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# Is the function f (x) = |x2 - x| differentiable at x = 2. If yes find it’s derivative.

Jitender Singh IIT Delhi
7 years ago
Ans:
Lets 1stcheck the differentiability at x = 2
Left Hand Derivative (LHD) = Right Hand Derivative (RHD)
$f(x) = |x^{2}-x|$
$LHD = \lim_{h\rightarrow 0}\frac{f(2-h)-f(2)}{-h}$
$f(x) = x^{2}-x, x < 0$
$= x-x^{2}, 0 < x < 1$
$= x^{2}-x, x > 1$
$LHD = \lim_{h\rightarrow 0}\frac{[(2-h)^{2}-(2-h)]-f(2)}{-h}$
$LHD = \lim_{h\rightarrow 0}\frac{h^{2}-3h}{-h} = 3$
$RHD = \lim_{h\rightarrow 0}\frac{[(2+h)^{2}-(2+h)]-f(2)}{h}$
$RHD = \lim_{h\rightarrow 0}\frac{h^{2}+3h}{h} = 3$
LHD = RHD
$\frac{d|x^{2}-x|}{dx} = \frac{d\sqrt{(x^{2}-x)^{2}}}{dx}$
Simply apply the chain rule, we have
$\frac{df(x)}{dx} = \frac{x(2x^{2}-3x+1)}{|x(x-1)|}$
Thanks & Regards
Jitender Singh
IIT Delhi