1. Home
  2. Calculus volume 3
  3. Differentiation of functions of
  4. The chain rule

4.5 The chain rule  (Page 4/9)

<< Chapter < Page
  Calculus volume 3     Page 4 / 9
Chapter >> Page >

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 ) .
Got questions? Get instant answers now!
Got questions? Get instant answers now!

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

Got questions? Get instant answers now!

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 .
Got questions? Get instant answers now!
Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!

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 .

Questions & Answers

Explain the following terms . (1) Abiotic factors in an ecosystem
Nomai Reply
Abiotic factors are non living components of ecosystem.These include physical and chemical elements like temperature,light,water,soil,air quality and oxygen etc
Qasim
what is biology
daniel Reply
what is diffusion
Emmanuel Reply
passive process of transport of low-molecular weight material according to its concentration gradient
AI-Robot
what is production?
Catherine
Pathogens and diseases
Reply
how did the oxygen help a human being
Achol Reply
how did the nutrition help the plants
Achol Reply
Biology is a branch of Natural science which deals/About living Organism.
Ahmedin Reply
what is phylogeny
Odigie Reply
evolutionary history and relationship of an organism or group of organisms
AI-Robot
ok
Deng
what is biology
Hajah Reply
cell is the smallest unit of the humanity biologically
Abraham
what is biology
Victoria Reply
what is biology
Abraham
HOW CAN MAN ORGAN FUNCTION
Alfred Reply
the diagram of the digestive system
Assiatu Reply
allimentary cannel
Ogenrwot
How does twins formed
William Reply
They formed in two ways first when one sperm and one egg are splited by mitosis or two sperm and two eggs join together
Oluwatobi
what is respiration
Deborah
what is genetics
Josephine Reply
Genetics is the study of heredity
Misack
how does twins formed?
Misack
What is manual
Hassan Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
<< Chapter < Page Page > Chapter >>
Practice Key Terms 3

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 3' conversation and receive update notifications?

Ask