By H.K. Dass

Bargains with partial differentiation, a number of integrals, functionality of a fancy variable, particular features, laplace transformation, complicated numbers, and facts.

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If w = f (u, v), where u = x + y and v = x – y, show that 2 dx dy du 2. If z = sin–1 (x – y), x = 3 t, y = 4 t3; show that 4. If u = xey z, where y = du 3 a 2 x 2 , z = sin x. Find d x 5. If u = x 2 + y2 + z2 – 2 xyz = 1, show that [Hint. du = x2 Ans. eyz 1 3x cot y dx dy dz 0 2 2 1 x 1 y 1 z2 x u u u dx dy dz 0 x y z = 2 (x – yz) dx + 2 (y – zx) dy + 2 (z – xy) dz = 0 But x2 + y2 + z2 – 2 xyz = 1, y2 – 2xyz = 1 – x2 – z2 2 2 2 2 2 2 2 y – 2xyz + x z = 1 + x z – x – z (y – xz)2 = (1 – x2) (1 – z2)] 2 2 6.

2u x 2 2 xy 2u 2u y 2 2 = g(u)[g(u) – 1] xy y sin 2u = g(u) 2u 2u y 2 2 = sin 2u (2 cos 2u – 1) = 2 sin 2u cos 2u – sin 2u xy y = sin 4u – sin 2u = 2 cos 3u sin u Proved. xy 1 Example 29. If u = sin x y 2 2u sin u cos 2u 2 u y . 2 2 xy x y 4 cos3 u x y 1 Solution. We have, u = sin x y y x 1 x y x x1/ 2 ( x) Let z = sin u = x y y x 1 x z = f (u) = sin u 1 z is a homogeneous function of degree . 2 By Euler’s deduction I u u f (u ) u u 1 sin u x y y = n x = x y f (u ) x y 2 cos u u u 1 x y tan u = x y 2 Prove that x 2 2u 2 xy Created with Print2PDF.

F Example 44. If f (x, y) = 0 and (y, z) = 0, show that y . z . d x x . y . Solution. t. (2) f f dy . t. ‘y’, we get 0 = d z . y z d y f dy x = f dx y dz y = dy z Multiplying (3) and (4), we get f x y dy dz = dx dy f y z f d z f . = . (4) f dz x y = f dx y z Proved. P. , Dec. 2005, Com. 2002) Example 45. If u = x log xy where x3 + y3 + 3 xy = 1. Find Solution.