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Question
If `log_10((x^3 - y^3)/(x^3 + y^3))` = 2, show that `dy/dx = -(99x^2)/(101y^2)`.
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Solution
`log_10((x^3 - y^3)/(x^3 + y^3))` = 2
∴ `(x^3 - y^3)/(x^3 + y^3)` = 102 = 100
∴ x3 – y3 = 100x3 + 100y3
∴ 101y3 = – 99x3
∴ y3 = `(-99)/(101)x^3`
Differentiating both sides w.r.t. x, we get
`3y^2 dy/dx = (-99)/(101) xx 3x^2`
∴ `dy/dx = -(99x^2)/(101y^2`
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