Applications Of Derivatives Question 213
Question: If $ I_{n}=\frac{d^{n}}{dx^{n}}(x^{n}\log x), $ then $ I_{n}-n{I_{n-1}}= $
[EAMCET 2003]
Options:
A) $ n $
B) $ n-1 $
C) $ n! $
D) $ (n-1)! $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I_{n}=\frac{{d^{n-1}}}{d{x^{n-1}}}[{x^{n-1}}+n{x^{n-1}}\log x] $
$ I_{n}=(n-1)!+n{I_{n-1}} $
therefore $ I_{n}-n{I_{n-1}}=(n-1)! $ .
BETA
