Find \( \frac{d y}{d x} \) by implicit differentiation. \[ y \cos (x)=5 x^{2}+3 y^{2} \] \( \frac{d y}{d x}=\square \)

To find \( \frac{d y}{d x} \) by implicit differentiation, we start with the given equation: \[ y \cos (x) = 5 x^{2} + 3 y^{2} \] Now, let's differentiate both sides with respect to \( x \): On the left side, we apply the product rule: \[ \frac{d}{dx}(y \cos(x)) = \frac{dy}{dx} \cos(x) + y(-\sin(x)) \] On the right side, we differentiate term by term: \[ \frac{d}{dx}(5x^2) = 10x \] \[ \frac{d}{dx}(3y^2) = 6y \frac{dy}{dx} \] (using the chain rule) Putting it all together, we have: \[ \frac{dy}{dx} \cos(x) - y \sin(x) = 10x + 6y \frac{dy}{dx} \] Now, we rearrange terms to solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \cos(x) - 6y \frac{dy}{dx} = 10x + y \sin(x) \] \[ \frac{dy}{dx} (\cos(x) - 6y) = 10x + y \sin(x) \] Finally, we can isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{10x + y \sin(x)}{\cos(x) - 6y} \] Thus, the result is \[ \frac{d y}{d x} = \frac{10x + y \sin(x)}{\cos(x) - 6y} \]