Question

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

• A. \(\frac{(n-1)!}{x^n}\)
• B. \(\frac{n!}{x^n}\)
• C. \(\frac{(n-2)!}{x^n}\)
• D. \((-1)^{n-1} \frac{(n-1)!}{x^n}\)

Hint:

For repeated differentiation of \(\log x\), memorize: \[ \frac{d^n}{dx^n}(\log x)=(-1)^{n-1}\frac{(n-1)!}{x^n} \] It is a standard result.

Answer 1

The Correct Option is D

Concept:
Differentiate \(\log x\) repeatedly and identify the pattern. ip

Step 1: Write the first few derivatives.
\[ \frac{d}{dx}(\log x)=\frac{1}{x} \] \[ \frac{d^2}{dx^2}(\log x)=-\frac{1}{x^2} \] \[ \frac{d^3}{dx^3}(\log x)=\frac{2!}{x^3} \] \[ \frac{d^4}{dx^4}(\log x)=-\frac{3!}{x^4} \] ip

Step 2: Observe the general pattern.
The sign alternates as: \[ +,-,+,-,\dots \] and the factorial pattern is: \[ (n-1)! \] So, \[ \frac{d^n}{dx^n}(\log x)=(-1)^{n-1}\frac{(n-1)!}{x^n} \] ip Hence, the correct answer is:
\[ \boxed{(D)\ (-1)^{n-1}\frac{(n-1)!}{x^n}} \]