Back to DIFFERENTIATION

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Analyzing +4 | -1

If

y(x)=x^{x}, x>0

, then

y^{\prime \prime}(2)-2 y^{\prime}

(2) is equal to:

View answer and explanation

Correct option: A

4\left(\log _{e} 2\right)^{2}-2

Option A is correct. By finding the first and second derivatives of

y(x) = x^x

and substituting

x=2

, we can evaluate

y''(2) - 2y'(2)

.

More Questions from DIFFERENTIATION

Practice All Chapter MCQs →

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

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

is equal to Let

x

y

be two real numbers such that

x>0

x y=1

. The minimum value of

x+y

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

[x]

denotes the integer part of

x

x=k, k

being an integer, the left derivative of

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}}