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4.5 The chain rule  (Page 4/9)

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In the next example we calculate the derivative of a function of three independent variables in which each of the three variables is dependent on two other variables.

Using the generalized chain rule

Calculate ∂ w / ∂ u and ∂ w / ∂ v using the following functions:

w = f ( x , y , z ) = 3 x 2 − 2 x y + 4 z 2 x = x ( u , v ) = e u sin v y = y ( u , v ) = e u cos v z = z ( u , v ) = e u .

The formulas for ∂ w / ∂ u and ∂ w / ∂ v are

∂ w ∂ u = ∂ w ∂ x · ∂ x ∂ u + ∂ w ∂ y · ∂ y ∂ u + ∂ w ∂ z · ∂ z ∂ u ∂ w ∂ v = ∂ w ∂ x · ∂ x ∂ v + ∂ w ∂ y · ∂ y ∂ v + ∂ w ∂ z · ∂ z ∂ v .

Therefore, there are nine different partial derivatives that need to be calculated and substituted. We need to calculate each of them:

∂ w ∂ x = 6 x − 2 y ∂ w ∂ y = −2 x ∂ w ∂ z = 8 z ∂ x ∂ u = e u sin v ∂ y ∂ u = e u cos v ∂ z ∂ u = e u ∂ x ∂ v = e u cos v ∂ y ∂ v = − e u sin v ∂ z ∂ v = 0 .

Now, we substitute each of them into the first formula to calculate ∂ w / ∂ u :

∂ w ∂ u = ∂ w ∂ x · ∂ x ∂ u + ∂ w ∂ y · ∂ y ∂ u + ∂ w ∂ z · ∂ z ∂ u = ( 6 x − 2 y ) e u sin v − 2 x e u cos v + 8 z e u ,

then substitute x ( u , v ) = e u sin v , y ( u , v ) = e u cos v , and z ( u , v ) = e u into this equation:

∂ w ∂ u = ( 6 x − 2 y ) e u sin v − 2 x e u cos v + 8 z e u = ( 6 e u sin v − 2 e u cos v ) e u sin v − 2 ( e u sin v ) e u cos v + 8 e 2 u = 6 e 2 u sin 2 v − 4 e 2 u sin v cos v + 8 e 2 u = 2 e 2 u ( 3 sin 2 v − 2 sin v cos v + 4 ) .

Next, we calculate ∂ w / ∂ v :

∂ w ∂ v = ∂ w ∂ x · ∂ x ∂ v + ∂ w ∂ y · ∂ y ∂ v + ∂ w ∂ z · ∂ z ∂ v = ( 6 x − 2 y ) e u cos v − 2 x ( − e u sin v ) + 8 z ( 0 ) ,

then we substitute x ( u , v ) = e u sin v , y ( u , v ) = e u cos v , and z ( u , v ) = e u into this equation:

∂ w ∂ v = ( 6 x − 2 y ) e u cos v − 2 x ( − e u sin v ) = ( 6 e u sin v − 2 e u cos v ) e u cos v + 2 ( e u sin v ) ( e u sin v ) = 2 e 2 u sin 2 v + 6 e 2 u sin v cos v − 2 e 2 u cos 2 v = 2 e 2 u ( sin 2 v + sin v cos v − cos 2 v ) .
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Calculate ∂ w / ∂ u and ∂ w / ∂ v given the following functions:

w = f ( x , y , z ) = x + 2 y − 4 z 2 x − y + 3 z x = x ( u , v ) = e 2 u cos 3 v y = y ( u , v ) = e 2 u sin 3 v z = z ( u , v ) = e 2 u .

∂ w ∂ u = 0 ∂ w ∂ v = 15 − 33 sin 3 v + 6 cos 3 v ( 3 + 2 cos 3 v − sin 3 v ) 2

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Drawing a tree diagram

Create a tree diagram for the case when

w = f ( x , y , z ) , x = x ( t , u , v ) , y = y ( t , u , v ) , z = z ( t , u , v )

and write out the formulas for the three partial derivatives of w .

