Power[ⅇ,9x]=Sin[x+3y] find Divide[dy,dx]

Problem: Find $\frac{dy}{dx}$ for the equation $e^{9x} = \sin(x + 3y)$.

Steps

1. Differentiate both sides with respect to $x$ (using the chain rule)
   $$\frac{d}{dx}(e^{9x}) = \frac{d}{dx}(\sin(x + 3y))$$
   $$9e^{9x} = \cos(x + 3y) \cdot \left(1 + 3\frac{dy}{dx}\right)$$
2. Divide by $\cos(x + 3y)$
   $$\frac{9e^{9x}}{\cos(x + 3y)} = 1 + 3\frac{dy}{dx}$$
3. Subtract 1 from both sides
   $$\frac{9e^{9x}}{\cos(x + 3y)} - 1 = 3\frac{dy}{dx}$$
4. Divide by 3
   $$\frac{dy}{dx} = \frac{1}{3} \left( \frac{9e^{9x}}{\cos(x + 3y)} - 1 \right)$$
5. Simplify the expression
   $$\frac{dy}{dx} = \frac{3e^{9x}}{\cos(x + 3y)} - \frac{1}{3}$$