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Second step I want to understand

Second step I want to understand 

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Grade:12

3 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
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)!
Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
7 years ago
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
milind
23 Points
7 years ago
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

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