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in Differentiation by (15 points)

If u=x^y, show that ð^3u/ðx^2ðy =ð^3y/ðx.ðy.ðz

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\(u = x^y\)

\(\frac{\partial u}{\partial x} = yx^{y-1}\)

\(\frac{\partial^2 u}{\partial x^2} = y(y-1)x^{y-2}\)

\(\frac{\partial^3 u}{\partial x^2\partial y} = (y^2 - y) x^{y-2} log x + (2y - 1)x^{y-2}\)

\(\frac{\partial ^2u}{\partial x\partial y} = yx^{y-1}logx + x^{y-1}\)

\(\frac{\partial ^2u}{\partial x\partial y \partial x} = y(y-1) x^{y-2} logx + \frac{yx^{y-1}}{x} + (y -1)x^{y-2}\)

\(= x^{y-2} (y^2 - y) logx + x^{y-2} (2y - 1)\)

\(\frac{\partial ^3u}{\partial x^2\partial y} = \frac{\partial ^2u}{\partial x\partial y \partial x} \)

Hence proved.

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