If {I_n} = frac{d^n}{dx^n}(x^n log x),then {I | vedclass

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(D) Given ${I_n} = \frac{d^n}{dx^n}(x^n \log x)$.Using the property of derivatives,we can write ${I_n} = \frac{d^{n-1}}{dx^{n-1}} \left( \frac{d}{dx}(

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$(n - 1)!$

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(D) Given ${I_n} = \frac{d^n}{dx^n}(x^n \log x)$.Using the property of derivatives,we can write ${I_n} = \frac{d^{n-1}}{dx^{n-1}} \left( \frac{d}{dx}(x^n \log x) \right)$.Applying the product rule: $\frac{d}{dx}(x^n \log x) = n x^{n-1} \log x + x^n \cdot \frac{1}{x} = n x^{n-1} \log x + x^{n-1}$.Thus,${I_n} = \frac{d^{n-1}}{dx^{n-1}}(n x^{n-1} \log x + x^{n-1})$.${I_n} = n \frac{d^{n-1}}{dx^{n-1}}(x^{n-1} \log x) + \frac{d^{n-1}}{dx^{n-1}}(x^{n-1})$.By definition,${I_{n-1}} = \frac{d^{n-1}}{dx^{n-1}}(x^{n-1} \log x)$.Also,the $(n-1)$-th derivative of $x^{n-1}$ is $(n-1)!$.Therefore,${I_n} = n{I_{n-1}} + (n-1)!$.Rearranging the terms,we get ${I_n} - n{I_{n-1}} = (n-1)!$.

If ${I_n} = \frac{d^n}{dx^n}(x^n \log x)$,then ${I_n} - n{I_{n - 1}} = $

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