Let g(x) be an antiderivative for f(x). Then | vedclass

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(B) Given that $g(x)$ is an antiderivative of $f(x)$,we have $g'(x) = f(x)$.We need to find the derivative of $\ln(1 + (g(x))^2)$ with respect to $x$.

Vedclass · Accepted Answer

$\frac{2f(x)g(x)}{1 + (g(x))^2}$

Vedclass · Answer

(B) Given that $g(x)$ is an antiderivative of $f(x)$,we have $g'(x) = f(x)$.We need to find the derivative of $\ln(1 + (g(x))^2)$ with respect to $x$.Using the chain rule,let $y = \ln(1 + (g(x))^2)$.Then $\frac{dy}{dx} = \frac{1}{1 + (g(x))^2} \cdot \frac{d}{dx}(1 + (g(x))^2)$.$\frac{dy}{dx} = \frac{1}{1 + (g(x))^2} \cdot (2g(x) \cdot g'(x))$.Since $g'(x) = f(x)$,we substitute this into the expression:$\frac{dy}{dx} = \frac{2g(x)f(x)}{1 + (g(x))^2}$.Thus,$\ln(1 + (g(x))^2)$ is an antiderivative for $\frac{2f(x)g(x)}{1 + (g(x))^2}$.

Let $g(x)$ be an antiderivative for $f(x)$. Then $\ln(1 + (g(x))^2)$ is an antiderivative for

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