1. Stating the problem: Find the derivative of the function $$f(x)=e^{3x} \sin(\ln(9+x))$$. 2. Use the product rule for differentiation: if $$f(x)=u(x)v(x)$$, then $$f'(x)=u'(x)v(x)+u(x)v'(x)$$. 3. Let $$u(x)=e^{3x}$$ and $$v(x)=\sin(\ln(9+x))$$. 4. Differentiate $$u(x)=e^{3x}$$ using the chain rule: $$u'(x) = e^{3x} \cdot 3 = 3e^{3x}$$. 5. Differentiate $$v(x)=\sin(\ln(9+x))$$ using the chain rule: - The outer function is $$\sin(z)$$ where $$z=\ln(9+x)$$, so $$\frac{d}{dx}[\sin(z)] = \cos(z) \cdot z'$$. - The inner function is $$z=\ln(9+x)$$, so $$z' = \frac{1}{9+x}$$. Thus, $$v'(x) = \cos(\ln(9+x)) \cdot \frac{1}{9+x}$$. 6. Substitute $$u', u, v'$$ into the product rule: $$f'(x) = 3e^{3x} \sin(\ln(9+x)) + e^{3x} \cdot \cos(\ln(9+x)) \cdot \frac{1}{9+x}$$. 7. Factor if desired: $$f'(x) = e^{3x} \left(3 \sin(\ln(9+x)) + \frac{\cos(\ln(9+x))}{9+x}\right)$$. Final answer: $$f'(x) = e^{3x} \left(3 \sin(\ln(9+x)) + \frac{\cos(\ln(9+x))}{9+x}\right)$$