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If log10(x3-y3x3+y3) = 2, show that dydx=-99x2101y2.

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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)`.

Sum
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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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Chapter 1: Differentiation - Exercise 1.3 [Page 40]

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