Sum

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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#### Solution

`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`

x^{3 }- y^{3 }= 100x^{3 }+ 100y^{3}

∴ - 99x^{3} - 101y^{3} = 0

- 101y^{3 }= 99x^{3}

Differentiat ing w.r.t. x on both sides

- 99(3x^{2}) - 101(3y^{2} `dy/dx`) = 0

∴ 99x^{2} + 101y^{2} `dy/dx`= 0

∴ 101y^{2} `dy/dx`= - 99x^{2}

∴ `dy/dx=(-99x^2)/(101y^2)`

Concept: Derivatives of Functions in Parametric Forms

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