Question

Is there a proof for \[\frac{d|x|}{dx}=\frac{x}{\sqrt x^2}\] and what do we do when x is negative?

Answer 1

I cant read properly !

Answer 2

Can u PLZ draw it

Answer 3

i think its supposed to be this.\[\frac{d}{dx}|x|=\frac{x}{\sqrt{x^2}}\]

Answer 4

its clear that the derivative of |x| is -1 if x < 0, 1 if x > 0, and doesnt exists if x = 0. Similarly, its clear that:\[\frac{x}{\sqrt{x^2}}=\frac{x}{|x|}\]which is -1 if x < 0, 1 if x > 0, and doesnt exists if x = 0. So they are the same.

Answer 5

use derivative by implicit

Answer 6

or you could write |x| = \(\sqrt x^2\)
and then diff.

Answer 7

by chain rule

Answer 8

Thanks, @joemath314159 and @hartnn

Answer 9

but, if u use by implicit derivative it will work to
let y=|x| = sqrt(x^2)
square both side,
y^2 = x^2
2y dy = 2x dx
dy/dx = 2x/2y = x/y = x/|x| = x/sqrt(x^2)

Answer 10

Thanks @RadEn