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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 (x1/2) ]+ [ a1/2 d/dx (1/x)1/2 ]
=(1/a)1/2 (1/2) (1/x)1/2 + a1/2(-1/2) (1/x)3/2
multiplying this eq by 2xy
2xydy/dx = (1/a)1/2 .x1/2. y - a1/2 .(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
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