If z=xy2 and x=e−t and y=sin(t), find the following derivative using the chain rule. Enter your answer as a function of t.

dzdt=

To find the derivative \(\frac{dz}{dt}\) of the function \(z = x y^2\) with respect to \(t\), we will use the chain rule.

Given:
- \(z = x y^2\),
- \(x = e^{-t}\),
- \(y = \sin(t)\),

The derivative of \(z\) with respect to \(t\) will require the product and chain rules.

### Step 1: Differentiate \(z = x y^2\) using the product rule
We apply the product rule:
\[
\frac{dz}{dt} = \frac{d}{dt}(x y^2) = \frac{dx}{dt} y^2 + x \frac{d}{dt}(y^2)
\]

### Step 2: Find \(\frac{dx}{dt}\)
Given \(x = e^{-t}\), we differentiate:
\[
\frac{dx}{dt} = -e^{-t}
\]

### Step 3: Find \(\frac{d}{dt}(y^2)\)
Given \(y = \sin(t)\), we apply the chain rule to differentiate \(y^2\):
\[
\frac{d}{dt}(y^2) = 2y \frac{dy}{dt}
\]
Since \(\frac{dy}{dt} = \cos(t)\) (derivative of \(\sin(t)\)):
\[
\frac{d}{dt}(y^2) = 2 \sin(t) \cos(t)
\]

### Step 4: Combine the results
Now substitute everything back into the expression for \(\frac{dz}{dt}\):
\[
\frac{dz}{dt} = (-e^{-t}) \sin^2(t) + e^{-t} \cdot 2 \sin(t) \cos(t)
\]

### Final answer:
\[
\frac{dz}{dt} = -e^{-t} \sin^2(t) + 2 e^{-t} \sin(t) \cos(t)
\]

Would you like further details or have any questions?

### Related questions:
1. How would the chain rule apply if \(x\) and \(y\) were both functions of another variable?
2. Can you rewrite the result using a trigonometric identity like \(\sin(2t)\)?
3. What happens if \(y = \cos(t)\) instead of \(\sin(t)\)?
4. How does the product rule work when differentiating three or more factors?
5. What is the second derivative \(\frac{d^2z}{dt^2}\)?

### Tip:
When using the chain rule, break down the derivative into simpler steps by focusing on one variable at a time.

If z=xy^2 and x=e^−t and y=sin(t), find the following derivative using the chain rule. Enter your answer as a function of t.
dz/dt=

Learn how to differentiate z=xy^2 where x=e^−t and y=sin(t) with respect to t using the chain and product rules. This step-by-step solution breaks down calculus concepts like the chain rule and product rule, suitable for college-level calculus students.

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