merely don't like piece-wise notation form, then follow the first commenter's note and use $sgn (x)$ representing $\lvert x\rvert'$ for $x\ne0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.