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Question : 121 of 160
Marks:
+1,
-0
Solution:
41. We have,
u(x,y)=y‌log‌x+x‌log‌yOn differentiating partially w.r.t.
x and
y respectively,
ux=‌+log‌y,uy=log‌x+‌Now,
uxuy−ux‌log‌x−uy‌log‌y+log‌x‌log‌y‌‌=(‌+log‌y)(log‌x+‌)−(‌+log‌y)‌log‌x‌‌−(log‌x+‌)‌log‌y+log‌x‌log‌y‌‌=1+‌‌log‌x+‌‌log‌y+log‌x‌log‌y‌‌−‌‌log‌x−log‌x‌log‌y−log‌x‌log‌y‌‌−‌‌log‌y+log‌x‌log‌y‌‌=1
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