Question

If `log_10 ((x^2 - y^2)/(x^2 + y^2))` = 2, then `dy/dx` is equal to ______.

Options

• `- (99x)/(101y)`

• `(99x)/(101y)`

• `- (99y)/(101x)`

• `(99y)/(101x)`

Answer 1

If `log_10 ((x^2 - y^2)/(x^2 + y^2))` = 2, then `dy/dx` is equal to `underlinebb(- (99x)/(101y))`.

Explanation:

Since, `log_10((x^2 - y^2)/(x^2 + y^2))` = 2

`\implies ((x^2 - y^2)/(x^2 + y^2))` = 10²

`\implies` x – y² = 100(x² + y²)

After differentiating on both sides, we get

`2x - 2y dy/dx = 100(2x + 2y dy/dx)`

`\implies x - y dy/dx = 100x + 100y dy/dx`

`\implies 101y dy/dx` = –99x

`\implies dy/dx = (-99x)/(101y)`