Q. If $I_{n} = \frac{d^{n}}{dx^{n}}(x^{n}\log{x),\ then\ I_{n} - nl_{n - 1}}$ is equal to
A) $n\$
B) $n - 1$
C) $n!$
D) $(n - 1)!$
Explanation
(d)
$I_{n} = \frac{d^{n - 1}}{dx^{n - 1}}\lbrack x^{n - 1} + nx^{n - 1}\log{x\rbrack}$
$I_{n} = (n - 1)\frac{d^{n - 2}}{dx^{n - 2}}x^{n - 2} + n\frac{d^{n - 1}}{dx^{n - 1}}(x^{n - 1}\log{x)}$
$I_{n} = (n - 1)! + nI_{n - 1}$
$\Rightarrow I_{n} - nI_{n - 1} = (n - 1)!$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.