If y=squareroot(x/a)+squareroot(a/x) , then show that 2xydy/dx=(x/a-a/x)

y^2 = x/a+a/x+2  ====================             a xy^2 = x^2+a^2      =====================           a2xydy/dx+ay^2 = 2x

2xydy/dx = x/a+x/a - y^2  ====================          x/a+x/a-x/a-a/x-2  =====================     x/a-a/x-2 = 2xydy/dx

so there may be any mistake in that qusetion

y = (x/a)^1/2 + (a/x)^1/2 ...............1

dy/dx = d/dx(x/a)^1/2 + d/dx(a/x)

         =[(1/a)^1/2  d/dx (x^1/2) ]+ [ a^1/2 d/dx (1/x)^1/2 ]

         =(1/a)^1/2 (1/2) (1/x)^1/2  +  a^1/2(-1/2) (1/x)^3/2

multiplying this eq by 2xy

2xydy/dx = (1/a)^1/2 .x^1/2. y  -   a^1/2 .(1/x)^1/2 y

             = y[ (x/a)^1/2 - (a/x)^1/2]         ..............2

from eq 1 putting y in eq 2

             =[(x/a)^1/2 + (a/x)^1/2].[(x/a)^1/2-(a/x)^1/2]

             =[x/a - a/x]

FIRST  FIND dy/dx.

MULTIPLY IT BY TWO.

ALSO Y=(x+a)/sqrt(ax).......1

NOW IF GO IN REVERSE THEN U HAVE TO PROVE THAT

2dy/dx=(x^2-a^2)/ayx^2

        now replace  y by equ 1

on simplifiation u would get the expression same as 2dy/dx