MathematicsQuestion 32022Mid SemesterMidsem 1hard4 marksa. Show that for the function f(x,y)={y(x2−y2)x2+y2,if (x,y)≠(0,0)0,if (x,y)=(0,0)f(x,y) = \begin{cases} \dfrac{y(x^2-y^2)}{x^2+y^2}, & \text{if } (x,y) \neq (0,0) \\ 0, & \text{if } (x,y) = (0,0) \end{cases}f(x,y)=⎩⎨⎧x2+y2y(x2−y2),0,if (x,y)=(0,0)if (x,y)=(0,0), fxy(0,0)≠fyx(0,0)f_{xy}(0,0) \neq f_{yx}(0,0)fxy(0,0)=fyx(0,0). (2 marks)b. If u=sin−1x1/3+y1/3x1/2+y1/2u = \sin^{-1}\sqrt{\dfrac{x^{1/3}+y^{1/3}}{x^{1/2}+y^{1/2}}}u=sin−1x1/2+y1/2x1/3+y1/3, then show thatx2uxx+2xyuxy+y2uyy=tanu12(1312+tan2u12).x^2u_{xx} + 2xyu_{xy} + y^2u_{yy} = \frac{\tan u}{12}\left(\frac{13}{12} + \frac{\tan^2 u}{12}\right).x2uxx+2xyuxy+y2uyy=12tanu(1213+12tan2u). (2 marks)Show solutionTry it yourself firstQuestion 11 of 13 in MathematicsTip: use ← → keysPrevious questionQ2 · 2022 · Midsem 1Next questionQ4 · 2022 · Midsem 1