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

Question

William Abdalla · Accepted Answer

#frac{-x^2y^2 - y cos y}{x cos y + 2x^2y^2 + xy sin y} = y'(x)# #-2 y' = 1 - frac{D_x cos y (xy) - cos y D_x (xy)}{x^2y^2}# #-2 y' = 1 - frac{- y' sin y (xy) - cos y (y + xy')}{x^2y^2}# #[-2x^2y^2] y' = x^2y^2 + y' [xy sin y] + y cos y + [x cos y]y'# #-x^2y^2 - y cos y = [x cos y + 2x^2y^2 + xy sin y]y'#

Charles Anctil · Answer

undefined 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:

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

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

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

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

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

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

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

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

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

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