1. **State the problem:** Find the derivative of the function $$f(x)=(4+x)^{6+x}, x>-4.$$ We need to express $$f'(x)$$ in terms of $$x$$. 2. **Rewrite the function for easier differentiation:** We use the fact that $$a^b = e^{b\ln(a)}$$ for $$a>0$$. So, $$f(x)=(4+x)^{6+x} = e^{(6+x)\ln(4+x)}.$$ 3. **Differentiate using chain rule:** Let $$g(x) = (6+x)\ln(4+x)$$, then $$f(x) = e^{g(x)}$$. By chain rule: $$f'(x) = e^{g(x)} \cdot g'(x) = (4+x)^{6+x} \cdot g'(x).$$ 4. **Find $$g'(x)$$:** $$g(x) = (6+x)\ln(4+x).$$ Use product rule: $$g'(x) = (6+x)' \cdot \ln(4+x) + (6+x) \cdot \frac{1}{4+x} (4+x)'.$$ Since $$(6+x)'=1$$ and $$(4+x)'=1,$$ $$g'(x) = 1 \cdot \ln(4+x) + (6+x) \cdot \frac{1}{4+x} = \ln(4+x) + \frac{6+x}{4+x}.$$ 5. **Write the final expression for $$f'(x)$$:** $$f'(x) = (4+x)^{6+x} \cdot \left[\ln(4+x) + \frac{6+x}{4+x}\right].$$