14.5 Videos Guide

14.5a

• The Chain Rule

  • For $z=f(x,\ y)$, where $x=x(t)$ and $y=y(t)$, $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

    • Exercise:
      Use the Chain Rule to find $\frac{dz}{dt}$.
      $z=\frac{x-y}{x+2y}$, $x=e^{\pi t}$, $y=e^{-\pi t}$

14.5b

• For $z=f(x,\ y)$, where $x=x(s,\ t)$ and $y=y(s,\ t)$,
  $\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$ and $\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$

Exercises:

• Use the Chain Rule to find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$.
  $z=\tan^{-1}{(x^{2}+y^{2})}$, $x=s\ln t$, $y=te^{s}$

14.5c

• Use the Chain Rule to find the indicated partial derivatives.
  $T=\frac{v}{2u+v}$, $u=pq\sqrt{r}$, $v=p\sqrt{q}r$;
  $\frac{\partial T}{\partial p}$, $\frac{\partial T}{\partial q}$, $\frac{\partial T}{\partial r}$ when $p=2$, $q=1$, $r=4$

14.5d

• Higher-order derivatives

14.5e

• Implicit differentiation

  • If $F(x,\ y)=0$, then $\frac{dy}{dx}=-\frac{F_{x}}{F_{y}}$, provided $F_{y}\neq0$

  • If $F(x,\ y,\ z)=0$, then $\frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}}$ and $\frac{\partial z}{\partial y}=-\frac{F_{y}}{F_{z}}$, provided $F_{z}\neq0$

Exercise:

• Use the above formula to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ for $x^{2}-y^{2}+z^{2}-2z=4$.