The Chain Rule

In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. In this section, we study extensions of the chain rule and learn how to take derivatives of compositions of functions of more than one variable.

Chain Rules for One or Two Independent Variables

Recall that the chain rule for the derivative of a composite of two functions can be written in the form

is a function of one variable. Or perhaps they are both functions of two variables, or even more. How would we calculate the derivative in these cases? The following theorem gives us the answer for the case of one independent variable.

Proof

The proof of this theorem uses the definition of differentiability of a function of two variables. Suppose that f is differentiable at the point

P (x_{0}, y_{0}),

where

lim_{(x, y) \to (x_{0}, y_{0})} \frac{E (x, y)}{\sqrt{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}} = 0 .

Since the first limit is equal to zero, we need only show that the second limit is finite:

both limits inside the last radical exist. Therefore, this value is finite. This proves the chain rule at

t = t_{0};

the rest of the theorem follows from the assumption that all functions are differentiable over their entire domains.

Closer examination of [link] reveals an interesting pattern. The first term in the equation is

\frac{\partial f}{\partial x} \cdot \frac{d x}{d t}

Recall that when multiplying fractions, cancelation can be used. If we treat these derivatives as fractions, then each product “simplifies” to something resembling

\partial f / d t .

that disappear in this simplification are often called intermediate variables: they are independent variables for the function

f,

is a function of two variables. This pattern works with functions of more than two variables as well, as we see later in this section.

Using the Chain Rule

Calculate $d z / d t$

for each of the following functions:

$z = f (x, y) = 4 x^{2} + 3 y^{2}, x = x (t) = sin t, y = y (t) = cos t$
$z = f (x, y) = \sqrt{x^{2} - y^{2}}, x = x (t) = e^{2 t}, y = y (t) = e^{− t}$

To use the chain rule, we need four quantities— $\partial z / \partial x, \partial z / \partial y, d x / d t,$
and
$d y / d t :$

$\begin{matrix} \frac{\partial z}{\partial x} = 8 x & \frac{\partial z}{\partial y} = 6 y \\ \frac{d x}{d t} = cos t & \frac{d y}{d t} = − sin t \end{matrix}$

Now, we substitute each of these into [link]:

$\begin{matrix} \frac{d z}{d t} & = \frac{\partial z}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial z}{\partial y} \cdot \frac{d y}{d t} \\ = (8 x) (cos t) + (6 y) (− sin t) \\ = 8 x cos t - 6 y sin t . \end{matrix}$

This answer has three variables in it. To reduce it to one variable, use the fact that
$x (t) = sin t and y (t) = cos t .$
We obtain

$\begin{matrix} \frac{d z}{d t} & = 8 x cos t - 6 y sin t \\ = 8 (sin t) cos t - 6 (cos t) sin t \\ = 2 sin t cos t . \end{matrix}$

This derivative can also be calculated by first substituting
$x (t)$
and
$y (t)$
into
$f (x, y),$
then differentiating with respect to
$t :$

$\begin{matrix} z & = f (x, y) \\ = f (x (t), y (t)) \\ = 4 {(x (t))}^{2} + 3 {(y (t))}^{2} \\ = 4 {sin}^{2} t + 3 {cos}^{2} t . \end{matrix}$

Then

$\begin{matrix} \frac{d z}{d t} & = 2 (4 sin t) (cos t) + 2 (3 cos t) (− sin t) \\ = 8 sin t cos t - 6 sin t cos t \\ = 2 sin t cos t, \end{matrix}$

which is the same solution. However, it may not always be this easy to differentiate in this form.
To use the chain rule, we again need four quantities— $\partial z / \partial x, \partial z / d y, d x / d t,$
and
$d y / d t :$

$\begin{matrix} \frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^{2} - y^{2}}} & \frac{\partial z}{\partial y} = \frac{− y}{\sqrt{x^{2} - y^{2}}} \\ \frac{d x}{d t} = 2 e^{2 t} & \frac{d x}{d t} = − e^{− t} . \end{matrix}$

We substitute each of these into [link]:

$\begin{matrix} \frac{d z}{d t} & = \frac{\partial z}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial z}{\partial y} \cdot \frac{d y}{d t} \\ = (\frac{x}{\sqrt{x^{2} - y^{2}}}) (2 e^{2 t}) + (\frac{− y}{\sqrt{x^{2} - y^{2}}}) (− e^{− t}) \\ = \frac{2 x e^{2 t} - y e^{− t}}{\sqrt{x^{2} - y^{2}}} . \end{matrix}$

