If f(x) = x + tan x and g(x) is the inverse o | vedclass

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(C) Given $f(x) = x + 	an x$ and $g(x)$ is the inverse of $f(x)$,so $f(g(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule

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

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(C) Given $f(x) = x + 	an x$ and $g(x)$ is the inverse of $f(x)$,so $f(g(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule:$f'(g(x)) \cdot g'(x) = 1$$g'(x) = \frac{1}{f'(g(x))}$.Since $f'(x) = 1 + \sec^2 x$,we have $f'(g(x)) = 1 + \sec^2(g(x))$.Using the identity $\sec^2 	heta = 1 + 	an^2 	heta$,we get:$f'(g(x)) = 1 + (1 + 	an^2(g(x))) = 2 + 	an^2(g(x))$.Thus,$g'(x) = \frac{1}{2 + 	an^2(g(x))}$.From the definition $f(g(x)) = x$,we have:$g(x) + 	an(g(x)) = x$$	an(g(x)) = x - g(x)$.Squaring both sides:$	an^2(g(x)) = (x - g(x))^2 = (g(x) - x)^2$.Substituting this into the expression for $g'(x)$:$g'(x) = \frac{1}{2 + (g(x) - x)^2}$.

If $f(x) = x + \tan x$ and $g(x)$ is the inverse of $f(x)$,then $g'(x)$ is equal to

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