By H.K. Dass
Offers with partial differentiation, a number of integrals, functionality of a posh variable, specified services, laplace transformation, complicated numbers, and statistics.
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U v dx du dv y u u 2 3 2 2 find , . 11. If u = x cos , x 3r 2 s, y 4 r 2 s , z 2r 3s z r s 10. If z = f (x, y) where x = uv, y = u y 4x y 4x y r y u y 6 x s2 y 6 xys y 6 r cos sin 2 sin , 2 cos sin 2 sin r z z z z s z z z z z z (x, y) where x = eu cos v, y = eu sin v. Prove that 2 2 2 f f 2u f e u y v Ans. 12. If z = f f x 2 cos sin ,y and z f ( x, y ) , then show that u u 13. If x = 2 z x2 2 z y2 1 14.
4) Created with Print2PDF. (7) [From (1)] xy x y On adding (6) and (7), we have 2 2 z z z 2 z y x y 2 2 x y x y = n(n – 1)u + n (n + 1) v + nu – nv x y = nu (n – 1 + 1) + nv (n + 1 – 1) = n2u + n2v = n2 (u + v) = n2z II. Deduction: Prove that x2 2 z 2 xy x2 2u x 2 2 xy Proved. 2u 2u y 2 2 = g(u) [g(u) – 1] (Nagpur University, Winter 2003) xy y f (u ) g(u) = n f (u ) Proof. By Euler’s deduction formula I u u f (u ) x y n. t. t. ‘y’, we have x 2u 2u u [ g (u ) 1] = 2 y x y y Multiplying (2) by x, (3) by y and adding, we get y x2 2u x 2 2 xy 2u y.
2 ( x 2 y 2 z 2 )2 1 x 1 2 x = 0 r r r 2 r 12. If x = r cos , y = r sin , prove that Hence deduce that (i) r x 1 x , r. . x r x r 2x cos 2u 2 (ii) 2 x 2 13. If v = (x2 – y2). f (xy), prove that 2v x 2 2 y 2 = 0 2v y 2 (c) 2r x 2 2r y 2 2 2 r 1 r r x y = (x4 – y4) f (xy) u v u v x 2 y 2 14. If ux + vy = 0 and x y = 1, show that x y y 2 x 2 15. If z = xy + yx, verify that 2 z 2z x y y x 16.