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

$f\left(x\right)={\log }_e(x^2+2x+4),\text{and}\mathrm{ g}\left(x\right)=\frac{4}{1+e^{-2x}}$

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

$\text{Then the value of the derivative of the composite function}f\circ g^{-1}\text{at}\mathrm{ x}=\mathrm{2 }\text{is \_\_\_\_\_\_}$

The correct answer is 0.25.

We are given two functions:
$f:R\to R\text{defined by}f\left(x\right)={\log }_e(x^2+2x+4)$
and
$g:R\to (0,4)\text{defined by}g\left(x\right)=\frac{4}{1+e^{-2x}}$

We need to find the value of the derivative of the composite function $h\left(x\right)=(f\circ g^{-1})\left(x\right)=f\left(g^{-1}\right(x\left)\right)$ at $x=2$.

Let $y=g^{-1}\left(x\right)$. Then $g\left(y\right)=x$.
For $x=2$, we have:
$g\left(y\right)=2\Rightarrow \frac{4}{1+e^{-2y}}=2$
Dividing both sides by 2:
$1+e^{-2y}=2\Rightarrow e^{-2y}=1$
This gives:
$-2y=0\Rightarrow y=0$
So, $g^{-1}\left(2\right)=0$.

By the chain rule, the derivative of $h\left(x\right)$ is:
$h^{\prime }\left(x\right)=f^{\prime }\left(g^{-1}\right(x\left)\right)·{\left(g^{-1}\right)}^{\prime }\left(x\right)$

Using the formula for the derivative of an inverse function, ${\left(g^{-1}\right)}^{\prime }\left(x\right)=\frac{1}{g^{\prime }\left(g^{-1}\right(x\left)\right)}$.
Thus, at $x=2$:
$h^{\prime }\left(2\right)=\frac{f^{\prime }\left(0\right)}{g^{\prime }\left(0\right)}$

Now we calculate $f^{\prime }\left(x\right)$:
$f^{\prime }\left(x\right)=\frac{d}{dx}\left[{\log }_e\right(x^2+2x+4\left)\right]=\frac{2x+2}{x^2+2x+4}$
Evaluating at $x=0$:
$f^{\prime }\left(0\right)=\frac{2}{4}$

Next, we calculate $g^{\prime }\left(x\right)$:
$g^{\prime }\left(x\right)=\frac{d}{dx}\left[4{(1+e^{-2x})}^{-1}\right]=-4{(1+e^{-2x})}^{-2}·(-2e^{-2x})=\frac{8e^{-2x}}{{(1+e^{-2x})}^2}$
Evaluating at $x=0$:
$g^{\prime }\left(0\right)=\frac{8·1}{{(1+1)}^2}=\frac{8}{4}=2$

Finally, we substitute the values back into the derivative of the composition:
$h^{\prime }\left(2\right)=\frac{f^{\prime }\left(0\right)}{g^{\prime }\left(0\right)}=\frac{1/2}{2}=\frac{1}{4}=0.25$