Given the equation \( 9 \sin 5 x y=1 \), evaluate \( \frac{d y}{d x} \). Assume that the equation implicitly defines \( y \) as a differentiable function of \( x \). If \( F(x, y)=9 \sin 5 x y-1 \), then \( F_{x}=\square \)

To find \( F_x \) for the function \( F(x, y) = 9 \sin(5xy) - 1 \), we need to differentiate \( F \) with respect to \( x \), treating \( y \) as a function of \( x \). Using the chain rule, we get: \[ F_x = \frac{\partial}{\partial x}(9 \sin(5xy)) = 9 \cos(5xy) \cdot (5y + 5x \frac{dy}{dx}) \] Putting it all together, we have: \[ F_x = 45y \cos(5xy) + 45xy \frac{dy}{dx} \] So the box will contain \( 45y \cos(5xy) + 45xy \frac{dy}{dx} \) for \( F_x \).