find \\( \\frac { d y } { d x } \\).\n\\( e ^ { 9 x } = \\sin ( x + 3 y ) \\)\n\\( \\frac { d y } { d x } =…

Explanation:

Step1: Differentiate both sides with respect to (x)

Differentiate (e^{9x}) using the chain rule: (\frac{d}{dx}(e^{9x}) = 9e^{9x}). Differentiate (\sin(x + 3y)) using the chain rule: (\frac{d}{dx}(\sin(x + 3y))=\cos(x + 3y)\left(1 + 3\frac{dy}{dx}\right)). So we have (9e^{9x}=\cos(x + 3y)\left(1 + 3\frac{dy}{dx}\right)).

Step2: Solve for (\frac{dy}{dx})

First, divide both sides by (\cos(x + 3y)): (\frac{9e^{9x}}{\cos(x + 3y)}=1 + 3\frac{dy}{dx}). Then, subtract (1) from both sides: (3\frac{dy}{dx}=\frac{9e^{9x}}{\cos(x + 3y)}-1). Finally, divide both sides by (3): (\frac{dy}{dx}=\frac{3e^{9x}}{\cos(x + 3y)}-\frac{1}{3}).

Answer:

(\frac{dy}{dx}=\frac{3e^{9x}}{\cos(x + 3y)}-\frac{1}{3})