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