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

```							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
```
8 years ago
```							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]
```
8 years ago
```							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

```
8 years ago
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