Question

If `tan ((x + y)/(x - y))` = k, then `dy/dx` is equal to ______.

Options

• `(-y)/x`

• `y/x`

• `sec^2 (y/x)`

• `-sec^2 (y/x)`

Answer 1

If `tan ((x + y)/(x - y))` = k, then `dy/dx` is equal to `underlinebb(y/x)`.

Explanation:

Given

`tan((x + y)/(x - y))` = k

`(x + y)/(x - y)` = tan^–1 k

On differentiating both sides, w.r.t. x, we get

`((x - y)d/dx(x + y) - (x + y)d/dx(x - y))/(x - y)^2 = d/dx [tan^-1 k]`

`\implies ((x - y)(1 + dy/dx) - (x + y)(1 - dy/dx))/(x - y)^2` = 0

`\implies (x - y)(1 + dy/dx) - (x + y)(1 - dy/dx)` = 0

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

`\implies [(x - y) + (x + y)] dy/dx` = (x + y) – (x – y)

`\implies 2x dy/dx` = 2y

`\implies dy/dx = y/x`.