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

ddx(f(g(x)))=f(g(x))g(x).

In this equation, both f(x)

and g(x)

are functions of one variable. Now suppose that f

is a function of two variables and g

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.

Chain Rule for One Independent Variable

Suppose that x=g(t)

and y=h(t)

are differentiable functions of t

and z=f(x,y)

is a differentiable function of xandy.

Then z=f(x(t),y(t))

is a differentiable function of t

and

dzdt=zx·dxdt+zy·dydt,

where the ordinary derivatives are evaluated at t

and the partial derivatives are evaluated at (x,y).

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(x0,y0),

where x0=g(t0)

and y0=h(t0)

for a fixed value of t0.

We wish to prove that z=f(x(t),y(t))

is differentiable at t=t0

and that [link] holds at that point as well.

Since f

is differentiable at P,

we know that

z(t)=f(x,y)=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0)+E(x,y),

where lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2=0.

We then subtract z0=f(x0,y0)

from both sides of this equation:

z(t)z(t0)=f(x(t),y(t))f(x(t0),y(t0))=fx(x0,y0)(x(t)x(t0))+fy(x0,y0)(y(t)y(t0))+E(x(t),y(t)).

Next, we divide both sides by tt0:

z(t)z(t0)tt0=fx(x0,y0)(x(t)x(t0)tt0)+fy(x0,y0)(y(t)y(t0)tt0)+E(x(t),y(t))tt0.

Then we take the limit as t

approaches t0:

limtt0z(t)z(t0)tt0=fx(x0,y0)limtt0(x(t)x(t0)tt0)+fy(x0,y0)limtt0(y(t)y(t0)tt0)+limtt0E(x(t),y(t))tt0.

The left-hand side of this equation is equal to dz/dt,

which leads to

dzdt=fx(x0,y0)dxdt+fy(x0,y0)dydt+limtt0E(x(t),y(t))tt0.

The last term can be rewritten as

limtt0E(x(t),y(t))tt0=limtt0(E(x,y)(xx0)2+(yy0)2(xx0)2+(yy0)2tt0)=limtt0(E(x,y)(xx0)2+(yy0)2)limtt0((xx0)2+(yy0)2tt0).

As t

approaches t0,

(x(t),y(t))

approaches (x(t0),y(t0)),

so we can rewrite the last product as

lim(x,y)(x0,y0)(E(x,y)(xx0)2+(yy0)2)lim(x,y)(x0,y0)((xx0)2+(yy0)2tt0).

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

lim(x,y)(x0,y0)((xx0)2+(yy0)2tt0)=lim(x,y)(x0,y0)((xx0)2+(yy0)2(tt0)2)=lim(x,y)(x0,y0)((xx0tt0)2+(yy0tt0)2)=(lim(x,y)(x0,y0)(xx0tt0))2+(lim(x,y)(x0,y0)(yy0tt0))2.

Since x(t)

and y(t)

are both differentiable functions of t,

both limits inside the last radical exist. Therefore, this value is finite. This proves the chain rule at t=t0;

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 fx·dxdt

and the second term is fy·dydt.

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

The variables xandy

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

but are dependent variables for the variable t.

Two terms appear on the right-hand side of the formula, and 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 dz/dt

for each of the following functions:

  1. z=f(x,y)=4x2+3y2,x=x(t)=sint,y=y(t)=cost
  2. z=f(x,y)=x2y2,x=x(t)=e2t,y=y(t)=et
  1. To use the chain rule, we need four quantities— z/x,z/y,dx/dt,

    and

    dy/dt:
    zx=8xzy=6ydxdt=costdydt=sint

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


    dzdt=zx·dxdt+zy·dydt=(8x)(cost)+(6y)(sint)=8xcost6ysint.

    This answer has three variables in it. To reduce it to one variable, use the fact that

    x(t)=sintandy(t)=cost.

    We obtain


    dzdt=8xcost6ysint=8(sint)cost6(cost)sint=2sintcost.

