The value of the derivative of $|x-1|+|x-3|$ at $x=2$ is

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

The value of the derivative of

|x-1|+|x-3|

x=2

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Correct option: B

Option B (0) is correct because at x=2, the function f(x) = |x-1| + |x-3| is differentiable, and its derivative is 0.

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f(x)=|x|^{|\tan x|}

f^{\prime}\left(\frac{-\pi}{6}\right)

x

y

x>0

x y=1

x+y

f(x)=[x] \sin (\pi x)

[x]

x

x=k, k

f(x)

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

\frac{{dy}}{{dx}}

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

z = y + f\left( v \right),

v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}