1. **State the problem:** Find the equation of the tangent line to the curve $y = \ln(x + 1)$ at the point $(0, 0)$. 2. **Recall the formula for the tangent line:** The equation of the tangent line to a curve $y = f(x)$ at $x = a$ is given by $$y = f(a) + f'(a)(x - a)$$ where $f'(a)$ is the derivative of $f(x)$ evaluated at $x = a$. 3. **Find the derivative of the function:** $$f(x) = \ln(x + 1)$$ Using the chain rule, $$f'(x) = \frac{1}{x + 1}$$ 4. **Evaluate the derivative at $x = 0$:** $$f'(0) = \frac{1}{0 + 1} = 1$$ 5. **Evaluate the function at $x = 0$:** $$f(0) = \ln(0 + 1) = \ln(1) = 0$$ 6. **Write the equation of the tangent line:** Using the point-slope form, $$y = f(0) + f'(0)(x - 0) = 0 + 1 \cdot x = x$$ **Final answer:** The equation of the tangent line is $$y = x$$