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

Step 1: Understanding the Concept:
We need to find the formula for the \(n\)-th derivative of \(\log x\). We can do this by observing the pattern of the first few derivatives.
Step 2: Key Formula or Approach:
Rule for differentiating powers: \(\frac{d}{dx} x^k = kx^{k-1}\).
Step 3: Detailed Explanation:
Let \(y = \log x\).
1. \(n=1\): \(\frac{dy}{dx} = \frac{1}{x} = x^{-1}\).
2. \(n=2\): \(\frac{d^2y}{dx^2} = -1 \cdot x^{-2} = \frac{-1}{x^2}\).
3. \(n=3\): \(\frac{d^3y}{dx^3} = (-1)(-2) \cdot x^{-3} = \frac{2}{x^3} = \frac{2!}{x^3}\).
4. \(n=4\): \(\frac{d^4y}{dx^4} = (-1)(-2)(-3) \cdot x^{-4} = \frac{-6}{x^4} = \frac{-3!}{x^4}\).
By induction, for the \(n\)-th derivative:
- The sign alternates as \((-1)^{n-1}\).
- The constant is \((n-1)!\).
- The power of \(x\) in the denominator is \(n\).
Thus, \(\frac{d^n}{dx^n} (\log x) = (-1)^{n-1} \frac{(n-1)!}{x^n}\).
Step 4: Final Answer:
The \(n\)-th derivative is \((-1)^{n-1} \frac{(n-1)!}{x^n}\).