Question

If x = `(t + 1)/(t - 1), y = (t - 1)/(t + 1), "then show that"  y^2 + "dy"/"dx"` = 0.

Answer 1

x = `("t" + 1)/("t" - 1)` .......(1)

y = `("t" - 1)/("t" + 1)` ......(2)

Multiplying equations (1) and (2)

x × y = `(("t" + 1))/(("t" - 1)) xx (("t" - 1))/(("t" + 1))`

x × y = 1

x = `1/"y"` ....(3)

Differentiating w.r.t. x,

`"x" xx "dy"/"dx" + "y" xx (1) = 0`

`1/"y" xx "dy"/"dx" + "y"/1 = 0` ....[From (3)]

`(1 xx "dy"/"dx" + "y"^2)/"y" = 0`

`"dy"/"dx" + "y"^2 = 0`

∴ `"y"^2 + "dy"/"dx" = 0`

Hence proved.