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Grade:12

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Hello student,
Please find answer to your question
I_{n} = \frac{d^{n}}{dx^{n}}(x^{n}log(x))
I_{1} = \frac{d}{dx}(xlog(x))
I_{1} = 1 + log(x)
I_{2} = \frac{d^{2}}{dx^{2}}(x^{2}log(x))
I_{2} = 3 + 2log(x)
I_{3} = \frac{d^{3}}{dx^{3}}(x^{3}log(x))
I_{3} = 11 + 6log(x)
I_{4} = \frac{d^{4}}{dx^{4}}(x^{4}log(x))
I_{4} = 50 + 24log(x)
I_{5} = \frac{d^{5}}{dx^{5}}(x^{5}log(x))
I_{5} = 274 + 120log(x)
I_{5}-5I_{4} = 24 = 4! = (5-1)!
I_{4}-4I_{3} = 6 = 3! = (4-1)!
I_{3}-3I_{2} = 2 = 2! = (3-1)!
Similarly, you can prove for n,
I_{n}-nI_{n-1} = (n-1)!
milind
23 Points
6 years ago
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  

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