Question

If $I_n=\frac{d^n}{d{x}^n}\left(x^n{\log }_{e}x\right)$, then $I_n-n{I}_{n-1}=$

• A. $n$
• B. $n-1$
• C. $n!$
• D. $(n-1)!$

Answer 1

The correct option is D $(n-1)!$

$In=\frac{d^n}{d{x}^n}\left(x^{n}log{e}^x\right)$

$I{n}^{n-1}=\frac{d^{n-1}}{d{x}^{n-1}}\left(x^{n-1}log{e}^x\right)$

$\Rightarrow lety=x^{n}log{e}^x$

$\Rightarrow y^1=n{x}^{n-1}logex+x^{n-1}$

$\Rightarrow y^2=n(x^{n-1}logx{)}^1+(n-1)x^{n-2}$

$\Rightarrow y^3=n(x^{n-1}logex{)}^{11}+(n-2\left)\right(n-1)x^{n-3}$

$\Rightarrow y^4=n\left(x^{n-1}log{e}^x\right)+(n-2)(n-1)(n-3)x^{x-4}$

$\Rightarrow y^n=n(x^{n-1}logex{)}^{n-1}+(n-1)!$

$\Rightarrow In-n{I}_{n-1}=(n-1)!$