JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

y = \mathrm{In}{\left( {\frac{x}{{a + bx}}} \right)^x},

then

{x^3}\frac{{{d^2}y}}{{d{x^2}}}

is equal to

View answer and explanation

Correct option: D

{\left( {x\frac{{dy}}{{dx}} - y} \right)^2}

The correct answer is D (

{\left( {x\frac{{dy}}{{dx}} - y} \right)^2}

) because it is derived through logarithmic differentiation and algebraic manipulation.

More Questions from DIFFERENTIATION

Practice All Chapter MCQs →

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}}