Question

$\frac{d^n}{d{x}^n}(\log x)$is equal to

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

Answer 1

The correct option is D

${(-1)}^{n-1}\frac{(n-1)!}{x}$

Explanation for the correct options:

Option(D): ${(-1)}^{n-1}\frac{(n-1)!}{x}$

Let $y=\log x$

$$\begin{gathered} \Rightarrow y^{\text{'}}=\frac{1}{x} \\ \Rightarrow y^{\text{'}\text{'}}=-\frac{1}{x^2} \\ \Rightarrow y^{\text{'}\text{'}\text{'}}=\frac{2}{x^3} \end{gathered}$$

Fourth Derivative$=-\frac{6}{x^4}$

Similarly $n^{th}$derivative $=$${(-1)}^{n-1}\frac{(n-1)!}{x^n}$

Hence, Option(D) is correct.