3.4. THE CHAIN RULE

3.4

241

The Chain Rule

You will recall from Calculus I that we apply the chain rule when we have the d f (g (x)). The composition of two functions, for example when computing dx chain rule applies in similar situations when dealing with functions of several variables. For example, f (x; y) is a function of two variables. However, we may be computing the partials of f (x; y) where both x and y are functions of one or more other variables. We look at di¤erent cases.

3.4.1

Partials of f (x; y) where x and y are functions of one other variable t

Before we start, let us remind the reader that if a variable depends on several variables then the derivatives are partial derivatives and we will use the partial @z but if a variable depends only on one other variable, derivative notation as in @x dx we will use the notation from calculus of functions of one variable as in . dt Now, suppose we have z = f (x; y), x = g (t), y = h (t). Then, in this case we can also think of z as a function of t. So, there is only one derivative to dz compute, . It can be computed as follows: dt dz @z dx @z dy = + dt @x dt @y dt

(3.1)

To remember this formula, we can use a tree structure shown in …gure 3.6 as follows. 1. Draw the tree from top to bottom. Assuming z = f (x; y), x = g (t), y = h (t), we start with z. 2. From z, we draw a branch for each variable z depends on, x and y in this case, so we draw two branches. 3. From each of these variables, we repeat the procedure, that is draw a branch for each variable it depends on. Both x and y only depend on t, so we draw one branch from each. The tree is interpreted as follows. Since z is ultimately a function of t, look at all the paths from z to t. 1. Each path is the partial derivative of the variable on top with respect to the variable at the bottom. 2. Multiply all the partials which appear on a path. 3. Add all these products collected on each path.

242

CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

z

x

y

t

t Figure 3.6: z = f (x (t) ; y (t))

In our case, we obtain the following: There are two paths from z to t. The @z dx . The second path is z ! y ! t which …rst one is z ! x ! t which gives @x dt @z dy @z dx @z dy dz gives . Adding these gives + which is . @y dt @x dt @y dt dt Similarly, if w = f (x; y; z) and x; y; z are functions of t, then the corresponding tree structure is shown in …gure 3.7. Again, w is ultimately a function of t. So, there is only one derivative to dw compute, . Using the interpretation outlines above, we obtain the following dt formula: @w dx @w dy @w dz dw = + + dt @x dt @y dt @z dt and so on.

Example 3.4.1 Find du dt

du if u = x2 dt

y 2 and x = t2

=

@u dx @u dy + @x dt @y dt 2x2t 2y3 cos t

=

4xt

=

= =

6 y cos t 2

4 t 4t

3

1, y = 3 sin t.

1 t 4t

6 (3 sin t) cos t

18 sin t cos t

3.4. THE CHAIN RULE

243

w

x

y

t

t

z

t

Figure 3.7: w = f (x (t) ; y (t) ; z (t))

3.4.2

Partials of f (x; y) where x and y are functions of two variables s and t

We can use the same tree structure as above to compute partial derivatives in this case. Suppose we have z = f (x; y), x = g (s; t), y = h (s; t). So, in this case, z is a function of s and t. So, there are two partial derivatives to compute: @z @z and . The corresponding tree structure is shown in …gure 3.8. @s @t The …rst partials of z with respect to x or t are computed as follows: @z @s @z @t

@z @x @z @y + @x @s @y @s @z @x @z @y + @x @t @y @t

= =

(3.2) (3.3)

@z Example 3.4.2 Let z = sin (x + y) where x = 2st and y = s2 + t2 . Find @s @z and . @t @z @s

=

@z @x @z @y + @x @s @y @s cos (x + y) (2t) + cos (x + y) (2s)

=

2t cos 2st + s2 + t2 + 2s cos st + s2 + t2

=

2 cos (s + t)

=

2

(s + t)

244

CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

z

x

y

s

t

s

t

Figure 3.8: z = f (x (s; t) ; y (s; t)) and @z @t

=

@z @x @z @y + @x @t @y @t cos (x + y) (2s) + cos (x + y) (2t)

=

2 cos (s + t)

=

Example 3.4.3 Let u = x2 @u @u and . @s @t @u @s

= =

@u @t

2

2xy + 2y 3 where x = s2 ln t and y = 2st3 . Find

@u @x @u @y + @x @s @y @s (2x 2y) (2s ln t) +

=

2s2 ln t

=

@u @x @u @y + @x @t @y @t

=

2s2 ln t

(s + t)

