JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

y=1+\frac{x}{\underline{1}}+\frac{x^2}{\underline{2}}+\frac{x^3}{\underline{3}}+\ldots, \frac{d y}{d x}

is equal to

View answer and explanation

Correct option: D

y

Option D (

y

) is correct because the given series is the expansion of

e^x

, and the derivative of

e^x

e^x

, which is represented by

y

in this case.

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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}}