    This derivative can also be calculated by first substituting

    x(t)

    and

    y(t)

    into

    f(x,y),

    then differentiating with respect to

    t:
    z=f(x,y)=f(x(t),y(t))=4(x(t))2+3(y(t))2=4sin2t+3cos2t.

    Then


    dzdt=2(4sint)(cost)+2(3cost)(sint)=8sintcost6sintcost=2sintcost,

    which is the same solution. However, it may not always be this easy to differentiate in this form.

  2. To use the chain rule, we again need four quantities— z/x,z/dy,dx/dt,

    and

    dy/dt:
    zx=xx2y2zy=yx2y2dxdt=2e2tdxdt=et.

    We substitute each of these into [link]:


    dzdt=zx·dxdt+zy·dydt=(xx2y2)(2e2t)+(yx2y2)(et)=2xe2tyetx2y2.

    To reduce this to one variable, we use the fact that

    x(t)=e2t

    and

    y(t)=et.

    Therefore,


    dzdt=2xe2t+yetx2y2=2(e2t)e2t+(et)ete4te−2t=2e4t+e−2te4te−2t.

    To eliminate negative exponents, we multiply the top by

    e2t

    and the bottom by

    e4t:
    dzdt=2e4t+e−2te4te−2t·e2te4t=2e6t+1e8te2t=2e6t+1e2t(e6t1)=2e6t+1ete6t1.

    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:
    z=f(x,y)=f(x(t),y(t))=(x(t))2(y(t))2=e4te−2t=(e4te−2t)1/2.

    Then


    dzdt=12(e4te−2t)1/2(4e4t+2e−2t)=2e4t+e−2te4te−2t.

    This is the same solution.

Calculate dz/dt

given the following functions. Express the final answer in terms of t.

z=f(x,y)=x23xy+2y2,x=x(t)=3sin2t,y=y(t)=4cos2t
dzdt=fxdxdt+fydydt=(2x3y)(6cos2t)+(−3x+4y)(−8sin2t)=−92sin2tcos2t72(cos22tsin22t)=−46sin4t72cos4t.
Hint

Calculate z/x,z/dy,dx/dt,

and dy/dt,

then use [link].

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.

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.

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 (z/x)×(dx/dt).

The bottom branch is similar: first the y

branch, then the t

branch. This branch is labeled (z/y)×(dy/dt).

To get the formula for dz/dt,

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 xandy,

and both x=g(u,v)

and y=h(u,v)

are functions of the independent variables uandv.

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 xandy.

Then, z=f(g(u,v),h(u,v))

is a differentiable function of uandv,

and

zu=zxxu+zyxu

and

zv=zxxv+zyyv.

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.

To derive the formula for z/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 z/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 uandv.

Using the Chain Rule for Two Variables

Calculate z/u

and z/v

using the following functions:

z=f(x,y)=3x22xy+y2,x=x(u,v)=3u+2v,y=y(u,v)=4uv.

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

and y/v:

zx=6x2yzy=−2x+2yxu=3xv=2yu=4yv=−1.

To find z/u,

we use [link]:

zu=zx·xu+zy·yu=3(6x2y)+4(−2x+2y)=10x+2y.

Next, we substitute x(u,v)=3u+2v

and y(u,v)=4uv:

zu=10x+2y=10(3u+2v)+2(4uv)=38u+18v.

To find z/v,

we use [link]:

zv=zxxv+zyyv=2(6x2y)+(−1)(−2x+2y)=14x6y.

Then we substitute x(u,v)=3u+2v

and y(u,v)=4uv:

zv=14x6y=14(3u+2v)6(4uv)=18u+34v.

Calculate z/u

and z/v

given the following functions:

z=f(x,y)=2xyx+3y,x(u,v)=e2ucos3v,y(u,v)=e2usin3v.
zu=0,zv=−21(3sin3v+cos3v)2
Hint

Calculate z/x,z/y,x/u,x/v,y/u,

and y/v,

then use [link] and [link].

