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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)`
Solution
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`.
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