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The first **derivative** **dy/dx** gives us the rate of change of **y** with respect to **x **which is $(e^{(-t)} – t * e^{(-t)})$ / $e^t$ and the second derivative **d²y/dx² **is [(- $e^{(-t)} + t * e^{(-t)}) * e^t – (e^{(-t)} – t * e^{(-t)}) * e^t] / e^{(2t)}$ = -2.

To find **dy/dx** and **d²y/dx²** at **x = $e^t$, y = t$e^{(-t)}$**, we’ll differentiate **y** with respect to **x** using the **chain rule**. By differentiating **y** with respect to **x** using the chain rule, we aim to uncover valuable insights into the slope and **curvature** of this **curve**.

## Expert Answer

First, we need to calculate **dy/dt** and **dx/dt**, which will allow us to find **dy/dx** using the chain rule.

Given: x = $e^t$ and y = t$e^{(-t)}$

Let’s start with the **derivatives**:

dx/dt: To find **dx/dt**, we differentiate **x** with respect to **t**:

dx/dt = d($e^t$)/dt = $e^t$

dy/dt: To find **dy/dt**, we **differentiate y** with respect to** t**:

dy/dt = d(te^(-t))/dt = e^(-t) – t * e^(-t)

Now that we have **dx/dt** and **dy/dt**, we can calculate **dy/dx**:

dy/dx = (dy/dt) / (dx/dt)

= $(e^{(-t)} – t * e^{(-t)})$ / $e^t$

This is the first **derivative**, which gives us the rate of change of **y** with respect to **x**.

Next, let’s find **d²y/dx²**, the **second derivative**. To do this, we’ll differentiate **dy/dx** with respect to** x**:

d²y/dx² = d/dx[$(e^{(-t)} – t * e^{(-t)})$ / $e^t$]

Using the quotient rule and chain rule, we **differentiate** each term separately:

d/dx$(e^{(-t)} – t * e^{(-t)})$ = – $e^{(-t)} – (-t * e^{(-t)})$ =

= – $e^{(-t)} + t * e^{(-t)} d/dx(e^t)$

= e^t

Now, we can apply the **quotient rule**:

d²y/dx² = [(- $e^{(-t)} + t * e^{(-t)}) * e^t – (e^{(-t)} – t * e^{(-t)}) * e^t] / (e^t)^2$

Simplifying the expression:

d²y/dx² = [(- $e^{(-t)} + t * e^{(-t)}) * e^t – (e^{(-t)} – t * e^{(-t)}) * e^t] / e^{(2t)}$

Simplify the expression:

d²y/dx² = -2

So, **d²y/dx²** is a constant value of **-2**.

## Numerical Result

The first derivative **dy/dx **is **$(e^{(-t)} – t * e^{(-t)})$ / $e^t$** and the second derivative **d²y/dx²** is **-2.**

To determine where the curve is **concave upward**, we need to find where **d²y/dx²** is positive. Since **d²y/dx²** is always -2, the curve is never concave upward. Therefore, the curve is concave downward for all values of **t**.

## Example

Calculate the derivatives for **x = $e^t$** and **y = t * $e^{(-t)}$** to find **dy/dx** and **d²y/dx² **at **t = 2**.

## Solution

First, we’ll find the values of **x** and **y** at **t = 2**:

x = $e^2$

≈7.389

y = 2 * $e^{(-2)}$

≈0.2707

Now, let’s calculate **dy/dx** at **t = 2** using the derivatives we previously derived:

dy/dx = ($e^{(-2)} – 2 * e^{(-2)})/e^2$

≈ (0.1353 – 0.5415)/7.389

≈ -0.0513

Next, let’s find **d²y/dx²** at **t = 2**:

d²y/dx² = [$(-e^{(-2)} + 2 * e^{(-2)}) * e^2 – (e^{(-2)} – 2 * e^{(-2)}) * e^2]/e^{(4)} $

≈ [(-0.1353 + 0.5415) * 7.389-(-0.1353 + 0.5415) * 7.389]/54.598

≈ 0