# What is the derivative of #e^y cos x = 3 + sin(xy)#?

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To find the derivative of the equation ( e^y \cos(x) = 3 + \sin(xy) ) with respect to ( x ), we'll use implicit differentiation.

Differentiating both sides of the equation with respect to ( x ):

[ \frac{d}{dx} \left( e^y \cos(x) \right) = \frac{d}{dx} \left( 3 + \sin(xy) \right) ]

Using the chain rule and product rule on the left side and the product rule on the right side:

[ \frac{d}{dx} \left( e^y \cos(x) \right) = e^y \frac{dy}{dx} \cos(x) - e^y \sin(x) ] [ \frac{d}{dx} \left( 3 + \sin(xy) \right) = \cos(xy) \frac{dy}{dx} + y\cos(xy) ]

Now, we'll rearrange the terms and solve for ( \frac{dy}{dx} ):

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

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

[ \left( e^y \cos(x) - \cos(xy) \right) \frac{dy}{dx} = e^y \sin(x) + y\cos(xy) ]

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

So, the derivative of ( e^y \cos(x) = 3 + \sin(xy) ) with respect to ( x ) is ( \frac{e^y \sin(x) + y\cos(xy)}{e^y \cos(x) - \cos(xy)} ).

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