Starting from the left, the function f has three independent variables: x , y , and z . Therefore, three branches must be emanating from the first node. Each of these three branches also has three branches, for each of the variables t , u , and v .

A diagram that starts with w = f(x, y, z). Along the first branch, it is written ∂w/∂x, then x = x(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then ∂w/∂x ∂x/∂t; the second subbranch says u and then ∂w/∂x ∂x/∂u; and the third subbranch says v and then ∂w/∂x ∂x/∂v. Along the second branch, it is written ∂w/∂y, then y = y(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then ∂w/∂y ∂y/∂t; the second subbranch says u and then ∂w/∂y ∂y/∂u; and the third subbranch says v and then ∂w/∂y ∂y/∂v. Along the third branch, it is written ∂w/∂z, then z = z(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then ∂w/∂z ∂z/∂t; the second subbranch says u and then ∂w/∂z ∂z/∂u; and the third subbranch says v and then ∂w/∂z ∂z/∂v.
Tree diagram for a function of three variables, each of which is a function of three independent variables.

The three formulas are

∂ w ∂ t = ∂ w ∂ x ∂ x ∂ t + ∂ w ∂ y ∂ y ∂ t + ∂ w ∂ z ∂ z ∂ t ∂ w ∂ u = ∂ w ∂ x ∂ x ∂ u + ∂ w ∂ y ∂ y ∂ u + ∂ w ∂ z ∂ z ∂ u ∂ w ∂ v = ∂ w ∂ x ∂ x ∂ v + ∂ w ∂ y ∂ y ∂ v + ∂ w ∂ z ∂ z ∂ v .
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Create a tree diagram for the case when

w = f ( x , y ) , x = x ( t , u , v ) , y = y ( t , u , v )

and write out the formulas for the three partial derivatives of w .

∂ w ∂ t = ∂ w ∂ x ∂ x ∂ t + ∂ w ∂ y ∂ y ∂ t ∂ w ∂ u = ∂ w ∂ x ∂ x ∂ u + ∂ w ∂ y ∂ y ∂ u ∂ w ∂ v = ∂ w ∂ x ∂ x ∂ v + ∂ w ∂ y ∂ y ∂ v
A diagram that starts with w = f(x, y). Along the first branch, it is written ∂w/∂x, then x = x(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then ∂w/∂x ∂x/∂t; the second subbranch says u and then ∂w/∂x ∂x/∂u; and the third subbranch says v and then ∂w/∂x ∂x/∂v. Along the second branch, it is written ∂w/∂y, then y = y(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then ∂w/∂y ∂y/∂t; the second subbranch says u and then ∂w/∂y ∂y/∂u; and the third subbranch says v and then ∂w/∂y ∂y/∂v.

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Implicit differentiation

Recall from Implicit Differentiation that implicit differentiation provides a method for finding d y / d x when y is defined implicitly as a function of x . The method involves differentiating both sides of the equation defining the function with respect to x , then solving for d y / d x . Partial derivatives provide an alternative to this method.

Consider the ellipse defined by the equation x 2 + 3 y 2 + 4 y − 4 = 0 as follows.

An ellipse with center near (0, –0.7), major axis horizontal and of length roughly 4.5, and minor axis of length roughly 3.
Graph of the ellipse defined by x 2 + 3 y 2 + 4 y − 4 = 0 .

This equation implicitly defines y as a function of x . As such, we can find the derivative d y / d x using the method of implicit differentiation:

d d x ( x 2 + 3 y 2 + 4 y − 4 ) = d d x ( 0 ) 2 x + 6 y d y d x + 4 d y d x = 0 ( 6 y + 4 ) d y d x = −2 x d y d x = − x 3 y + 2 .

We can also define a function z = f ( x , y ) by using the left-hand side of the equation defining the ellipse. Then f ( x , y ) = x 2 + 3 y 2 + 4 y − 4 . The ellipse x 2 + 3 y 2 + 4 y − 4 = 0 can then be described by the equation f ( x , y ) = 0 . Using this function and the following theorem gives us an alternative approach to calculating d y / d x .

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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