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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
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Solution
`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`.
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