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Dydx of xy+y2=tanx+y is -

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Question

`(dy)/(dx)` of `xy + y^2 = tan x + y` is

Options

  • `(sec^2 x - y)/(x + 2y - 1)`

  • `(sec^2x)/(x + 2y)`

  • `y/(sec^2x)`

  • `(x + 2y)/(sec^2x)`

MCQ

Solution

`(sec^2 x - y)/(x + 2y - 1)`

Explanation:

`xy + y^2 = tan x + y`

Differentiating w.r.t x

`(1 * y + x (dy)/(dx)) + (2y (dy)/(dx)) = sec^2x + (dy)/(dx)`

or `(x + y - 1) (dy)/(dx) = sec^2x - y`

⇒ `(dy)/(dx) = (sec^2x - y)/(x + 2y - 1)`

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