Question

Use the chain rule to find dz/dt for the given equation z = x² y² xy, where x = sin(t) and y = 9eᵗ.

Answer 1

`\(\frac{{dz}}{{dt}} = 2\sin(t) \cdot 9e^t + \sin(t) \cdot 2 \cdot 9e^t\)`

Step-by-step explanation:

`To find \(\frac{{dz}}{{dt}}\) using the chain rule for the given equation \(z = x^2y^2xy\), where \(x = \sin(t)\) and \(y = 9e^t\), let's break down the process.`

`First, we calculate the partial derivatives of \(z\) with respect to \(x\) and \(y\):`

`\[ \frac{{dz}}{{dx}} = 2xy^2y + y^2 \cdot 2x = 2xy^3 + 2x^2y^2 \]`

`\[ \frac{{dz}}{{dy}} = x^2 \cdot 2xy + x^2y^2 \cdot 2 = 2x^2y^2 + 2x^2y \]`

`Next, we substitute \(x = \sin(t)\) and \(y = 9e^t\) into these expressions:`

`\[ \frac{{dz}}{{dx}} = 2\sin(t) \cdot (9e^t)^3 + 2(\sin(t))^2 \cdot (9e^t)^2 \]`

Now, we apply the chain rule formula:

`\[ \frac{{dz}}{{dt}} = \frac{{dz}}{{dx}} \cdot \frac{{dx}}{{dt}} + \frac{{dz}}{{dy}} \cdot \frac{{dy}}{{dt}} \]`

`\[ \frac{{dz}}{{dt}} = [2\sin(t) \cdot (9e^t)^3 + 2(\sin(t))^2 \cdot (9e^t)^2] \cdot \cos(t) + [(\sin(t))^2 \cdot 2 \cdot 9e^t + (\sin(t))^2 \cdot (9e^t)^2] \cdot 9e^t \]`

Simplifying further gives the final result:

`\[ \frac{{dz}}{{dt}} = 2\sin(t) \cdot 9e^t + \sin(t) \cdot 2 \cdot 9e^t \]`

This detailed calculation ensures a step-by-step understanding of how the chain rule is applied to find the derivative with respect to time
`(\(\frac{{dz}}{{dt}}\)).`