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(x1,x2,…,xm)

be a differentiable function of m

independent variables, and for each i{1,…,m},

let xi=xi(t1,t2,…,tn)

be a differentiable function of n

independent variables. Then

wtj=wx1x1tj+wx2x2tj++wxmxmtj

for any j{1,2,…,n}.

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)=3x22xy+4z2x=x(u,v)=eusinvy=y(u,v)=eucosvz=z(u,v)=eu.

The formulas for w/u

and w/v

are

wu=wx·xu+wy·yu+wz·zuwv=wx·xv+wy·yv+wz·zv.

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

wx=6x2ywy=−2xwz=8zxu=eusinvyu=eucosvzu=euxv=eucosvyv=eusinvzv=0.

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

wu=wx·xu+wy·yu+wz·zu=(6x2y)eusinv2xeucosv+8zeu,

then substitute x(u,v)=eusinv,y(u,v)=eucosv,

and z(u,v)=eu

into this equation:

wu=(6x2y)eusinv2xeucosv+8zeu=(6eusinv2eucosv)eusinv2(eusinv)eucosv+8e2u=6e2usin2v4e2usinvcosv+8e2u=2e2u(3sin2v2sinvcosv+4).

Next, we calculate w/v:

wv=wx·xv+wy·yv+wz·zv=(6x2y)eucosv2x(eusinv)+8z(0),

then we substitute x(u,v)=eusinv,y(u,v)=eucosv,

and z(u,v)=eu

into this equation:

wv=(6x2y)eucosv2x(eusinv)=(6eusinv2eucosv)eucosv+2(eusinv)(eusinv)=2e2usin2v+6e2usinvcosv2e2ucos2v=2e2u(sin2v+sinvcosvcos2v).

Calculate w/u

and w/v

given the following functions:

w=f(x,y,z)=x+2y4z2xy+3zx=x(u,v)=e2ucos3vy=y(u,v)=e2usin3vz=z(u,v)=e2u.
wu=0wv=1533sin3v+6cos3v(3+2cos3vsin3v)2
Hint

Calculate nine partial derivatives, then use the same formulas from [link].

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,andz.

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,andv.

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.

The three formulas are

wt=wxxt+wyyt+wzztwu=wxxu+wyyu+wzzuwv=wxxv+wyyv+wzzv.

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.

wt=wxxt+wyytwu=wxxu+wyyuwv=wxxv+wyyv

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.

Hint

Determine the number of branches that emanate from each node in the tree.

Implicit Differentiation

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

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 dy/dx.

Partial derivatives provide an alternative to this method.

Consider the ellipse defined by the equation x2+3y2+4y4=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.

This equation implicitly defines y

as a function of x.

As such, we can find the derivative dy/dx

using the method of implicit differentiation:

ddx(x2+3y2+4y4)=ddx(0)2x+6ydydx+4dydx=0(6y+4)dydx=−2xdydx=x3y+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)=x2+3y2+4y4.

The ellipse x2+3y2+4y4=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 dy/dx.

Implicit Differentiation of a Function of Two or More Variables

Suppose the function z=f(x,y)

defines y

implicitly as a function y=g(x)

of x

via the equation f(x,y)=0.

Then

dydx=f/xf/y

provided fy(x,y)0.

If the equation f(x,y,z)=0

defines z

implicitly as a differentiable function of xandy,

then

zx=f/xf/zandzy=f/yf/z

as long as fz(x,y,z)0.

[link] is a direct consequence of [link]. In particular, if we assume that y

is defined implicitly as a function of x

via the equation f(x,y)=0,

we can apply the chain rule to find dy/dx:

ddxf(x,y)=ddx(0)fx·dxdx+fy·dydx=0fx+fy·dydx=0.

Solving this equation for dy/dx

gives [link]. [link] can be derived in a similar fashion.

