14.5 The Chain Rule - Michigan State University

¡ì14.5 The Chain Rule

The Chain Rule

The Chain Rule (Case 1)

The Chain Rule (Case 2)

The Chain Rule (General Version)

Implicit Differentiation

Let us first recall the chain rule for functions of a single variable.

Given y = f (x) and z = g (y ), the derivative of the composition

z = g ? f (x) = g (f (x)) is

dz dy

dz

=

dx

dy dx

or

[g ? f (x)]0 = g 0 (f (x))f 0 (x)

Diagram:

d

d

z ¡úy ¡úx

The Chain Rule (Case 1)

Suppose z = f (x, y ), x = g (t) and y = h(t), and assume that all

functions are differentiable. Then z is a differentiable function of t

given by the composition z = f (g (t), h(t)) with derivative

dz

?z dx

?z dy

=

+

dt

?x dt

?y dt

Tree Diagram:

z

.

&

&

.

x

y

t

Example

dz

If z = x 2 y + 3xy 4 , where x = sin(2t) and y = cos t, find

when

dt

t = 0.

Solution.

dz

dt

?z dx

?z dy

+

?x dt

?y dt

= (2xy + 3y 4 ) cos(2t) ¡¤ 2 + (x 2 + 3x ¡¤ 4y 3 )(? sin t)

=

= 2(2xy + 3y 4 ) cos(2t) ? (x 2 + 12xy 3 ) sin t

Example

dz

If z = x 2 y + 3xy 4 , where x = sin(2t) and y = cos t, find

when

dt

t = 0.

Solution.

dz

dt

?z dx

?z dy

+

?x dt

?y dt

= (2xy + 3y 4 ) cos(2t) ¡¤ 2 + (x 2 + 3x ¡¤ 4y 3 )(? sin t)

=

= 2(2xy + 3y 4 ) cos(2t) ? (x 2 + 12xy 3 ) sin t

When t = 0, x = sin(2 ¡¤ 0) = 0 and y = cos 0 = 1, so

dz

|t=0 = 2(2 ¡¤ 0 ¡¤ 1 + 3 ¡¤ 14 ) cos(2 ¡¤ 0) ? (02 + 12 ¡¤ 0 ¡¤ 13 ) sin 0 = 6.

dt

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download