diffenciate:

y = |x| = √x²
=> y + ∂y = √(x + ∂x)²
=> ∂y = √(x + ∂x)² - √x²

So ∂y∂x = √(x + ∂x)² - √x²∂x
=> dydx = lim_{∂x→0} √(x + ∂x)² - √x²∂x
Rationalizing,

dydx = lim_{∂x→0} (x + ∂x)² - x²∂x(√(x + ∂x)² + √x²)

dydx = lim_{∂x→0}2x∂x - ∂x²∂x(√(x + ∂x)² + √x²)
= lim_{∂x→0}2x - ∂x(√(x + ∂x)² + √x²)

Apply the limit.
= 2x2√x²
= x|x|

|x| can be either +x or -x .......therefore why is the answer not +1 or -1 straight away??????

x/|x| gives the same answer +1 or -1 by my above point......can someone clear my doubt????

Ur right..but the pink post most elegantly uses the first principle..

oh yeah first principle......actually I do not know first principle.....it is in the ncert but my bad I did not read the ncert....and differnciation of tan^2 (x) came in my final exams of 4 marks......and i knew only the answer and nothing else..... :( :(