1. **State the problem:** We need to find the derivative of the function $$f(x) = x^{x+1}$$ using the definition of the derivative. 2. **Recall the definition of the derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Apply the definition:** $$f'(x) = \lim_{h \to 0} \frac{(x+h)^{(x+h)+1} - x^{x+1}}{h} = \lim_{h \to 0} \frac{(x+h)^{x+h+1} - x^{x+1}}{h}$$ 4. **Rewrite the function using exponentials and logarithms:** Since $$x^{x+1} = e^{(x+1)\ln x}$$ for $$x>0$$, we write $$f(x) = e^{(x+1)\ln x}$$ 5. **Use the chain rule on the exponent:** Let $$g(x) = (x+1)\ln x$$, then $$f(x) = e^{g(x)}$$ 6. **Find $$g'(x)$$:** $$g'(x) = \frac{d}{dx}[(x+1)\ln x] = (x+1) \cdot \frac{1}{x} + \ln x \cdot 1 = \frac{x+1}{x} + \ln x = 1 + \frac{1}{x} + \ln x$$ 7. **Apply the chain rule:** $$f'(x) = f(x) \cdot g'(x) = x^{x+1} \left(1 + \frac{1}{x} + \ln x\right)$$ 8. **Simplify the expression:** $$f'(x) = x^{x+1} \left(1 + \frac{1}{x} + \ln x\right) = x^{x+1} \left(1 + \frac{1}{x} + \ln x\right)$$ This is the derivative of $$f(x) = x^{x+1}$$ using the definition and properties of logarithms and exponentials. **Final answer:** $$\boxed{f'(x) = x^{x+1} \left(1 + \frac{1}{x} + \ln x\right)}$$