Question

$y={\log }_{e}x$ and $x$ is a positive integer, then $\frac{d^{n}y}{d{x}^n}$ is equal to

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

Answer 1

Answer: (D)

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

$y={\log }_{e}x$

$y^{\prime }=\frac{1}{x}$ ....... first derivative
$y^{\prime \prime }=\frac{-1}{x^2}$ ....... second derivative
$y^{\prime \prime \prime }=\frac{2}{x^3}$ ....... third derivative
Thus we can generalize the nth differential as

$\frac{d^{n}y}{d{x}^n}=(-1{)}^{n-1}\ast (n-1)!\ast x^{-n}$
Option D