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