To find $\frac{dy}{dt}$, we need to use the chain rule. The given function is:

$y = \sqrt{e^{-2t^2} + 4}$

Step 1: Apply the chain rule

The derivative of $y = \sqrt{u}$ with respect to $t$ is:

$\frac{dy}{dt} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dt}$

Here, $u = e^{-2t^2} + 4$. Let's find $\frac{du}{dt}$.

Step 2: Differentiate $u$ with respect to $t$

$u = e^{-2t^2} + 4$

The derivative of $e^{-2t^2}$ with respect to $t$ is found using the chain rule:

[ \frac{d}{dt}[e^{-2t^2}] = e^{-2t^2} \cdot \frac{d}{dt}[-2t^2] = e^{-2t^2} \cdot (-4t) ]

The derivative of the constant $4$ is $0$. Thus:

$\frac{du}{dt} = -4t \cdot e^{-2t^2}$

Step 3: Combine results

Now substitute $u$ and $\frac{du}{dt}$ back into the chain rule formula:

$\frac{dy}{dt} = \frac{1}{2\sqrt{e^{-2t^2} + 4}} \cdot (-4t \cdot e^{-2t^2})$

Simplify:

$\frac{dy}{dt} = \frac{-4t \cdot e^{-2t^2}}{2\sqrt{e^{-2t^2} + 4}}$

$\frac{dy}{dt} = \frac{-2t \cdot e^{-2t^2}}{\sqrt{e^{-2t^2} + 4}}$

Final Answer:

$\frac{dy}{dt} = \frac{-2t \cdot e^{-2t^2}}{\sqrt{e^{-2t^2} + 4}}$

Let me know if you'd like further clarification or details on any step!

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Relative Questions:

1. How does the chain rule apply in similar cases involving square roots?
2. What is the derivative of $e^{kt^n}$ for any constants $k$ and $n$?
3. How would the answer change if the square root was replaced by a cube root?
4. What is the significance of the $e^{-2t^2}$ term in determining the behavior of $\frac{dy}{dt}$?
5. How do we handle higher-order derivatives for this function?

Tip: Always simplify intermediate results before substituting back to avoid unnecessary complexity!