2x + 6y 2

4st3 (2s ln t) +

4st3

s2 t

+

2t3

2s2 ln t + 24s2 t6

2s2 ln t + 24s2 t6

Example 3.4.4 Given z = f (x; y), x = r2 + s2 and y = 2rs …nd

2t3

6st2 @z @2z and @r @r2

3.4. THE CHAIN RULE Computation of

245

@z . @r @z @r

= =

Computation of @2z @r2

@z @x @z @y + @x @r @y @r @z @z 2r + 2s @x @y

@2z @r2

@ @r @ @r

@z @r @z @z 2r + 2s = @x @y @r @z @ @z = 2 + 2r +2 @r @x @r @x @z @ @z @ @z = 2 + 2r + 2s @x @r @x @r @y =

@s @r

@z @y

+ 2s

@ @r

@z @y (3.4)

We compute separately @ @r

@z @x

= =

@ @z @x @ + @x @x @r @y @2z @2z 2r 2 + 2s @x @y@x

@z @x

@ @x

@z @y

@y @r (3.5)

and @ @r

@z @y

@z @x @ + @y @r @y @2z @2z = 2r + 2s 2 @x@y @y =

@y @r (3.6)

If f has continuous second partial derivatives, then combining Equations 3.4, 3.5 and 3.6 gives 2 2 @2z @z @2z 2@ z 2@ z = 2 + 4r + 8rs + 4s @r2 @x @x2 @y@x @y 2

If f does not have continuous partial derivatives, then we cannot combine the mixed partials, and we get an expression slightly more complicated: @2z @z @2z @2z @2z @2z =2 + 4r2 2 + 4rs + 4rs + 4s2 2 2 @r @x @x @y@x @x@y @y

246

CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

F

x

y

x

x Figure 3.9: F (x; f (x))

3.4.3

General Case

Suppose f is a function of n variables x1 , x2 , :::, xn and each xi is in turn a function of m variables t1 , t2 , :::, tm . f is then a function of m variables. The m …rst partials of f are: @f @f @x1 @f @x2 @f @xn = + + ::: + @ti @x1 @ti @x2 @ti @xn @ti for each i = 1; 2; :::; m.

3.4.4

Implicit Di¤erentiation

The chain rule can be used to derive a simpler method for …nding the derivative of an implicitly de…ned function. y de…ned implicitly as a function of x in a relation of the form F (x; y) = 0 Suppose that F (x; y) = 0 de…nes y as an implicit function of x we will call dy . We do so by di¤erentiating both sides of y = f (x). We wish to …nd dx F (x; y) = 0 with respect to x. The right side is easy to di¤erentiate and we will skip the details here! To di¤erentiate the left side with respect to x, F (x; y), we will use the chain rule, remembering that F (x; y) = F (x; f (x)). So, F is ultimately a function of x. The corresponding tree structure is shown in …gure 3.9.We obtain:

3.4. THE CHAIN RULE

247

F

x

y

x

z

y

x

y

Figure 3.10: F (x; y; f (x; y)) @F dx @F + @x dx @y @F @F + @x @y

dy dx dy dx dy dx

=

0

=

0

=

@F @x @F @y

=

Fx Fy

dy if 2xy y 3 + 1 x 2y = 0. dx Using the notation above, F (x; y) = 2xy y 3 + 1 x 2y

Example 3.4.5 Find

dy dx

= =

Fx Fy 2x

2y 1 3y 2 2

z de…ned implicitly as a function of x and y in a relation of the form F (x; y; z) = 0 Suppose that F (x; y; z) = 0 de…nes z as an implicit function of x and y that is F (x; y; z) = F (x; y; f (x; y)). We can use a tree structure to …nd the partial derivatives. It is shown in …gure 3.10. The dotted lines indicate that if x were a function of y, it is where we would have drawn a branch. Since x and y are the independent variables, they do not depend on each other. So, x is not a function of y and y is not a function of x. If we di¤erentiate each side with respect to x, using the tree structure, we

248

CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

get @F dx @F @z + =0 @x dx @z @x Now,

@x = 1 so, we get @x

@F @z @F + =0 @x @z @x

That is @z = @x

@F @x @F @z

=

Fx Fz

@z = @y

@F @y @F @z

=

Fy Fz

Similarly, we can show that

Example 3.4.6 Find

@z if z is de…ned implicitly by @x x3 + y 3 + z 3 + 6xyz = 1

Using the formula above with F (x; y; z) = x3 + y 3 + z 3 + 6xyz @z @x

= = =

1, we see that

Fx Fz 3x2 + 6yz 3z 2 + 6xy x2 + 2yz z 2 + 2xy

Compare this solution with the solution of example 3.3.10 on page 229.

