Chain rule in 3 dimensions

Question

How would I approach setting up this problem?

Q: Use the Chain Rule to find $\frac{dz}{dt}$ at $t=1$. Where $z=\cos{x+3y}$, $x=2t^4$, $y=-\frac{2}{t}$.

Answer 1

$z$ is a function of $x(t)$ and $y(t)$, if you set $z$ in terms of $t$ you can take the total derivative of $z$ which is the derivative for one variable function. Otherwise you'll need to apply the chain rule $\frac{dz}{dt}=\frac{dz}{dx}\frac{dx}{dt}+\frac{dz}{dy}\frac{dy}{dt}$ $$\frac{dz}{dt}=-sin(x)(8t^3)+3(\frac{2}{t^2})=-8t^3sin(2t^4)+\frac{6}{t^2}$$

Answer 2

See here for general higher-dimensional chain rule. For your problem it becomes $$\frac{dz(x(t),y(t))}{dt}=\frac{dz(x,y)}{dx}\frac{dx(t)}{dt}+\frac{dz(x,y)}{dy}\frac{dy(t)}{dt}$$ where on the left we are considering $z$ as a function of $t$ because $z$ depends on $x$ and $y$ which in turn depend on $t$; on the right we consider $z$ just as a function of $x$ and $y$ independent of $t$.

Answer 3

$$z=\cos{x+3y}$$

$$\frac {dz}{dx}=-\sin x, \frac {dz}{dy}=3$$

$$ x=2t^4, y=-\frac{2}{t}$$

$$\frac {dx}{dt}=8t^3, \frac {dy}{dt}=\frac {2}{t^2}$$

$$\frac {dz}{dt}= \frac {dz}{dx}\frac {dx}{dt}+\frac {dz}{dy}\frac {dy}{dt}=$$

$$\frac{dz}{dt}=-\sin(x)(8t^3)+3(\frac{2}{t^2})=-8t^3\sin(2t^4)+\frac{6}{t^2}=6-8\sin(2)$$