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Inverse of a Function

If f x =x+t...

Question

If $f (x) = x + tan x$ and $g^{- 1} = f$ then $g^{'} (x)$ equals

\frac{1}{2 + {[g (x) + x]}^{2}}

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

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

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\frac{1}{2 - {[g (x) - x]}^{2}}

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Solution

The correct option is C $\frac{1}{2 + {[g (x) - x]}^{2}}$
$f (x) = x + tan x ⇢ (i) g^{- 1} (x) = f (x) ⇢ (i i) f (g (x)) = g (x) + tan g (x) (f r o m e q u a t i o n (i)) g^{- 1} (g (x)) = g (x) + tan g (x) (f r o m e q u a t i o n (i i)) x = g (x) + tan g (x) ⇢ (i i i) ({∵ g}^{- 1} (g (x)) = x)$
Differentiating the above equation w.r.t x,
$1 = g^{^{'}} (x) + ({sec}^{2} g (x)) g^{^{'}} (x) g^{^{'}} (x) = \frac{1}{1 + {sec}^{2} g (x)} = \frac{1}{2 + {tan}^{2} g (x)} (∵ {sec}^{2} g (x) = 1 + {tan}^{2} g (x)) f r o m e q n (i i i), tan g (x) = x - g (x) ∴ g^{^{'}} (x) = \frac{1}{2 + {(x - g (x))}^{2}}$

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