If ${I_n} = \frac{{{d^n}}}{{d{x^n}}}({x^n}\log x),$ then ${I_n} - n{l_{n - 1}}$ is equal to

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Application +4 | -1

{I_n} = \frac{{{d^n}}}{{d{x^n}}}({x^n}\log x),

then

{I_n} - n{l_{n - 1}}

is equal to

View answer and explanation

Correct option: D

\left( {n - 1} \right)!

Option D (

(n - 1)!

) is correct because the expression

{I_n} - n{l_{n - 1}}

simplifies to

(n - 1)!

after applying the given recurrence relation.

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