Using chain rule to find $~\frac{dz}{dt}~$, leaving answer in terms of $~t~$ Can anyone tell me if I have done this correctly
Using the chain rule Find $~   \frac{dz}{dt} ~$
if $$~ z=xy^2,~~~~ x=e^{-3t} , ~~~~y=-\sin(2t) ~ $$
Leaving answer in terms of $~t~$

$$   \frac{dz}{dt} =\frac{\partial z}{\partial x}  \cdot  \frac{d x}{d t} + \frac{\partial z}{\partial y}  \cdot  \frac{dy}{d t}$$
$$   \frac{dz}{dt} = y^2 \cdot (-3)e^{-3t}+2xy \cdot (-2)\cos(2t)$$
$$   \frac{dz}{dt} = \sin^2(2t)\cdot(-3)e^{-3t}+2e^{-3t}\cdot(-)\sin(2t) \cdot(-2)\cos(2t)$$
Many thanks in advance
 A: We need to find:
$$\text{z}'(t)=\frac{\text{d}\text{z}(t)}{\text{d}t}=\frac{\text{d}}{\text{d}t}\left(e^{-3t}\cdot\left(-\sin(2t)\right)^2\right)=$$
$$\frac{\text{d}}{\text{d}t}\left(e^{-3t}\sin^2(2t)\right)=e^{-3t}\sin(2t)\left(4\cos(2t)-3\sin(2t)\right)$$

Using:


*

*The product rule:
$$\frac{\text{d}}{\text{d}t}\left(f(t)\cdot y(t)\right)=f(t)\cdot\frac{\text{d}}{\text{d}t}\left(y(t)\right)+y(t)\cdot\frac{\text{d}}{\text{d}t}\left(f(t)\right)=y(t)\cdot f'(t)+f(t)\cdot y'(t)$$

*$$\frac{\text{d}}{\text{d}t}\left(e^{x(t)}\right)=e^{x(t)}\cdot\frac{\text{d}}{\text{d}t}\left(x(t)\right)=x'(t)\cdot e^{x(t)}$$

*When $\text{C}$ is a constant:
$$\frac{\text{d}}{\text{d}t}\left(\text{C}\cdot q(t)\right)=\text{C}\cdot\frac{\text{d}}{\text{d}t}\left(q(t)\right)=\text{C}\cdot q'(t)$$

*When $\text{n}$ is a constant, using the chain rule:
$$\frac{\text{d}}{\text{d}t}\left(w(t)^\text{n}\right)=\text{n}\cdot w(t)^{\text{n}-1}\cdot\frac{\text{d}}{\text{d}t}\left(w(t)\right)=\text{n}\cdot w(t)^{\text{n}-1}\cdot w'(t)$$

*When $\text{m}$ is a constant, using the chain rule:
$$\frac{\text{d}}{\text{d}t}\left(v(\text{m}t)\right)=\frac{\text{d}}{\text{d}t}\left(\text{m}t\right)\cdot v'(\text{m}t)=\text{m}\cdot v'(\text{m}t)$$