Let’s now return to the problem that we started before the previous theorem. Using [link] and the function f(x,y)=x2+3y2+4y4,

we obtain

fx=2xfy=6y+4.

Then [link] gives

dydx=f/xf/y=2x6y+4=x3y+2,

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

Implicit Differentiation by Partial Derivatives
  1. Calculate dy/dx

    if

    y

    is defined implicitly as a function of

    x

    via the equation

    3x22xy+y2+4x6y11=0.

    What is the equation of the tangent line to the graph of this curve at point

    (2,1)?
  2. Calculate z/x

    and

    z/y,

    given

    x2eyyzex=0.
  1. Set f(x,y)=3x22xy+y2+4x6y11=0,

    then calculate

    fx

    and

    fy: fx=6x2y+4fy=−2x+2y6.

    The derivative is given by


    dydx=f/xf/y=6x2y+4−2x+2y6=3xy+2xy+3.

    The slope of the tangent line at point

    (2,1)

    is given by


    dydx\|(x,y)=(2,1)=3(2)1+221+3=74.

    To find the equation of the tangent line, we use the point-slope form ([link]):


    yy0=m(xx0)y1=74(x2)y=74x72+1y=74x52.

    A rotated ellipse with equation 3x2 – 2xy + y2 + 4x – 6y – 11 = 0 and with tangent at (2, 1). The equation for the tangent is given by y = 7/4 x – 5/2. The ellipse’s major axis is parallel to the tangent line.

  2. We have f(x,y,z)=x2eyyzex.

    Therefore,


    fx=2xeyyzexfy=x2eyzexfz=yex.

    Using [link],


    zx=f/xf/y=2xeyyzexyex=2xeyyzexyexandzy=f/yf/z=x2eyzexyex=x2eyzexyex.

Find dy/dx

if y

is defined implicitly as a function of x

by the equation x2+xyy2+7x3y26=0.

What is the equation of the tangent line to the graph of this curve at point (3,−2)?

dydx=2x+y+72yx+3\|(3,−2)=2(3)+(−2)+72(−2)(3)+3=114

Equation of the tangent line: y=114x+254

Hint

Calculate f/dx

and f/dy,

then use [link].

Key Concepts

Key Equations


dzdv=zx·xv+zy·yv

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

Let w(x,y,z)=xycosz,

where x=t,y=t2,

and z=arcsint.

Find dwdt.

dwdt=ycosz+xcosz(2t)xysinz1t2

Let w(t,v)=etv

where t=r+s

and v=rs.

Find wr

and ws.

If w=5x2+2y2,x=−3s+t,

and y=s4t,

find ws

and wt.

ws=−30x+4y, wt=10x16y

If w=xy2,x=5cos(2t),

and y=5sin(2t),

find wt.

If f(x,y)=xy,x=rcosθ,

and y=rsinθ,

find fr

and express the answer in terms of r

and θ.

fr=rsin(2θ)

Suppose f(x,y)=x+y,u=exsiny,x=t2,

and y=πt,

where x=rcosθ

and y=rsinθ.

Find fθ.

For the following exercises, find dfdt

using the chain rule and direct substitution.

f(x,y)=x2+y2, x=t,y=t2
dfdt=2t+4t3
f(x,y)=x2+y2,y=t2,x=t
f(x,y)=xy,x=1t,y=1+t
dfdt=−1
f(x,y)=xy,x=et,y=2et
f(x,y)=ln(x+y), x=et,y=et
dfdt=1
f(x,y)=x4, x=t,y=t

Let w(x,y,z)=x2+y2+z2,

x=cost,y=sint,

and z=et.

Express w

as a function of t

and find dwdt

directly. Then, find dwdt

using the chain rule.

dwdt=2e2t

in both cases

Let z=x2y,

where x=t2

and y=t3.

Find dzdt.

Let u=exsiny,

where x=t2

and y=πt.

Find dudt

when x=ln2

and y=π4.

