Question:

Step 1: Define the function to be differentiated. We are given that $g(x)$ is an antiderivative for $f(x)$, which means $g'(x) = f(x)$. We need to find the function for which $\ln(1+(g(x))^2)$ is an antiderivative. This requires us to differentiate the given expression with respect to $x$. Let $y = \ln(1+(g(x))^2)$. Step 2: Differentiate the function using the chain rule. To find the derivative of $y$ with respect to $x$, we apply the chain rule. The derivative of $\ln(u)$ is $\frac{1}{u} \frac{du}{dx}$. Here, $u = 1+(g(x))^2$. $$ \frac{dy}{dx} = \frac{1}{1+(g(x))^2} \cdot \frac{d}{dx}(1+(g(x))^2) $$ Step 3: Continue differentiating and apply the property of antiderivatives. Now, we need to differentiate the term $(1+(g(x))^2)$ with respect to $x$. The derivative of a constant (1) is 0. For $(g(x))^2$, we apply the chain rule again: $\frac{d}{dx}(u^2) = 2u \frac{du}{dx}$. Thus, $\frac{d}{dx}((g(x))^2) = 2g(x) \cdot g'(x)$. Substituting these derivatives back into the expression for $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{1}{1+(g(x))^2} \cdot (0 + 2g(x) \cdot g'(x)) $$ $$ \frac{dy}{dx} = \frac{2g(x) \cdot g'(x)}{1+(g(x))^2} $$ Since $g(x)$ is an antiderivative for $f(x)$, we know that $g'(x) = f(x)$. Substituting this into the equation: $$ \frac{dy}{dx} = \frac{2g(x) \cdot f(x)}{1+(g(x))^2} $$ Step 4: State the final answer. The derivative of $\ln(1+(g(x))^2)$ is $\frac{2f(x)g(x)}{1+(g(x))^2}$. Therefore, $\ln(1+(g(x))^2)$ is an antiderivative for $\frac{2f(x)g(x)}{1+(g(x))^2}$. This matches Option 2. The final answer is $\frac{2f(x)g(x)}{1+(g(x))^2}$.