For N {N}, {d^{{n}}}{d x^{{n}}}( x)=

Correct Answer is : (−1)n−1(n−1)!xn(-1)ⁿ-1} {(n-1)!}{xⁿ}(−1)n−1xn(n−1)!

Correct Answer is : (−1)n−1(n−1)!xn(-1)ⁿ-1} {(n-1)!}{xⁿ}(−1)n−1xn(n−1)!

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MHTCET 22 Apr 2025 Shift 1 Paper

Question : 138 of 150

Marks: +1, -0

For

N \in \mathbb{N}, \frac{d^{\mathrm{n}}}{d x^{\mathrm{n}}}(\log x)=

Solution:

Concept:

Repeated differentiation of

\log x

produces a pattern of alternating signs and factorial terms.

Explanation:

Let

y = \log x

First derivative:

\frac{dy}{dx} = \frac{1}{x}

Second derivative:

\frac{d^2y}{dx^2} = -\frac{1}{x^2}

Third derivative:

\frac{d^3y}{dx^3} = \frac{2}{x^3}

Fourth derivative:

\frac{d^4y}{dx^4} = -\frac{6}{x^4}

The sign alternates and the numerator is

(n-1)!

Therefore,

\frac{d^n}{dx^n}(\log x) = (-1)^{n-1}\frac{(n-1)!}{x^n}

Answer:

\frac{d^n}{dx^n}(\log x) = (-1)^{n-1}\frac{(n-1)!}{x^n}

, so the correct option is D.

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