Question

If \[ x \cos y = \sin(x + y), \] find \(\frac{dy}{dx}\).

Hint:

Use implicit differentiation and apply product and chain rules carefully when variables are mixed in functions.

Answer 1

Differentiate both sides with respect to \(x\), using implicit differentiation: \[ \frac{d}{dx} \left( x \cos y \right) = \frac{d}{dx} \left( \sin(x + y) \right). \] Using product and chain rules on the left: \[ \cos y + x (-\sin y) \frac{dy}{dx} = \cos(x + y) \left( 1 + \frac{dy}{dx} \right). \] Rearranged: \[ \cos y - x \sin y \frac{dy}{dx} = \cos(x + y) + \cos(x + y) \frac{dy}{dx}. \] Group terms involving \(\frac{dy}{dx}\): \[ - x \sin y \frac{dy}{dx} - \cos(x + y) \frac{dy}{dx} = \cos(x + y) - \cos y. \] Factor \(\frac{dy}{dx}\): \[ \frac{dy}{dx} \left( - x \sin y - \cos(x + y) \right) = \cos(x + y) - \cos y. \] Therefore, \[ \frac{dy}{dx} = \frac{\cos(x + y) - \cos y}{- x \sin y - \cos(x + y)}. \]