To reduce this to one variable, we use the fact that
$x (t) = e^{2 t}$
and
$y (t) = e^{− t} .$
Therefore,

$\begin{matrix} \frac{d z}{d t} & = \frac{2 x e^{2 t} + y e^{− t}}{\sqrt{x^{2} - y^{2}}} \\ = \frac{2 (e^{2 t}) e^{2 t} + (e^{− t}) e^{− t}}{\sqrt{e^{4 t} - e^{−2 t}}} \\ = \frac{2 e^{4 t} + e^{−2 t}}{\sqrt{e^{4 t} - e^{−2 t}}} . \end{matrix}$

To eliminate negative exponents, we multiply the top by
$e^{2 t}$
and the bottom by
$\sqrt{e^{4 t}} :$

$\begin{matrix} \frac{d z}{d t} & = \frac{2 e^{4 t} + e^{−2 t}}{\sqrt{e^{4 t} - e^{−2 t}}} \cdot \frac{e^{2 t}}{\sqrt{e^{4 t}}} \\ = \frac{2 e^{6 t} + 1}{\sqrt{e^{8 t} - e^{2 t}}} \\ = \frac{2 e^{6 t} + 1}{\sqrt{e^{2 t} (e^{6 t} - 1)}} \\ = \frac{2 e^{6 t} + 1}{e^{t} \sqrt{e^{6 t} - 1}} . \end{matrix}$

Again, this derivative can also be calculated by first substituting
$x (t)$
and
$y (t)$
into
$f (x, y),$
then differentiating with respect to
$t :$

$\begin{matrix} z & = f (x, y) \\ = f (x (t), y (t)) \\ = \sqrt{{(x (t))}^{2} - {(y (t))}^{2}} \\ = \sqrt{e^{4 t} - e^{−2 t}} \\ = {(e^{4 t} - e^{−2 t})}^{1 / 2} . \end{matrix}$

Then

$\begin{matrix} \frac{d z}{d t} & = \frac{1}{2} {(e^{4 t} - e^{−2 t})}^{− 1 / 2} (4 e^{4 t} + 2 e^{−2 t}) \\ = \frac{2 e^{4 t} + e^{−2 t}}{\sqrt{e^{4 t} - e^{−2 t}}} . \end{matrix}$

This is the same solution.

It is often useful to create a visual representation of [link] for the chain rule. This is called a tree diagram for the chain rule for functions of one variable and it provides a way to remember the formula ([link]). This diagram can be expanded for functions of more than one variable, as we shall see very shortly.

has two independent variables, there are two lines coming from this corner. The upper branch corresponds to the variable

x

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; therefore, it is labeled

(\partial z / \partial x) \times (d x / d t) .

add all the terms that appear on the rightmost side of the diagram. This gives us [link].

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,

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

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.

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 $\partial w / \partial u$

and $\partial w / \partial v$

using the following functions:

\begin{matrix} 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} . \end{matrix}

The formulas for $\partial w / \partial u$

and $\partial w / \partial v$

are

\begin{matrix} \frac{\partial w}{\partial u} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial u} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial u} \\ \frac{\partial w}{\partial v} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial v} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial v} . \end{matrix}

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

\begin{matrix} \frac{\partial w}{\partial x} & = & 6 x - 2 y & \frac{\partial w}{\partial y} & = & −2 x & \frac{\partial w}{\partial z} & = & 8 z \\ \frac{\partial x}{\partial u} & = & e^{u} sin v & \frac{\partial y}{\partial u} & = & e^{u} cos v & \frac{\partial z}{\partial u} & = & e^{u} \\ \frac{\partial x}{\partial v} & = & e^{u} cos v & \frac{\partial y}{\partial v} & = & − e^{u} sin v & \frac{\partial z}{\partial v} & = & 0 . \end{matrix}

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

\begin{matrix} \frac{\partial w}{\partial u} & = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial u} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial u} \\ = (6 x - 2 y) e^{u} sin v - 2 x e^{u} cos v + 8 z e^{u}, \end{matrix}

