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 \( F_{x}=45 y \cos (5 x y) \). If \( F(x, y)=9 \sin 5 x y-1 \), then \( F_{y}=45 x \cos (5 x y) \). \( \frac{d y}{d x}= \)

To evaluate \( \frac{dy}{dx} \) using implicit differentiation for the equation \( 9 \sin(5xy) = 1 \), we start with the function \( F(x, y) = 9 \sin(5xy) - 1 \). First, we apply implicit differentiation to both sides of the equation: 1. Differentiate \( F(x, y) \) with respect to \( x \): \[ F_x = \frac{\partial}{\partial x}(9 \sin(5xy)) = 9 \cos(5xy) \cdot \frac{d}{dx}(5xy) = 9 \cos(5xy)(5y + 5x \frac{dy}{dx}) = 45 \cos(5xy)(y + x \frac{dy}{dx}) \] 2. Differentiate \( F(x, y) \) with respect to \( y \): \[ F_y = \frac{\partial}{\partial y}(9 \sin(5xy)) = 9 \cos(5xy) \cdot 5x = 45x \cos(5xy) \] 3. By the implicit differentiation rule, we have: \[ \frac{dy}{dx} = -\frac{F_x}{F_y} \] 4. Substituting \( F_x \) and \( F_y \) gives: \[ \frac{dy}{dx} = -\frac{45 \cos(5xy)(y + x \frac{dy}{dx})}{45x \cos(5xy)} = -\frac{y + x\frac{dy}{dx}}{x} \] 5. Rearranging terms to solve for \( \frac{dy}{dx} \): \[ x \frac{dy}{dx} = - (y + x \frac{dy}{dx}) \implies x \frac{dy}{dx} + y + x \frac{dy}{dx} = 0 \] \[ (x + x) \frac{dy}{dx} = -y \implies 2x \frac{dy}{dx} = -y \implies \frac{dy}{dx} = -\frac{y}{2x} \] Thus, the result is: \[ \frac{dy}{dx} = -\frac{y}{2x} \]