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प्रश्न
If `log_10((x^3 - y^3)/(x^3 + y^3)) = 2, "show that" "dy"/"dx" = -(99x^2)/(101y^2)`
उत्तर
`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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