39y = |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→02x∂x - ∂x²∂x(√(x + ∂x)² + √x²)
= lim∂x→02x - ∂x(√(x + ∂x)² + √x²)
Apply the limit.
= 2x2√x²
= x|x|
1thank you,Pritish.
I only wanted to know the first step ie. |x|=√x.
1|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????
39They asked for differentiation via first principle or ab initio as its called...
21Ur right..but the pink post most elegantly uses the first principle..
1oh 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..... :( :(
