JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

y = \left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{2}{x}} \right)\left( {1 + \frac{3}{x}} \right) \ldots \left( {1 + \frac{n}{x}} \right)

and

x \ne 0,

then

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

when

x = - 1

View answer and explanation

Correct option: C

{\left( { - 1} \right)^n}\left( {n - 1} \right)!

Option D (

{\left( { - 1} \right)^n}\left( {n - 1} \right)!

) is correct because it is obtained by differentiating the given product and substituting

x = -1

More Questions from DIFFERENTIATION

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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}}