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:

\begin{matrix} \frac{\partial w}{\partial 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) . \end{matrix}

Next, we calculate $\partial w / \partial v :$

\begin{matrix} \frac{\partial w}{\partial v} & = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial v} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial v} \\ = (6 x - 2 y) e^{u} cos v - 2 x (− e^{u} sin v) + 8 z (0), \end{matrix}

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:

\begin{matrix} \frac{\partial w}{\partial 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) . \end{matrix}

Implicit Differentiation

Recall from Implicit Differentiation that implicit differentiation provides a method for finding

d y / d x

The method involves differentiating both sides of the equation defining the function with respect to

x,

by using the left-hand side of the equation defining the ellipse. Then

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

Using this function and the following theorem gives us an alternative approach to calculating

d y / d x .

Let’s now return to the problem that we started before the previous theorem. Using [link] and the function

f (x, y) = x^{2} + 3 y^{2} + 4 y - 4,

which is the same result obtained by the earlier use of implicit differentiation.

Key Concepts

Key Equations

For the following exercises, use the information provided to solve the problem.

Let $w (x, y, z) = x y cos z,$

where $x = t, y = t^{2},$

and $z = arcsin t .$

Find $\frac{d w}{d t} .$

\frac{d w}{d t} = y cos z + x cos z (2 t) - \frac{x y sin z}{\sqrt{1 - t^{2}}}

Let $w (t, v) = e^{t v}$

where $t = r + s$

and $v = r s .$

Find $\frac{\partial w}{\partial r}$

and $\frac{\partial w}{\partial s} .$

If $w = 5 x^{2} + 2 y^{2}, x = −3 s + t,$

and $y = s - 4 t,$

find $\frac{\partial w}{\partial s}$

and $\frac{\partial w}{\partial t} .$

\frac{\partial w}{\partial s} = −30 x + 4 y,

\frac{\partial w}{\partial t} = 10 x - 16 y

If $w = x y^{2}, x = 5 cos (2 t),$

and $y = 5 sin (2 t),$

find $\frac{\partial w}{\partial t} .$

If $f (x, y) = x y, x = r cos θ,$

and $y = r sin θ,$

find $\frac{\partial f}{\partial r}$

and express the answer in terms of $r$

and $θ .$

\frac{\partial f}{\partial r} = r sin (2 θ)

Suppose $f (x, y) = x + y, u = e^{x} sin y, x = t^{2},$

and $y = π t,$

where $x = r cos θ$

and $y = r sin θ .$

Find $\frac{\partial f}{\partial θ} .$

For the following exercises, find $\frac{d f}{d t}$

using the chain rule and direct substitution.

f (x, y) = x^{2} + y^{2},

x = t, y = t^{2}

\frac{d f}{d t} = 2 t + 4 t^{3}

f (x, y) = \sqrt{x^{2} + y^{2}}, y = t^{2}, x = t

f (x, y) = x y, x = 1 - \sqrt{t}, y = 1 + \sqrt{t}

\frac{d f}{d t} = −1

f (x, y) = \frac{x}{y}, x = e^{t}, y = 2 e^{t}

f (x, y) = ln (x + y),

x = e^{t}, y = e^{t}

\frac{d f}{d t} = 1

f (x, y) = x^{4},

x = t, y = t

Let $w (x, y, z) = x^{2} + y^{2} + z^{2},$

x = cos t, y = sin t,

and $z = e^{t} .$

Express $w$

as a function of $t$

and find $\frac{d w}{d t}$

directly. Then, find $\frac{d w}{d t}$

using the chain rule.

\frac{d w}{d t} = 2 e^{2 t}

in both cases

Let $z = x^{2} y,$

where $x = t^{2}$

and $y = t^{3} .$

Find $\frac{d z}{d t} .$

Let $u = e^{x} sin y,$

where $x = t^{2}$

and $y = π t .$

Find $\frac{d u}{d t}$

when $x = ln 2$

and $y = \frac{π}{4} .$

2 \sqrt{2} t + \sqrt{2} π = \frac{d u}{d t}

For the following exercises, find $\frac{d y}{d x}$

using partial derivatives.

