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JEE Advanced · Mathematics · 24. Differentiation

Let \(\mathbb{R}\) denote the set of all real numbers. Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) and \(g: \mathbb{R} \rightarrow(0,4)\) be functions defined by \(f(x)=\log _e\left(x^2+2 x+4\right)\), and \(g(x)=\frac{4}{1+e^{-2 x}}\)
Define the composite function \(f \circ g^{-1}\) by \(\left(f \circ g^{-1}\right)(x)=\left(g^{-1}(x)\right)\), 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 ________ .

  1. A 0.23
  2. B 0.94
  3. C 0.25
  4. D 0.12
Verified Solution

Answer & Solution

Correct Answer

(C) 0.25

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \text { Let } \mathrm{h}(\mathrm{x})=\mathrm{f}\left(\mathrm{g}^{-1}(\mathrm{x})\right) \text { and } \mathrm{g}(0)=2 \\ & \begin{aligned} \mathrm{h}^{\prime}(\mathrm{x}) & =\mathrm{f}^{\prime}\left(\mathrm{g}^{-1}(\mathrm{x})\right) \cdot\left(\mathrm{g}^{-1}(\mathrm{x})\right)^{\prime} \\ \mathrm{h}^{\prime}(2) & =\mathrm{f}^{\prime}\left(\mathrm{g}^{-1}(2) \cdot\left(\mathrm{g}^{-1}\right)^{\prime}(2)\right. \\ & =\mathrm{f}^{\prime}(0) \cdot\left(\mathrm{g}^{-1}\right)^{\prime}(2)\end{aligned}\end{aligned}\)
Now
\(\begin{aligned}
& f(x)=\log _0\left(x^2+2 x+4\right) \\
& f^{\prime}(x)=\frac{2 x+2}{x^2+2 x+4} \\
& f^{\prime}(0)=\frac{1}{2} \\
& g(x)=\frac{4}{1+e^{-2 x}}, g(0)=2 \\
& g^{-1}(g(x))=x
\end{aligned}\)
\begin{array}{l|l}\left(\left(\mathrm{g}^{-1}\right)^{\prime}(\mathrm{g}(\mathrm{x}))\right) \mathrm{g}^{\prime}(\mathrm{x})=1 & \mathrm{~g}(\mathrm{x})=\frac{4}{1+\mathrm{e}^{-2 \mathrm{x}}} \\\left(\mathrm{g}^{-1}\right)^{\prime}(2)=\frac{1}{\mathrm{~g}^{\prime}(0)} & \mathrm{g}^{\prime}(\mathrm{x})=\frac{8 \mathrm{e}^{-2 \mathrm{x}}}{\left(1+\mathrm{e}^{-2 \mathrm{x}}\right)^2} \\=\frac{1}{2} & \mathrm{~g}^{\prime}(0)=\frac{8}{4}=2 \\So \mathrm{h}(2)=\frac{1}{4}=0.25 &\end{array}
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