Use the Chain Rule (Calculus 2)

Question

asked Sep 25, 2021 126k views

1 Answer

Jesseplymale · Answer 1 · 2021-09-28T21:15:00+0000

1. By the chain rule,

$(\mathrm dz)/(\mathrm dt)=(\partial z)/(\partial x)(\mathrm dx)/(\mathrm dt)+(\partial z)/(\partial y)(\mathrm dy)/(\mathrm dt)$

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

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
, compute its gradient:

$\\abla w=\langle w_x,w_y,w_z\rangle=(\langle1,-1,1\rangle)/(√(1-(x-y+z)^2))}=(\langle1,-1,1\rangle)/(√(-2t-t^2))$

$\implies w_t=\frac1{√(-2t-t^2)}$

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