# How do you differentiate #-2y=x-cosy/(xy)#?

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To differentiate the given function ( -2y = x - \frac{\cos(y)}{xy} ) with respect to (x), we'll use the quotient rule and the chain rule:

[ \frac{d}{dx}(-2y) = \frac{d}{dx}\left(x - \frac{\cos(y)}{xy}\right) ]

[ -2\frac{dy}{dx} = 1 - \frac{d}{dx}\left(\frac{\cos(y)}{xy}\right) ]

[ -2\frac{dy}{dx} = 1 - \frac{xy\frac{d}{dx}(\cos(y)) - \cos(y)\frac{d}{dx}(xy)}{(xy)^2} ]

[ -2\frac{dy}{dx} = 1 - \frac{xy(-\sin(y)\frac{dy}{dx}) - \cos(y)(y\frac{dx}{dx} + x\frac{dy}{dx})}{(xy)^2} ]

[ -2\frac{dy}{dx} = 1 - \frac{-xy\sin(y)\frac{dy}{dx} - xy\cos(y) - xy^2\frac{dy}{dx}}{(xy)^2} ]

[ -2\frac{dy}{dx} = 1 - \frac{-xy\sin(y)\frac{dy}{dx} - xy\cos(y) - xy^2\frac{dy}{dx}}{x^2y^2} ]

[ -2\frac{dy}{dx} = 1 + \frac{xy\sin(y)\frac{dy}{dx} + xy\cos(y) + xy^2\frac{dy}{dx}}{x^2y^2} ]

[ -2\frac{dy}{dx} = 1 + \frac{xy(\sin(y)\frac{dy}{dx} + \cos(y) + y^2\frac{dy}{dx})}{x^2y^2} ]

[ -2\frac{dy}{dx} = 1 + \frac{xy(\sin(y) + y^2)\frac{dy}{dx} + xy\cos(y)}{x^2y^2} ]

[ \frac{dy}{dx} = \frac{1 + \frac{xy\cos(y)}{x^2y^2}}{-2 - \frac{xy(\sin(y) + y^2)}{x^2y^2}} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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