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JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

If

f(x)=(1+x)^n

, then the value of

f(0)+

f^{\prime}(0)+\frac{1}{2 !} f^{\prime \prime}(0)+\cdots+\frac{1}{n !} f^n(0)

is

View answer and explanation

Correct option: B

2 n

Option D (none of these) is correct because the expression simplifies to

2^n

, not

2n

.

More Questions from DIFFERENTIATION

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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}}