Let denote the set of all real numbers. Let f : and g : ( 0 , 4 ) be functions defined by f ( x ) = log e ( x 2 + 2 x + 4 ) and g ( x ) = 4 1 + e - 2 x . Define the composite function f ? g - 1 by ( f ? g - 1 ) ( x ) = f ( g - 1 ( x ) ) , w

Question

Let $ℝ$ denote the set of all real numbers. Let $f : ℝ \to ℝ$ and $g : ℝ \to (0, 4)$ be functions defined by

$f (x) = \log_{e} (x^{2} + 2 x + 4)$ and $g (x) = \frac{4}{1 + e^{- 2 x}} .$

Define the composite function $f \circ g^{- 1}$ by $(f \circ g^{- 1}) (x) = f (g^{- 1} (x)),$ where $g^{- 1}$ is the inverse of the function $g$ .

Then the value of the derivative of the composite function $f \circ g^{- 1}$ at $x = 2$ is ________. [2025]

Smart Achievers · Accepted Answer

(0.25)

$Let h (x) = f (g^{- 1} (x)) and g (0) = 2$

$\Rightarrow h' (x) = f' (g^{- 1} (x)) \cdot (g^{- 1} (x))'$

$\Rightarrow h' (2) = f' (g^{- 1} (2)) \cdot (g^{- 1})' (2)$ $(put x = 2) = f' (0) \cdot (g^{- 1})' (2)$

Now, $f (x) = \log_{e} (x^{2} + 2 x + 4)$

$f' (x) = \frac{2 x + 2}{x^{2} + 2 x + 4} \Rightarrow f' (0) = \frac{1}{2}$

$g (x) = \frac{4}{1 + e^{- 2 x}},$ $g (0) = 2$

$∴$ $g' (x) = \frac{8 e^{- 2 x}}{{(1 + e^{- 2 x})}^{2}}$

$g' (0) = \frac{8}{4} = 2$ $∵$ $g^{- 1} (g (x)) = x$

$(g^{- 1})' (g (x)) g' (x) = 1$

$(g^{- 1})' (2) = \frac{1}{g' (0)} = \frac{1}{2}$

$∴$ $h' (2) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25$

Clearly, $f (x) = g (x)$ has 3 solutions.

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