4.5 The chain rule (Page 3/9)

Calculus volume 3 Page 3 / 9

A diagram that starts with z = f(x, y). Along the first branch, it is written ∂z/∂x, then x = x(t), then dx/dt, then t, and finally it says ∂z/∂x dx/dt. Along the other branch, it is written ∂z/∂y, then y = y(t), then dy/dt, then t, and finally it says ∂z/∂y dy/dt. — Tree diagram for the case $\frac{d z}{d t} = \frac{\partial z}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial z}{\partial y} \cdot \frac{d y}{d t} .$

In this diagram, the leftmost corner corresponds to $z = f (x, y) .$ Since $f$ has two independent variables , there are two lines coming from this corner. The upper branch corresponds to the variable $x$ and the lower branch corresponds to the variable $y .$ Since each of these variables is then dependent on one variable $t,$ one branch then comes from $x$ and one branch comes from $y .$ Last, each of the branches on the far right has a label that represents the path traveled to reach that branch. The top branch is reached by following the $x$ branch, then the $t$ branch; therefore, it is labeled $(\partial z / \partial x) \times (d x / d t) .$ The bottom branch is similar: first the $y$ branch, then the $t$ branch. This branch is labeled $(\partial z / \partial y) \times (d y / d t) .$ To get the formula for $d z / d t,$ add all the terms that appear on the rightmost side of the diagram. This gives us [link] .

In [link] , $z = f (x, y)$ is a function of $x and y,$ and both $x = g (u, v)$ and $y = h (u, v)$ are functions of the independent variables $u and v .$

Chain rule for two independent variables

Suppose $x = g (u, v)$ and $y = h (u, v)$ are differentiable functions of $u$ and $v,$ and $z = f (x, y)$ is a differentiable function of $x and y .$ Then, $z = f (g (u, v), h (u, v))$ is a differentiable function of $u and v,$ and

\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial x}{\partial u}

and

\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v} .

We can draw a tree diagram for each of these formulas as well as follows.

A diagram that starts with z = f(x, y). Along the first branch, it is written ∂z/∂x, then x = g(u, v), at which point it breaks into another two branches: the first subbranch says ∂x/∂u, then u, and finally it says ∂z/∂x ∂x/∂u; the second subbranch says ∂x/∂v, then v, and finally it says ∂z/∂x ∂x/∂v. Along the other branch, it is written ∂z/∂y, then y = h(u, v), at which point it breaks into another two branches: the first subbranch says ∂y/∂u, then u, and finally it says ∂z/∂y ∂y/∂u; the second subbranch says ∂y/∂v, then v, and finally it says ∂z/∂y ∂y/∂v. — Tree diagram for $\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial u}$ and $\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial v} .$

To derive the formula for $\partial z / \partial u,$ start from the left side of the diagram, then follow only the branches that end with $u$ and add the terms that appear at the end of those branches. For the formula for $\partial z / \partial v,$ follow only the branches that end with $v$ and add the terms that appear at the end of those branches.

There is an important difference between these two chain rule theorems. In [link] , the left-hand side of the formula for the derivative is not a partial derivative, but in [link] it is. The reason is that, in [link] , $z$ is ultimately a function of $t$ alone, whereas in [link] , $z$ is a function of both $u and v .$

Using the chain rule for two variables

Calculate $\partial z / \partial u$ and $\partial z / \partial v$ using the following functions:

z = f (x, y) = 3 x^{2} - 2 x y + y^{2}, x = x (u, v) = 3 u + 2 v, y = y (u, v) = 4 u - v .

To implement the chain rule for two variables, we need six partial derivatives— $\partial z / \partial x, \partial z / \partial y, \partial x / \partial u, \partial x / \partial v, \partial y / \partial u,$ and $\partial y / \partial v :$

\begin{matrix} \frac{\partial z}{\partial x} = 6 x - 2 y & \frac{\partial z}{\partial y} = −2 x + 2 y \\ \frac{\partial x}{\partial u} = 3 & \frac{\partial x}{\partial v} = 2 \\ \frac{\partial y}{\partial u} = 4 & \frac{\partial y}{\partial v} = −1 . \end{matrix}

To find $\partial z / \partial u,$ we use [link] :

\begin{matrix} \frac{\partial z}{\partial u} & = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial u} \\ = 3 (6 x - 2 y) + 4 (−2 x + 2 y) \\ = 10 x + 2 y . \end{matrix}

Next, we substitute $x (u, v) = 3 u + 2 v$ and $y (u, v) = 4 u - v :$

\begin{matrix} \frac{\partial z}{\partial u} & = 10 x + 2 y \\ = 10 (3 u + 2 v) + 2 (4 u - v) \\ = 38 u + 18 v . \end{matrix}

To find $\partial z / \partial v,$ we use [link] :

\begin{matrix} \frac{\partial z}{\partial v} & = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v} \\ = 2 (6 x - 2 y) + (−1) (−2 x + 2 y) \\ = 14 x - 6 y . \end{matrix}

Then we substitute $x (u, v) = 3 u + 2 v$ and $y (u, v) = 4 u - v :$

\begin{matrix} \frac{\partial z}{\partial v} & = 14 x - 6 y \\ = 14 (3 u + 2 v) - 6 (4 u - v) \\ = 18 u + 34 v . \end{matrix}

Got questions? Get instant answers now!

Calculate $\partial z / \partial u$ and $\partial z / \partial v$ given the following functions:

z = f (x, y) = \frac{2 x - y}{x + 3 y}, x (u, v) = e^{2 u} cos 3 v, y (u, v) = e^{2 u} sin 3 v .

$\frac{\partial z}{\partial u} = 0, \frac{\partial z}{\partial v} = \frac{−21}{{(3 sin 3 v + cos 3 v)}^{2}}$

Got questions? Get instant answers now!

The generalized chain rule

Now that we’ve see how to extend the original chain rule to functions of two variables, it is natural to ask: Can we extend the rule to more than two variables? The answer is yes, as the generalized chain rule states.

Generalized chain rule

Let $w = f (x_{1}, x_{2},…, x_{m})$ be a differentiable function of $m$ independent variables, and for each $i \in {1,…, m},$ let $x_{i} = x_{i} (t_{1}, t_{2},…, t_{n})$ be a differentiable function of $n$ independent variables. Then

\frac{\partial w}{\partial t_{j}} = \frac{\partial w}{\partial x_{1}} \frac{\partial x_{1}}{\partial t_{j}} + \frac{\partial w}{\partial x_{2}} \frac{\partial x_{2}}{\partial t_{j}} + ⋯ + \frac{\partial w}{\partial x_{m}} \frac{\partial x_{m}}{\partial t_{j}}

for any $j \in {1, 2,…, n} .$

<< Chapter < Page Page > Chapter >>

Practice Key Terms 3

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

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

	40 Biology 40 The Circulatory System MCQ By OpenStax Start Quiz
	Business Law Ethics By Kevin Moquin Start Quiz
	7 Java Messaging Service By JavaChamp Team Start Quiz
	17 Dr. Avery GI Bio Path quiz By Brooke Delaney Start Exam
©flickr: eLife	Mycobacterial Infections (Mandell 8th Ch 251-254) By Cath Yu Start Quiz
	1 Understanding Societies 1 By Jessica Collett Start Exam
	21 Kidney/Liver Biopath By Brooke Delaney Start Exam
	English Proficiency Test By Anindyo Mukhopadhyay Start Quiz
	30 Biology 30 Plant Form and Physiology MCQ By OpenStax Start Quiz
	Are you a 5sos fan? By Rylee Minllic Start Quiz