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JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

If

f(x)=x^n

, then the value of

f(1)+\frac{f^{\prime}(1)}{\lfloor\underline{1}}+\frac{f^{\prime \prime}(1)}{\underline{2}}+\frac{f^{\prime \prime \prime}(1)}{\underline{3}}+\cdots+\frac{f^{(n)}(1)}{\underline{n} \underline{n}}

is

View answer and explanation

Correct option: B

2^n

Option B (

2^n

) is correct because the given expression is the binomial expansion of

(1+1)^n

, which equals

2^n

.

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y

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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}}

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v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}