Question

If `sin^-1((x^5 - y^5)/(x^5 + y^5)) = pi/(6), "show that" "dy"/"dx" = x^4/(3y^4)`

Answer 1

`sin^-1((x^5 - y^5)/(x^5 + y^5)) = pi/(6)`

∴ `(x^5 - y^5)/(x^5 + y^5) = sin  pi/(6) = (1)/(2)`
∴ 2x⁵ – 2y⁵ = x⁵ + y⁵
∴ 3y⁵ = x⁵
Differentiating both sides w.r.t. x, we get
`3 xx 5y^4"dy"/"dx"` = 5x⁴

∴ `"dy"/"dx" = x^4/(3y^4)`.