Question

Please answer this question !!



It is solo urgent!!

Points:-35!!☺☺

Answer 12 and 13 questions alone!!!

Answer 1

i hope you will understand it

the basic is that while doing partial differentiation wrt y take x as const. and whild doing wrt x take y as constant
i don't have much space so left these on final step so pls solve it.

I hope you will understand this close texting.

Answer 2

12. x.du/dx + y.du/dy =

u = sin(x + y)/(√x + √y)
sin-¹u = (x + y) /(√x + √y)
now differentiate , with respect to x take y constant .

1/√( 1- u²) du/dx = {(√x + √y).1 - (x + y).1/2√x }/(√x + √y)² -----(1)

again,
differentiate with respect to y , take x constant .
1/√(1 - u²).du/dy = {(√x + √y).1 - (x + y)/2√y}/(√x + √y)² -----------(2)

multiply x in eqn (1) and y in eqn (2) then add .

1/√(1 - u²).{x.du/dx + y.du/dx } = {x(√x + √y) -√x/2(x + y) + y(√x + √y) - √y/2(x + y)}/(√x + √y)²

= {(x + y)(√x + √y) -(x + y)(√x + √y)/2 }/(√x + √y)²
=(x + y)(√x + √y)/2(√x + √y)²
=1/2(x + y)/(√x + √y)

now,
(1 - u²) = (1 - sin²P ) = cos²P
where , P = (x + y)/(√x + √y)
so, 1/√(1 -u²) = 1/cosP
hence,
xdu/dx + ydu/dy = 1/2(x + y)/(√x + √y)cos(x + y)/(√x + √y)
hence, proved ///





13. u = x²/y = 2y²/x = K ( let )
u = Ky/x²
u = Kx/2y²
y/x² = x/2y²
2y³ = x³
differentiate
6y².dy = 3x².dx
2y².dx.dy = x².(dx)²
dx.dy = x²/2y²(dx)²


LHS = d²u/dx.dy =

sorry , i never got 2nd , its too much lengthy .