Question

If , `y=sqrt(x/a)-sqrt(a/x)` So, Prove that ?? `2xy(dy/dx)= x/a-a/x`

If , `y=sqrt(x/a)-sqrt(a/x)` So, Prove that `?`
`2xydy/dx= x/a-a/x`

Answer 1

Given: `y=sqrt(x/a)-sqrt(a/x)`

Differentiating:

`dy/dx = (sqrt(a/x) + sqrt(x/a))/(2 x)`

Multiply both sides by2xy:

`2xydy/dx = 2xy(sqrt(a/x) + sqrt(x/a))/(2 x)`

`(2x)/(2x)` becomes 1:

`2xydy/dx = y(sqrt(a/x) + sqrt(x/a))`

Substitute `y=sqrt(x/a)-sqrt(a/x)` on the right side only:

`2xydy/dx = (sqrt(x/a)-sqrt(a/x))(sqrt(a/x) + sqrt(x/a))`

Please observe that the right side is the pattern `(a - b)(a+b) = a^2-b^2` where `a = sqrt(x/a)` and `b = sqrt(a/x)`

`2xydy/dx = x/a-a/x" Q.E.D."`

Answer 2

Kindly see a Proof in Explanation.

Explanation:

We have, `y=sqrt(x/a)-sqrt(a/x)`.

Diff.ing w.r.t. `x, dy/dx=1/sqrta*(1/2x^(-1/2))-sqrta*(-1/2x^(-3/2))`.

`:. dy/dx=1/(2sqrt(ax))+sqrta/(2xsqrtx)`.

Multiplying by `2x, 2xdy/dx=sqrt(x/a)+sqrt(a/x)`.

Again multiplying by `y, 2xydy/dx=y(sqrt(x/a)+sqrt(a/x))`,

`i.e., 2xydy/dx=(sqrt(x/a)-sqrt(a/x))(sqrt(x/a)+sqrt(a/x))`.

`rArr 2xydy/dx=x/a-a/x`, as desired!

Answer 3

We seek to show that:

`2xydy/dx= x/a-a/x` where `y=sqrt(x/a)-sqrt(a/x)`

Squaring the expression, and then expanding, we get:

`y^2 = (sqrt(x/a)-sqrt(a/x))^2`
`\ \ \ \= (sqrt(x/a))^2 - 2(sqrt(x/a))(sqrt(a/x)) + (sqrt(a/x))^2`
`\ \ \ \= x/a - 2 + a/x`

Differentiating Implicitly, we get

`2y dy/dx = 1/a -a/x^2`

Finally, multiplying by `x` we get the desired result:

`2xy dy/dx = x/a -a/x \ \ \` QED