If g is the inverse of the function f(x) and | vedclass

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(B) Given that $g(x)$ is the inverse of $f(x)$,we have $f(g(x)) = x$. Differentiating both sides with respect to $x$ using the chain rule,we get $f^{\

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

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(B) Given that $g(x)$ is the inverse of $f(x)$,we have $f(g(x)) = x$. Differentiating both sides with respect to $x$ using the chain rule,we get $f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$. Therefore,$f^{\prime}(g(x)) = \frac{1}{g^{\prime}(x)}$. Given $g(x) = x + 	an x$,we find $g^{\prime}(x) = 1 + \sec^2 x$. Substituting this into the expression,we get $f^{\prime}(g(x)) = \frac{1}{1 + \sec^2 x}$. Let $y = g(x)$,then $x = f(y)$. Substituting $x = f(y)$ into the equation,we get $f^{\prime}(y) = \frac{1}{1 + \sec^2(f(y))}$. Replacing $y$ with $x$,we obtain $f^{\prime}(x) = \frac{1}{1 + \sec^2(f(x))}$.

If $g$ is the inverse of the function $f(x)$ and $g(x) = x + \tan x$,then $f^{\prime}(x) = $

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