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.