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प्रश्न
If `log_10((x^3-y^3)/(x^3+y^3))=2 "then show that" dy/dx = [-99x^2]/[101y^2]`
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उत्तर
`log_10((x^3-y^3)/(x^3+y^3))=2`
Convert logarithmic form into exponential form,
`((x^3-y^3)/(x^3+y^3))=10^2 `
`((x^3-y^3)/(x^3+y^3))=100`
x3 - y3 = 100x3 + 100y3
∴ - 99x3 - 101y3 = 0
- 101y3 = 99x3
Differentiat ing w.r.t. x on both sides
- 99(3x2) - 101(3y2 `dy/dx`) = 0
∴ 99x2 + 101y2 `dy/dx`= 0
∴ 101y2 `dy/dx`= - 99x2
∴ `dy/dx=(-99x^2)/(101y^2)`
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