If I_n = d^n d x^n ( x^n x), then I_n - n I_ n - 1 = - Vidyadip

MCQ

If ${I_n} = {{{d^n}} \over {d{x^n}}}({x^n}\log x),$ then ${I_n} - n{I_{n - 1}} = $

A
$n$
B
$n - 1$
C
$n!$
✓
$(n - 1)!$

✓

Answer

Correct option: D.

$(n - 1)!$

d
(d) ${I_n} = \frac{{{d^{n - 1}}}}{{d{x^{n - 1}}}}[{x^{n - 1}} + n{x^{n - 1}}\log x]$

${I_n} = (n - 1)! + n{I_{n - 1}}$ ==> ${I_n} - n{I_{n - 1}} = (n - 1)!$.

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