1. **Problem Statement:** We are given the function $y = \ln(1 + x)$ and asked to find the first differential $dy$ and the second differential $d^2y$ at $x = 10$. Then, we explain how the changing differential affects feature sensitivity in linear models. 2. **Formula and Rules:** - The first differential $dy$ is given by $dy = y' dx$, where $y' = \frac{dy}{dx}$. - The second differential $d^2y$ is given by $d^2y = y'' (dx)^2$, where $y'' = \frac{d^2y}{dx^2}$. - For $y = \ln(1 + x)$, the derivatives are: $$y' = \frac{1}{1 + x}$$ $$y'' = -\frac{1}{(1 + x)^2}$$ 3. **Calculate the first differential $dy$ at $x=10$:** - Compute $y'$ at $x=10$: $$y'(10) = \frac{1}{1 + 10} = \frac{1}{11}$$ - Thus, the first differential is: $$dy = y'(10) dx = \frac{1}{11} dx$$ 4. **Calculate the second differential $d^2y$ at $x=10$:** - Compute $y''$ at $x=10$: $$y''(10) = -\frac{1}{(1 + 10)^2} = -\frac{1}{121}$$ - Thus, the second differential is: $$d^2y = y''(10) (dx)^2 = -\frac{1}{121} (dx)^2$$ 5. **Explanation of changing differential and feature sensitivity:** - The first derivative $y'$ shows how sensitive $y$ is to changes in $x$. At $x=10$, $y'$ is small ($\frac{1}{11}$), indicating that changes in $x$ produce smaller changes in $y$. - The negative second derivative $y''$ indicates the function is concave down, meaning the rate of change of sensitivity decreases as $x$ increases. - In linear models, this means that for large $x$, the log transform reduces the impact of large feature values, stabilizing variance and reducing skewness. - This helps the model by preventing large values from dominating and improving feature sensitivity balance. **Final answers:** - $dy = \frac{1}{11} dx$ - $d^2y = -\frac{1}{121} (dx)^2$