Differential Calculus> Question is attached in image. Sir explai...

Question is attached in image. Sir explain this solution. In = dn-1/dxn-1 [xn-1 + nxn-1 log x ] In = (n-1) dn-2 / dx n-2 xn-2 + n dn-1 / dxn-1 (xn-1 logx)
In =(n-1 ) ! + nIn-1
In – nIn-1 =(n-1)! Proved

milind , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Hello student,
Please find answer to your question
$I_{n} = \frac{d^{n}}{dx^{n}}(x^{n}log(x))$
$I_{n} = \frac{d^{n-1}}{dx^{n-1}}(x^{n}.\frac{1}{x} + log(x).nx^{n-1})$
$I_{n} = \frac{d^{n-1}}{dx^{n-1}}(x^{n-1} + n.x^{n-1}log(x))$
$I_{n} = \frac{d^{n-2}}{dx^{n-2}}((n-1)x^{n-2} + n(x^{n-1}.\frac{1}{x} + log(x).(n-1)x^{n-2})$
$I_{n} = (n-1)\frac{d^{n-2}}{dx^{n-2}}(x^{n-2}) + n\frac{d^{n-2}}{dx^{n-2}}(x^{n-2} + .(n-1)x^{n-2}log(x))$
$I_{n} = (n-1)! + n\frac{d^{n-2}}{dx^{n-2}}(x^{n-2} + .(n-1)x^{n-2}log(x))$ …..(1)
$I_{n-1} = \frac{d^{n-1}}{dx^{n-1}}(x^{n-1}log(x))$
$I_{n-1} = \frac{d^{n-2}}{dx^{n-2}}(x^{n-1}.\frac{1}{x} + log(x).(n-1)x^{n-2})$
$I_{n-1} = \frac{d^{n-2}}{dx^{n-2}}(x^{n-2} + (n-1)x^{n-2}log(x))$
Put this in equation (1), we have
$I_{n} = (n-1)! + nI_{n-1}$
$I_{n} - nI_{n-1} = (n-1)!$