Eind the derivative: \[ \text { 10) } x^{2} y+x=\sin (x y) \]

To find the derivative of the equation \( x^{2} y + x = \sin(xy) \), we'll use implicit differentiation. First, differentiate both sides with respect to \( x \). 1. For the left side, apply the product rule to \( x^2y \): \[ \frac{d}{dx}(x^2y) = 2xy + x^2 \frac{dy}{dx}. \] Then differentiate \( x \): \[ \frac{d}{dx}(x) = 1. \] 2. So, for the left side, we have: \[ 2xy + x^2 \frac{dy}{dx} + 1. \] 3. For the right side \( \sin(xy) \), apply the chain rule: \[ \frac{d}{dx}(\sin(xy)) = \cos(xy) \cdot \left( y + x \frac{dy}{dx} \right). \] Now, equate both sides: \[ 2xy + x^2 \frac{dy}{dx} + 1 = \cos(xy) \cdot \left( y + x \frac{dy}{dx} \right). \] Next, collect all terms involving \( \frac{dy}{dx} \) on one side: \[ x^2 \frac{dy}{dx} - \cos(xy) \cdot x \frac{dy}{dx} = \cos(xy) \cdot y - 2xy - 1. \] Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} (x^2 - \cos(xy) \cdot x) = \cos(xy) \cdot y - 2xy - 1. \] Finally, solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\cos(xy) \cdot y - 2xy - 1}{x^2 - \cos(xy) \cdot x}. \]