Question

Let $f(x)=2 x+e^x-e^{-x}$ and $g(x)$ is inverse of $f(x)$, then $g^{\prime}(x)$ is

• A. $\frac{1}{2+\sqrt{(2 g(x)-x)^2-4}}$
• B. $\frac{1}{2-\sqrt{(2 g(x)-x)^2+4}}$
• C. $\frac{1}{2+\sqrt{(2 g(x)-x)^2+4}}$
• D. None of these

Answer 1

Answer: (C)

$ g^{\prime}(f(x))=\frac{1}{2+e^x+e^{-x}}=\frac{1}{2+\sqrt{\left(e^x-e^{-x}\right)^2+4}}$
$x \rightarrow g(x) $
$g^{\prime}(x)=\frac{1}{2+\sqrt{(x-2 g(x))^2+4}} $