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Differential Calculus> Second step I want to understand...

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Second step I want to understand

milind , 11 Years ago

Grade 12

3 Answers

Jitender Singh

Ans:
There is some little mistake in the expression
$I_{n} = \frac{d^{n}}{dx^{n}}(x^{n}log(x))$
Differentiete one time
& apply the chain rule
$I_{n} = \frac{d^{n-1}}{dx^{n-1}}(x^{n}.\frac{1}{x}+log(x).nx^{n-1})$
$I_{n} = \frac{d^{n-1}}{dx^{n-1}}(x^{n-1})+n\frac{d^{n-1}}{dx^{n-1}}(x^{n-1}log(x))$
$I_{n} = (n-1)\frac{d^{n-2}}{dx^{n-2}}(x^{n-2})+n\frac{d^{n-1}}{dx^{n-1}}(x^{n-1}log(x))$ ….....(1)
$I_{n-1} = \frac{d^{n-1}}{dx^{n-1}}(x^{n-1}log(x))$
Put in (1)
$I_{n} = (n-1)\frac{d^{n-2}}{dx^{n-2}}(x^{n-2})+nI_{n-1}$
$\frac{d^{n}}{dx^{n}}(x^{n}) = n!$
$I_{n} = (n-1)(n-2)!+nI_{n-1}$
$I_{n} = (n-1)!+nI_{n-1}$
$I_{n}-nI_{n-1} = (n-1)!$

Approved

Last Activity: 11 Years ago

Sumit Majumdar

Dear student,
This is a formula for nth order differentiation. Can you please elaborate what is the second step that you want to be clarified. You need to specify whether you like to do an nth order integration of the differential equation or nth order differntiation.
Regards
Sumit

Last Activity: 11 Years ago

milind

sumit sir ….….…i understood this question ….…..................but i have another question can u please tell me my problem ….….…....the question is

please explain fully ….if it has 0/0 form …..then please tell why ….….…..question i posted in differential calculation ..see there

Last Activity: 11 Years ago

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