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If y = xn-1 ln x, then the nth order derivative of y with respect to x at \(x=\dfrac{1}{2}\) is:1. 3 ⋅ [n!]2. 2 ⋅ [(n + 1)!]3. 3(n - 1) ⋅ [n!]4. 2 ⋅ [(n - 1)!] |
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Answer» Correct Answer - Option 4 : 2 ⋅ [(n - 1)!] Given: y = xn – 1 log x ----(1) Diff. (1) w.r.t. ‘x’ we get; \({y_1} = \left( {n - 1} \right){x^{n - 2}}\log x + {x^{n - 1}}.\frac{1}{x}\) xy1 = (n - 1) xn - 1 log x + xn – 1 xy1 = (n - 1) y + xn – 1 ----(2) ∵ y = xn – 1 log x Diff. (2) (n - 1) times by Leibinitz’s theorem, Xyn + (n - 1) yn – 1 = (n - 1) yn – 1 + (n - 1)! xyn = (n - 1)! \({y_n} = \frac{{\left( {n - 1} \right)!}}{x}\) nth derivative with respect to \(x = \frac{1}{2}\) is; yn = 2(n - 1)! |
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