JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} {{x^3}}&{{x^4}}&{3{x^2}}\\ 1&{ - 6}&4\\ p&{{p^2}}&{{p^3}} \end{array}} \right|

, here

p

is a constant, then

\frac{{{d^4}f\left( x \right)}}{{d{x^4}}}

View answer and explanation

Correct option: D

A constant

The correct answer is 'A constant' because after taking the fourth derivative of the determinant, the resulting expression is a constant value.

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}}