Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
88 LECTURE 12. THE CHAIN RULE Consequently f(t + ∆t) − f(t) ∆t = = = = = f(x + ∆x, y + ∆y) − f(x, y + ∆y) + f(x, y + ∆y) − f(x, y) ∆t f(x + ∆x, y + ∆y) − f(x, y + ∆y) ∆t f(x, y + ∆y) − f(x, y) + ∆t f(x + ∆x, y + ∆y) − f(x, y + ∆y) ∆x ∆t ∆x f(x, y + ∆y) − f(x, y) ∆y + ∆t ∆y f(x + ∆x, y + ∆y) − f(x, y + ∆y) ∆x ∆x ∆t f(x, y + ∆y) − f(x, y) ∆y + ∆y ∆t f(x + ∆x, y + ∆y) − f(x, y + ∆y) x(t + ∆t) − x(t) ∆x ∆t f(x, y + ∆y) − f(x, y) y(t + ∆t) − y(t) + ∆y ∆t Therefore df dt f(t + ∆t) − f(t) = lim ∆t→0 ∆t = lim ∆t→0 = ∂f dx ∂x dt + ∂f ∂y f(x + ∆x, y + ∆y) − f(x, y + ∆y) ∆x f(x, y + ∆y) − f(x, y) + lim ∆t→0 ∆y dy dt lim ∆t→0 x(t + ∆t) − x(t) lim ∆t→0 ∆t y(t + ∆t) − y(t) ∆t Implicit Differentiation using the Chain Rule. Suppose that f(x, y) = 0 and we want to find dy/dx, but we can’t solve for y as a function of x. We can solve this problem using implicit differentiation, as we did for a function of a single variable. Example 12.6 Find dy/dx for x 3 + 2x 2 y − 10y 5 = 0. Revised December 6, 2006. Math 250, Fall 2006
LECTURE 12. THE CHAIN RULE 89 Solution. Using implicit differentiation, we differentiate both sides of the equation with respect to x and solve for dy/dx x 3 + 2x 2 y − 10y 5 = 0 ⇒ d dx (x3 + 2x 2 y − 10y 5 ) = d dx (0) ⇒ d dx (x3 ) + 2 d dx (x2 y) − 10 d dx (y5 ) = 0 [ ⇒ 3x 2 + 2 x 2 dy dx + y d ] dx x2 − 10(5)y 4 dy dx = 0 ⇒ 3x 2 + 2x 2 dy dy + 4xy − 50y4 dx dx = 0 Now bring all the terms that have a dy/dx in them to one side of the equation: Factor dy/dx on the left: Solving for dy/dx 2x 2 dy dy − 50y4 = −4xy − 3x2 dx dx dy dx (2x2 − 50y 4 ) = −4xy − 3x 2 dy −4xy − 3x2 = dx 2x 2 − 50y 4 Now consider the same problem of finding dy/dx when f is a function of two parameterized variables f(x(t), y(t). Since dx/dx = 1 0 = df(x(t), y(t)) dx Hence we have the following result. 0 = ∂f ∂x + ∂f dy ∂y dx ⇒ ∂f dy ∂y dx = −∂f ∂x = ∂f dx ∂x dx + ∂f dy ∂y dx Theorem 12.1 Implicit Differentiation If f(x, y) = 0 then By a similar argument, we also have dy dx = −∂f/∂x ∂f/∂y Theorem 12.2 Implicit Differentiation If f(x, y, z) = 0 then ∂z ∂x = −∂f/∂x ∂f/∂z and ∂z ∂y = −∂f/∂y ∂f/∂z Math 250, Fall 2006 Revised December 6, 2006.
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LECTURE 12. THE CHAIN RULE 89<br />
Solution. Using implicit differentiation, we differentiate both sides of the equation<br />
with respect to x and solve for dy/dx<br />
x 3 + 2x 2 y − 10y 5 = 0<br />
⇒<br />
d<br />
dx (x3 + 2x 2 y − 10y 5 ) = d<br />
dx (0)<br />
⇒<br />
d<br />
dx (x3 ) + 2 d<br />
dx (x2 y) − 10 d<br />
dx (y5 ) = 0<br />
[<br />
⇒ 3x 2 + 2 x 2 dy<br />
dx + y d ]<br />
dx x2 − 10(5)y 4 dy<br />
dx = 0<br />
⇒<br />
3x 2 + 2x 2 dy<br />
dy<br />
+ 4xy − 50y4<br />
dx dx = 0<br />
Now bring all the terms that have a dy/dx in them to one side of the equation:<br />
Factor dy/dx on the left:<br />
Solving for dy/dx<br />
2x 2 dy dy<br />
− 50y4 = −4xy − 3x2<br />
dx dx<br />
dy<br />
dx (2x2 − 50y 4 ) = −4xy − 3x 2<br />
dy −4xy − 3x2<br />
=<br />
dx 2x 2 − 50y 4<br />
Now consider the same problem of finding dy/dx when f is a function of two<br />
parameterized variables f(x(t), y(t).<br />
Since dx/dx = 1<br />
0 =<br />
df(x(t), y(t))<br />
dx<br />
Hence we have the following result.<br />
0 = ∂f<br />
∂x + ∂f dy<br />
∂y dx<br />
⇒ ∂f dy<br />
∂y dx = −∂f ∂x<br />
<br />
= ∂f dx<br />
∂x dx + ∂f dy<br />
∂y dx<br />
Theorem 12.1 Implicit Differentiation If f(x, y) = 0 then<br />
By a similar argument, we also have<br />
dy<br />
dx = −∂f/∂x ∂f/∂y<br />
Theorem 12.2 Implicit Differentiation If f(x, y, z) = 0 then<br />
∂z<br />
∂x = −∂f/∂x ∂f/∂z<br />
and<br />
∂z<br />
∂y = −∂f/∂y ∂f/∂z<br />
Math 250, Fall 2006 Revised December 6, 2006.
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