22t+2π=dudt

For the following exercises, find dydx

using partial derivatives.

sin(6x)+tan(8y)+5=0
x3+y2x3=0
dydx=3x2+y22xy
sin(x+y)+cos(xy)=4
x22xy+y4=4
dydx=yxx+2y3
xey+yex2x2y=0
x2/3+y2/3=a2/3
dydx=yx3
xcos(xy)+ycosx=2
exy+yey=1
dydx=yexyxexy+ey(1+y)
x2y3+cosy=0

Find dzdt

using the chain rule where z=3x2y3,x=t4,

and y=t2.

dzdt=42t13

Let z=3cosxsin(xy),x=1t,

and y=3t.

Find dzdt.

Let z=e1xy,x=t1/3,

and y=t3.

Find dzdt.

dzdt=103t7/3×e1t10/3

Find dzdt

by the chain rule where z=cosh2(xy),x=12t,

and y=et.

Let z=xy,x=2cosu,

and y=3sinv.

Find zu

and zv.

zu=−2sinu3sinv

and zv=−2cosucosv3sin2v

Let z=ex2y,

where x=uv

and y=1v.

Find zu

and zv.

If z=xyex/y,

x=rcosθ,

and y=rsinθ,

find zr

and zθ

when r=2

and θ=π6.

zr=3e3, zθ=(243)e3

Find ws

if w=4x+y2+z3,x=ers2,y=ln(r+st),

and z=rst2.

If w=sin(xyz),x=13t,y=e1t,

and z=4t,

find wt.

wt=cos(xyz)×yz×(−3)cos(xyz)xze1t+cos(xyz)xy×4

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

is said to be homogeneous of degree n

if f(tx,ty)=tnf(x,y).

For all homogeneous functions of degree n,

the following equation is true: xfx+yfy=nf(x,y).

Show that the given function is homogeneous and verify that xfx+yfy=nf(x,y).

f(x,y)=3x2+y2
f(x,y)=x2+y2
f(tx,ty)=t2x2+t2y2=t1f(x,y), fy=x12(x2+y2)1/2×2x+y12(x2+y2)1/2×2y=1f(x,y)
f(x,y)=x2y2y3

The volume of a right circular cylinder is given by V(x,y)=πx2y,

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=12t

and y=13t

so that xandy

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

and y=5?

34π3

The pressure P

of a gas is related to the volume and temperature by the formula PV=kT,

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

and T.

Find dPdt

when k=1,

dVdt=2

cm3/min, dTdt=12

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.

dVdt=1066π3cm3/min

The volume of a frustum of a cone is given by the formula V=13πz(x2+y2+xy),

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=10in.,y=12in.,andz=18in.

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,andz.

(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=2in.,y=3in.,andz=1in.

dAdt=12in.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)=xyzyz+xz+xy.

Suppose at a given time the x

resistance is 100Ω,

the y resistance is 200Ω,

and the z

resistance is 300Ω.

Also, suppose the x

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

the y

resistance is changing at the rate of 1Ω/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=1+t

and y=2+13t,

where xandy

are measured in centimeters. The temperature function satisfies Tx(2,3)=4

and Ty(2,3)=3.

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

sec?

2°C/sec

The xandy

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

and v(x,y)=−2x;

x0;y0.

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

is s(x,y)=u(x,y)2+v(x,y)2.

Find sx

and sy

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),andt=t(r,s).

Use a tree diagram and the chain rule to find an expression for ur.

ur=ux(xwwr+xttr)+uy(ywwr+yttr)+uz(zwwr+zttr)

Glossary

generalized chain rule
the chain rule extended to functions of more than one independent variable, in which each independent variable may depend on one or more other variables
intermediate variable
given a composition of functions (e.g., f(x(t),y(t))),

the intermediate variables are the variables that are independent in the outer function but dependent on other variables as well; in the function

f(x(t),y(t)),

the variables

xandy

are examples of intermediate variables

tree diagram
illustrates and derives formulas for the generalized chain rule, in which each independent variable is accounted for

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