Question

Use the Chain Rule (Calculus 2)

Answer 1

1. By the chain rule,

$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{\partial z}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial z}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}$

I'm going to switch up the notation to save space, so for example, $z_x$ is shorthand for $\frac{\partial z}{\partial x}$.

$z_t=z_xx_t+z_yy_t$

We have

$x=e^{-t}\implies x_t=-e^{-t}$

$y=e^t\implies y_t=e^t$

$z=\tan(xy)\implies\begin{cases}z_x=y\sec^2(xy)=e^t\sec^2(1)\\z_y=x\sec^2(xy)=e^{-t}\sec^2(1)\end{cases}$

$\implies z_t=e^t\sec^2(1)(-e^{-t})+e^{-t}\sec^2(1)e^t=0$

Similarly,

$w_t=w_xx_t+w_yy_t+w_zz_t$

where

$x=\cosh^2t\implies x_t=2\cosh t\sinh t$

$y=\sinh^2t\implies y_t=2\cosh t\sinh t$

$z=t\implies z_t=1$

To capture all the partial derivatives of $w$, compute its gradient:

$\nabla w=\langle w_x,w_y,w_z\rangle=\dfrac{\langle1,-1,1\rangle}{\sqrt{1-(x-y+z)^2}}}=\dfrac{\langle1,-1,1\rangle}{\sqrt{-2t-t^2}}$

$\implies w_t=\dfrac1{\sqrt{-2t-t^2}}$

2. The problem is asking for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$. But $z$ is already a function of $x,y$, so the chain rule isn't needed here. I suspect it's supposed to say "find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$" instead.

If that's the case, then

$z_s=z_xx_s+z_yy_s$

$z_t=z_xx_t+z_yy_t$

as the hint suggests. We have

$z=\sin x\cos y\implies\begin{cases}z_x=\cos x\cos y=\cos(s+t)\cos(s^2t)\\z_y=-\sin x\sin y=-\sin(s+t)\sin(s^2t)\end{cases}$

$x=s+t\implies x_s=x_t=1$

$y=s^2t\implies\begin{cases}y_s=2st\\y_t=s^2\end{cases}$

Putting everything together, we get

$z_s=\cos(s+t)\cos(s^2t)-2st\sin(s+t)\sin(s^2t)$

$z_t=\cos(s+t)\cos(s^2t)-s^2\sin(s+t)\sin(s^2t)$