if (z = xy^{3}) and (x = e^{-t}) and (y=sin(t)), find the following derivative using the chain - rule. enter…

Explanation:

Step1: Apply the chain - rule

The chain - rule for $z = f(x,y)$ where $x = g(t)$ and $y = h(t)$ is $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$. First, find the partial derivatives of $z = xy^{3}$ with respect to $x$ and $y$. $\frac{\partial z}{\partial x}=y^{3}$ and $\frac{\partial z}{\partial y}=3xy^{2}$.

Step2: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

Given $x = e^{-t}$, then $\frac{dx}{dt}=-e^{-t}$. Given $y=\sin(t)$, then $\frac{dy}{dt}=\cos(t)$.

Step3: Substitute into the chain - rule formula

$\frac{dz}{dt}=y^{3}(-e^{-t})+3xy^{2}\cos(t)$.

Step4: Replace $x$ and $y$ with functions of $t$

Substitute $x = e^{-t}$ and $y=\sin(t)$ into the above expression: $\frac{dz}{dt}=\sin^{3}(t)(-e^{-t})+3e^{-t}\sin^{2}(t)\cos(t)$. Factor out $-e^{-t}\sin^{2}(t)$: $\frac{dz}{dt}=-e^{-t}\sin^{2}(t)(\sin(t) - 3\cos(t))$.

Answer:

$-e^{-t}\sin^{2}(t)(\sin(t)-3\cos(t))$