[Solved] If Y = Xn-1 Ln X, Then The Nth Order Derivative Of Y Wi

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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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