Question

X^2-3x+1=0 and y^2-4y +1=0 then find the value of (x+y)(xy+1)/xy

Answer 1

SOLUTION 13 : Begin with x2 + xy + y2 = 1 . Differentiate both sides of the equation, getting

D ( x2 + xy + y2 ) = D ( 1 ) ,

2x + ( xy' + (1)y ) + 2 y y' = 0 ,

so that (Now solve for y' .)

xy' + 2 y y' = - 2x - y ,

(Factor out y' .)

y' [ x + 2y ] = - 2 x - y ,

and the first derivative as a function of x and y is

(Equation 1)

$ y' = \displaystyle{ - 2 x - y \over x + 2y } $ .

To find y'' , differentiate both sides of this equation, getting

$ y'' = \displaystyle{ (x + 2y) D(-2x - y) - (-2x - y) D(x + 2y) \over (x + 2y)^2 } $

$ = \displaystyle{ (x + 2y)(-2 - y') - (-1)(2x + y) (1 + 2y') \over (x + 2y)^2 } $

$ = \displaystyle{ -2x - 4y -xy' - 2yy' + 2x + y + 4xy' + 2yy' \over (x + 2y)^2 } $

$ = \displaystyle{ 3xy' - 3y \over (x + 2y)^2 } $ .

Use Equation 1 to substitute for y' , getting

$ y'' = \displaystyle{ 3x \Big( \displaystyle{ - 2 x - y \over x + 2y } \Big) - 3y \over (x + 2y)^2 } $