**Solve the following.**

If x = `"4t"/(1 + "t"^2), "y" = 3((1 - "t"^2)/(1 + "t"^2))` then show that `"dy"/"dx" = (-9"x")/"4y"`.

#### Solution

x = `"4t"/(1 + "t"^2)`

Differentiating both sides w.r.t. ‘t’, we get

`"dx"/"dt" = ((1 + "t"^2)*"d"/"dx" ("4t") - "4t" * "d"/"dx" (1 + "t"^2))/(1 + "t"^2)^2`

`= ((1 + "t"^2)(4) - 4"t"(0 + 2"t"))/(1 + "t"^2)^2`

`= (4 + 4"t"^2 - 8"t"^2)/(1 + "t"^2)^2`

`= (4 - 4"t"^2)/(1 + "t"^2)^2`

`= (4(1 - "t"^2))/(1 + "t"^2)^2`

y = `3((1 - "t"^2)/(1 + "t"^2))`

Differentiating both sides w.r.t. ‘t’, we get

`"dx"/"dt" = 3 "d"/"dx" ((1 - "t"^2)/(1 + "t"^2))`

`= 3 [((1 + "t"^2) "d"/"dt" (1 - "t"^2) - (1 - "t"^2) * "d"/"dt" (1 + "t"^2))/(1 + "t"^2)^2]`

`= 3[((1 + "t"^2)(0 - 2"t") - (1 - "t"^2)(0 + 2"t"))/(1 + "t"^2)^2]`

`= 3 [(-2"t" (1 + "t"^2) - 2"t"(1 - "t"^2))/(1 + "t"^2)^2]`

`= 3(- 2"t") [(1 + "t"^2 + 1 - "t"^2)/(1 + "t"^2)^2]`

`= - 6"t" xx 2/(1 + "t"^2)^2`

`= "- 12t"/(1 + "t"^2)^2`

∴ `"dy"/"dx" = (("dy"/"dt"))/(("dx"/"dt")) = ("- 12t"/(1 + "t"^2)^2)/((4(1 - "t"^2))/(1 + "t"^2)^2)`

∴ `"dy"/"dx" = (- 3"t")/(1 - "t"^2)` ....(i)

Also `(- 9"x")/"4y" = (- 9(("4t")/(1 + "t"^2)))/(4xx3 ((1 - "t"^2)/(1 + "t"^2))) = (- 3"t")/(1 - "t"^2)` ....(ii)

From (i) and (ii), we get

`"dy"/"dx" = (- 9"x")/"4y"`