sin (6 x) + tan (8 y) + 5 = 0

x^{3} + y^{2} x - 3 = 0

\frac{d y}{d x} = - \frac{3 x^{2} + y^{2}}{2 x y}

sin (x + y) + cos (x - y) = 4

x^{2} - 2 x y + y^{4} = 4

\frac{d y}{d x} = \frac{y - x}{− x + 2 y^{3}}

x e^{y} + y e^{x} - 2 x^{2} y = 0

x^{2 / 3} + y^{2 / 3} = a^{2 / 3}

\frac{d y}{d x} = − \sqrt[3]{\frac{y}{x}}

x cos (x y) + y cos x = 2

e^{x y} + y e^{y} = 1

\frac{d y}{d x} = - \frac{y e^{x y}}{x e^{x y} + e^{y} (1 + y)}

x^{2} y^{3} + cos y = 0

Find $\frac{d z}{d t}$

using the chain rule where $z = 3 x^{2} y^{3}, x = t^{4},$

and $y = t^{2} .$

\frac{d z}{d t} = 42 t^{13}

Let $z = 3 cos x - sin (x y), x = \frac{1}{t},$

and $y = 3 t .$

Find $\frac{d z}{d t} .$

Let $z = e^{1 - x y}, x = t^{1 / 3},$

and $y = t^{3} .$

Find $\frac{d z}{d t} .$

\frac{d z}{d t} = - \frac{10}{3} t^{7 / 3} \times e^{1 - t^{10 / 3}}

Find $\frac{d z}{d t}$

by the chain rule where $z = {cosh}^{2} (x y), x = \frac{1}{2} t,$

and $y = e^{t} .$

Let $z = \frac{x}{y}, x = 2 cos u,$

and $y = 3 sin v .$

Find $\frac{\partial z}{\partial u}$

and $\frac{\partial z}{\partial v} .$

\frac{\partial z}{\partial u} = \frac{−2 sin u}{3 sin v}

and $\frac{\partial z}{\partial v} = \frac{−2 cos u cos v}{3 {sin}^{2} v}$

Let $z = e^{x^{2} y},$

where $x = \sqrt{u v}$

and $y = \frac{1}{v} .$

Find $\frac{\partial z}{\partial u}$

and $\frac{\partial z}{\partial v} .$

If $z = x y e^{x / y},$

x = r cos θ,

and $y = r sin θ,$

find $\frac{\partial z}{\partial r}$

and $\frac{\partial z}{\partial θ}$

when $r = 2$

and $θ = \frac{π}{6} .$

\frac{\partial z}{\partial r} = \sqrt{3} e^{\sqrt{3}},

\frac{\partial z}{\partial θ} = (2 - 4 \sqrt{3}) e^{\sqrt{3}}

Find $\frac{\partial w}{\partial s}$

if $w = 4 x + y^{2} + z^{3}, x = e^{r s^{2}}, y = ln (\frac{r + s}{t}),$

and $z = r s t^{2} .$

If $w = sin (x y z), x = 1 - 3 t, y = e^{1 - t},$

and $z = 4 t,$

find $\frac{\partial w}{\partial t} .$

\frac{\partial w}{\partial t} = cos (x y z) \times y z \times (−3) - cos (x y z) x z e^{1 - t} + cos (x y z) x y \times 4

For the following exercises, use this information: A function $f (x, y)$

is said to be homogeneous of degree $n$

if $f (t x, t y) = t^{n} f (x, y) .$

For all homogeneous functions of degree $n,$

the following equation is true: $x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f (x, y) .$

Show that the given function is homogeneous and verify that $x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f (x, y) .$

f (x, y) = 3 x^{2} + y^{2}

f (x, y) = \sqrt{x^{2} + y^{2}}

f (t x, t y) = \sqrt{t^{2} x^{2} + t^{2} y^{2}} = t^{1} f (x, y),

\frac{\partial f}{\partial y} = x \frac{1}{2} {(x^{2} + y^{2})}^{− 1 / 2} \times 2 x + y \frac{1}{2} {(x^{2} + y^{2})}^{− 1 / 2} \times 2 y = 1 f (x, y)

f (x, y) = x^{2} y - 2 y^{3}

The volume of a right circular cylinder is given by $V (x, y) = π x^{2} y,$

where $x$

is the radius of the cylinder and y is the cylinder height. Suppose $x$

and $y$

are functions of $t$

given by $x = \frac{1}{2} t$

and $y = \frac{1}{3} t$

so that $x and y$

are both increasing with time. How fast is the volume increasing when $x = 2$

and $y = 5 ?$

\frac{34 π}{3}

The pressure $P$

of a gas is related to the volume and temperature by the formula $P V = k T,$

where temperature is expressed in kelvins. Express the pressure of the gas as a function of both $V$

