Answer
$$
y=\frac{1}{e^{x^{2}}+1},\quad \quad [\frac{dx}{dt}=3,\quad x=1]
$$
$$
\frac{d y}{d t}= -2
$$
Work Step by Step
$$
y=\frac{1}{e^{x^{2}}+1},\quad \quad [\frac{dx}{dt}=3,\quad x=1]
$$
Now, we can calculate $\frac{dy}{dt}$ by using the quotient rule. as follows:
$$
\begin{aligned} \frac{d y}{d t} &=\frac{\left[\begin{array}{c}{(1-\sqrt{x})\left(\frac{1}{2} x^{-1 / 2} \frac{d x}{d t}\right)} \\ {-1(1+\sqrt{x})\left(-\frac{1}{2}\right)\left(x^{-1 / 2} \frac{d x}{d t}\right)}\end{array}\right]}{(1-\sqrt{x})^{2}} \\ \text { when } [\frac{dx}{dt}=-4,\quad x=4]
& \\ \frac{d y}{d t} &=\frac{\left[\begin{array}{c}{(1-\sqrt{4})\left(\frac{1}{2} 4^{-1 / 2} (-4)\right)} \\ {-1(1+\sqrt{4})\left(-\frac{1}{2}\right)\left(4^{-1 / 2} (-4)\right)}\end{array}\right]}{(1-\sqrt{4})^{2}}\\
&=\frac{\left[\begin{array}{c}{(1-2)\left(\frac{1}{2 \cdot 2}\right)(-4)} \\ {-(1+2)\left(\frac{-1}{2-2}\right)(-4)}\end{array}\right]}{(1-2)^{2}}\\
&=\frac{1-3}{1}\\
&=-2
\end{aligned}
$$
