Question

If  `log_10((x^3-y^3)/(x^3+y^3))=2 "then show that"  dy/dx = [-99x^2]/[101y^2]`

Answer 1

`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)`