If $V= x^nlogx$ then prove that $V_n = nV_{n-1} + (n-1)!$ where $V_n $ is the nth derivative of $V$

Asked Oct 06 '18 at 15:47

Active Oct 06 '18 at 18:11

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If $V= x^nlogx$ then prove that $V_n = nV_{n-1} + (n-1)!$ where $V_n $ is the nth derivative of $V$

My attempt :

$V=x^nlogx$

$\therefore \, V_1 = nx^{n-1}logx + x^{n-1}$

$\therefore \, V_1= \dfrac{nV}{x}+x^{n-1}$

$\therefore \, xV_1=nV+x^n$

Differentiating $(n-1)$ times,

$\therefore \, xV_n+(0)V_1=nV_{n-1}\,+ (n-1)!$

$\therefore \, xV_n=nV_{n-1}\,+ (n-1)!$

I don't know how an extra $x$ appears in the LHS in my answer.

edited Oct 06 '18 at 18:11

paulplusx

asked Oct 06 '18 at 15:47

Anish Koulgi

1

https://math.stackexchange.com/questions/94018/finding-the-n-th-derivatives-of-xn-ln-x-and-frac-ln-xx – lab bhattacharjee Oct 06 '18 at 17:34
I don't think that your relation $V_n = nV_{n-1} + (n-1)!$ is correct. Try putting smaller values for $n$ and observe. – paulplusx Oct 07 '18 at 04:59

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