To find the derivative of \( v_y(t) = v_0 e^{-\frac{(t - t_0)^2}{2 \sigma^2}} \), we will use the chain rule and the properties of exponential functions. Here's the step-by-step process:

### Step 1: Identify the structure
The function has the form:
\[
v_y(t) = v_0 e^{u(t)}
\]
where:
\[
u(t) = -\frac{(t - t_0)^2}{2 \sigma^2}
\]

### Step 2: Differentiate the outer function
The derivative of \( v_y(t) \) with respect to \( t \) is:
\[
\frac{d}{dt} v_y(t) = v_0 \cdot \frac{d}{dt} \left( e^{u(t)} \right) = v_0 e^{u(t)} \cdot \frac{du(t)}{dt}
\]

### Step 3: Differentiate the exponent (inner function)
Now, we find the derivative of \( u(t) \):
\[
u(t) = -\frac{(t - t_0)^2}{2 \sigma^2}
\]
Using the chain rule:
\[
\frac{du(t)}{dt} = -\frac{1}{2 \sigma^2} \cdot 2 (t - t_0) = -\frac{(t - t_0)}{\sigma^2}
\]

### Step 4: Combine the results
Now, substitute \( \frac{du(t)}{dt} \) into the derivative of \( v_y(t) \):
\[
\frac{d}{dt} v_y(t) = v_0 e^{-\frac{(t - t_0)^2}{2 \sigma^2}} \cdot \left( -\frac{(t - t_0)}{\sigma^2} \right)
\]

### Final Answer:
\[
\frac{d}{dt} v_y(t) = -v_0 \frac{(t - t_0)}{\sigma^2} e^{-\frac{(t - t_0)^2}{2 \sigma^2}}
\]

This is the derivative of the function \( v_y(t) \).

Would you like further details or any clarification?

Here are 5 related questions:
1. How is the chain rule applied in differentiating exponential functions?
2. What is the significance of the Gaussian function in this context?
3. How do you handle higher-order derivatives of this function?
4. What are the physical interpretations of \( \sigma \), \( t_0 \), and \( v_0 \)?
5. How would the result change if the exponent was a different quadratic term?

**Tip:** When differentiating exponentials, always focus on identifying the outer function (exponential) and the inner function (the exponent).

Learn how to differentiate the function vy(t) = v0 e^(-(t-t0)^2 / (2σ^2)) using the chain rule. This solution provides a step-by-step process involving Gaussian functions, exponential differentiation, and the chain rule. Ideal for university-level calculus students.

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