and $T .$

Find $\frac{d P}{d t}$

when $k = 1,$

\frac{d V}{d t} = 2

cm³/min, $\frac{d T}{d t} = \frac{1}{2}$

K/min, $V = 20$

cm³, and $T = 20 ° F .$

The radius of a right circular cone is increasing at $3$

cm/min whereas the height of the cone is decreasing at $2$

cm/min. Find the rate of change of the volume of the cone when the radius is $13$

cm and the height is $18$

cm.

\frac{d V}{d t} = \frac{1066 π}{3} {cm}^{3} / min

The volume of a frustum of a cone is given by the formula $V = \frac{1}{3} π z (x^{2} + y^{2} + x y),$

where $x$

is the radius of the smaller circle, $y$

is the radius of the larger circle, and $z$

is the height of the frustum (see figure). Find the rate of change of the volume of this frustum when $x = 10 in ., y = 12 in., and z = 18 in .$

A conical frustum (that is, a cone with the pointy end cut off) with height x, larger radius y, and smaller radius x.

A closed box is in the shape of a rectangular solid with dimensions $x, y, and z .$

(Dimensions are in inches.) Suppose each dimension is changing at the rate of $0.5$

in./min. Find the rate of change of the total surface area of the box when $x = 2 in ., y = 3 in., and z = 1 in .$

\frac{d A}{d t} = 12 in .^{2} / min

The total resistance in a circuit that has three individual resistances represented by $x, y,$

and $z$

is given by the formula $R (x, y, z) = \frac{x y z}{y z + x z + x y} .$

Suppose at a given time the $x$

resistance is $100 Ω,$

the y resistance is $200 Ω,$

and the $z$

resistance is $300 Ω .$

Also, suppose the $x$

resistance is changing at a rate of $2 Ω / min,$

the $y$

resistance is changing at the rate of $1 Ω / min,$

and the $z$

resistance has no change. Find the rate of change of the total resistance in this circuit at this time.

The temperature $T$

at a point $(x, y)$

is $T (x, y)$

and is measured using the Celsius scale. A fly crawls so that its position after $t$

seconds is given by $x = \sqrt{1 + t}$

and $y = 2 + \frac{1}{3} t,$

where $x and y$

are measured in centimeters. The temperature function satisfies $T_{x} (2, 3) = 4$

and $T_{y} (2, 3) = 3 .$

How fast is the temperature increasing on the fly’s path after $3$

sec?

2 ° C/sec

The $x and y$

components of a fluid moving in two dimensions are given by the following functions: $u (x, y) = 2 y$

and $v (x, y) = −2 x;$

x \geq 0; y \geq 0 .

The speed of the fluid at the point $(x, y)$

is $s (x, y) = \sqrt{u {(x, y)}^{2} + v {(x, y)}^{2}} .$

Find $\frac{\partial s}{\partial x}$

and $\frac{\partial s}{\partial y}$

using the chain rule.

Let $u = u (x, y, z),$

where $x = x (w, t), y = y (w, t), z = z (w, t), w = w (r, s), and t = t (r, s) .$

Use a tree diagram and the chain rule to find an expression for $\frac{\partial u}{\partial r} .$

\begin{matrix} \frac{\partial u}{\partial r} & = \frac{\partial u}{\partial x} (\frac{\partial x}{\partial w} \frac{\partial w}{\partial r} + \frac{\partial x}{\partial t} \frac{\partial t}{\partial r}) + \frac{\partial u}{\partial y} (\frac{\partial y}{\partial w} \frac{\partial w}{\partial r} + \frac{\partial y}{\partial t} \frac{\partial t}{\partial r}) \\ + \frac{\partial u}{\partial z} (\frac{\partial z}{\partial w} \frac{\partial w}{\partial r} + \frac{\partial z}{\partial t} \frac{\partial t}{\partial r}) \end{matrix}

Chain Rules for One or Two Independent Variables

Proof

The Generalized Chain Rule

Implicit Differentiation

Key Concepts

Key Equations

Glossary