Find \( \frac{d y}{d x} \) 6) \( \cos (x y)+x^{2}=x y^{2} \)

To find \(\frac{dy}{dx}\) for the equation \(\cos(xy) + x^2 = xy^2\), we will use implicit differentiation. 1. Differentiate both sides of the equation with respect to \(x\): \[ \frac{d}{dx} [\cos(xy)] + \frac{d}{dx} [x^2] = \frac{d}{dx} [xy^2] \] Using the chain rule on \(\cos(xy)\): \[ -\sin(xy) \left( y + x\frac{dy}{dx} \right) + 2x = y^2 + x \frac{dy}{dx} \cdot 2y \] 2. Now simplify and isolate \(\frac{dy}{dx}\): \[ -\sin(xy)(y + x \frac{dy}{dx}) + 2x = y^2 + 2xy \frac{dy}{dx} \] Rearranging this gives: \[ -\sin(xy) y - \sin(xy) x \frac{dy}{dx} + 2x - y^2 = 2xy \frac{dy}{dx} \] 3. Collect all terms containing \(\frac{dy}{dx}\) on one side: \[ -\sin(xy) x \frac{dy}{dx} - 2xy \frac{dy}{dx} = y^2 + \sin(xy) y - 2x \] 4. Factor out \(\frac{dy}{dx}\): \[ \frac{dy}{dx} \left(-\sin(xy) x - 2xy\right) = y^2 + \sin(xy) y - 2x \] 5. Finally, solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{y^2 + \sin(xy) y - 2x}{-\sin(xy) x - 2xy} \] And that's the derivative of \(y\) with respect to \(x\).