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Differential Calculus> Question is attached in image ….jitendra ...

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Question is attached in image ….jitendra Sir explain this solution. I will follow your method but I want this too 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

4 Answers

Jitender Singh

Last Activity: 10 Years ago

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

milind

Last Activity: 10 Years ago

Jitendra Sir Can u explain me second step ....

milind

Last Activity: 10 Years ago

Second of answer ….please explain …..please

milind

Last Activity: 10 Years ago

Second step of answer please explain

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