If (x+y) sin u = x2y2, then x ∂u/∂x + y ∂u/∂y is equal to

(1) sin u

(2) cosec u

(3) tan u

(4) 3 tan u

Solution:

Given (x+y) sin u = x2y2

sin u = x2y2/(x+y)

= x4(y2/x2)/x(1 + (y/x))

= x3(y/x)2/(1 + (y/x))

= x3 f(y/x)

This is a homogeneous function of degree 3.

So by Euler’s theorem

x∂z/∂x + y ∂z/∂y = nz

Here n = 3

Let z = sin u

x∂z/∂x + y ∂z/∂y = 3z

x (∂/∂x) sin u + y (∂/∂y) sin u = 3 sin u

x cos u ∂u/∂x + y cos u ∂u/∂y = 3 sin u

x ∂u/∂x + y ∂u/∂y = 3 sin u/cos u

x ∂u/∂x + y ∂u/∂y = 3 tan u

Hence option (4) is the answer.

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