Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1

Marks : 3 M

Year : May 2015

u $ \; = \; x^y \\ \; \\ Differentiating \; u \; partially \; w. r. t. \; x, \\ \therefore \frac{\partial u}{\partial x} \; = \; yx^{y-1} \; \; \ldots (i) \\ \; \\ Differentiating \; equation \; (i) \; partially \; w. r. t. \; y, \\ \frac{\partial^2 u}{\partial x \partial y } \; = \; yx^{y-1} log x \; + \; x^{y-1} \\ \; \\ Now \; differentiating \; again \; w. r. t. \; x \; partially, \\ \therefore \frac{\partial^3 u}{\partial x \partial y \partial x} \; = \; y \bigg[ x^{y-1} \frac{1}{x} \; + \; logx (y\;-\;1)x^{y-2} \bigg] \; + \; (y\;-\;1)x^{y-2} \\ \; \\ \therefore \frac{\partial^3 u}{\partial x \partial y \partial x} \; = \; x^{y-2} [ y \; + \; y(y-1)logx \; + \; y-1] \; = \; x^{y-2} [ 2y \; - \; 1 \; + \; y(y-1)logx ] $