Question

Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : `sec((x^5 + y^5)/(x^5 - y^5))` = a² 

Answer 1

`sec((x^5 + y^5)/(x^5 - y^5))` = a² 

∴ `(x^5 + y^5)/(x^5 - y^5) = sec^-1(a^2)` = k
∴ x⁵ + y⁵ = kx⁵ – ky⁵
∴ (1 + k)y⁵ = (k – 1)x5

∴  `y^5/x^5 = (k - 1)/(k + 1)`

∴  `y/x = ((k - 1)/(k + 1))^(1/5)`, a constant
Differentiating both sides w.r.t. x, we get
`"d"/"dx"(y/x)` = 0

∴ `(x."dy"/"dx" - y."d"/"dx"(x))/(x^2)` = 0

∴ `x"dy"/"dx" - y xx 1` = 0

∴ `"dy"/"dx" = y/x.`
Alternative Method :
`sec((x^5 + y^5)/(x^5 - y^5))` = a²

∴ `(x^5 + y^5)/(x^5 - y^5)` = sec^–1a² = k    ...(Say)
∴ x⁵ + y⁵ = kx⁵ – ky⁵
∴ (1 + k)y⁵ = (k – 1)x⁵
∴ `y^5/x^5 = (k - 1)/(k + 1)`                     ...(1)
∴ y⁵ = k'x⁵, where k' = `(k - 1)/(k + 1)`
Differentiating both sides w.r.t. x, we get
`5y^4"dy"/"dx"` = k' x 5x4

∴ `"dy"/"dx" = k'.x^4/y^4`

∴ `"dy"/"dx" = ((k - 1)/(k + 1)).x^4/y^4`

= `y^5/x^5 xx x^4/y^4`                         ...[By (1)]

∴ `"dy"/"dx" = y/x`.