3.4.5

Problems

For the problems below, do the following: Find

dw using the chain rule. dt

Find

dw by …rst expressing w as a function of t. dt

Evaluate

dw at the given value of t. dt

1. w = x2 + y 2 , x = cos t, y = sin t, t = . 2. w =

x y 1 + , x = cos2 t, y = sin2 t, z = , t = 3. z z t

3.4. THE CHAIN RULE 3. w = 2yex

249

ln z, x = ln t2 + 1 , y = tan

1

t, z = et , t = 1.

For the problems below, do the following: Find

@z @z and using the chain rule. @u @v

Find

@z @z and by …rst expressing z as a function of u and v. @u @v

Evaluate

@z @z and at the given point (u; v). @u @v

4. z = 4ex ln y, x = ln (u cos v), y = u sin v, (u; v) = 2;

4

.

For the problems below, do the following: Find

@w @w and using the chain rule. @u @v

Find

@w @w and by …rst expressing w as a function of u and v. @u @v

Evaluate

@w @w and at the given point (u; v). @u @v

5. w = xy + yz + xz, x = u + v, y = u

v, z = uv, (u; v) =

1 ;1 . 2

For the problems below, do the following: Find

@u @u @u , and using the chain rule. @x @y @z

Find

@u @u @u , and by …rst expressing u as a function of u and v. @x @y @z

Evaluate 6. u =

p q

@u @u @u , and at the given point (x; y; z). @x @y @z

q , p = x + y + z, q = x r

y + z, r = x + y

z, (x; y; z) =

p

3; 2; 1 .

Draw a tree diagram and write a formula for the each derivative below. 7.

dz for z = f (x; y), x = g (t), y = h (t). dt

8.

@w @w and for w = h (x; y; z), x = f (u; v), y = g (u; v) and z = k (u; v). @u @v

9.

@w @w and for w = g (x; y), x = g (u; v) and y = k (u; v). @u @v

250

CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

10.

@z @z and for z = f (x; y), x = g (t; s) and y = h (t; s). @t @s

11.

@w @w and for w = g (u), u = h (s; t). @s @t

12.

@w @w and for w = f (x; y), x = g (r) and y = h (s). @r @s

dy Assuming the relations below de…ne y as an implicit function of x, …nd dx at the given point. 13. x3

2y 2 + xy = 0, (1; 1)

14. x2 + xy + y 2

7 = 0, (1; 2)

Assuming the relations below de…ne z as an implicit function of x and y, @z @z …nd and at the given point. @x @y 15. z 3

xy + yz + y 3

2 = 0, (1; 1; 1).

16. sin (x + y) + sin (y + z) + sin (x + z) = 0, ( ; ; ). @w when r = 1, s = @r z = sin (r + s).

17. Find

2

1, if w = (x + y + z) , x = r s, y = cos (r + s),

Additional problems. 18. If z = f (x; y) where f is di¤erentiable, x = g (t), y = h (t), g (3) = 2, dz g 0 (3) = 5, h (3) = 7, h0 (3) = 4, fx (2; 7) = 6, fy (2; 7) = 8, …nd dt when t = 3. 19. The pressure of 1 mole of perfect gas is increasing at a rate of 0.05 kPa/s and the temperature is increasing at a rate of 0.15 K/s. Knowing that dV P V = 8:31T …nd when P = 20kP a and T = 320K. dt 20. If z = f (x; y) where x = r cos and y = r sin @z @z and . @r @ 2 @z 2. Show that + @x 1. Find

@z @y

2

=

@z @r

2

+

1 r2

@z @

@z @z + = 0 (hint: let u = x @x @y chain rule to compute the partial derivatives).

21. If z = f (x

y), show that

2

y then use the

3.4. THE CHAIN RULE

3.4.6

251

Answers

For the problems below, do the following: Find

dw using the chain rule. dt

Find

dw by …rst expressing w as a function of t. dt

Evaluate

dw at the given value of t. dt

1. w = x2 + y 2 , x = cos t, y = sin t, t = .

dw dt

= =

w = cos2 t + sin2 t = 1 hence dw dt 2. w =

@w dx @w dy + @x dt @y dt 0 dw = 0. dt

=0 t=

x y 1 + , x = cos2 t, y = sin2 t, z = , t = 3. z z t dw dt

= =

@w dx @w dy @w dz + + @x dt @y dt @z dt 1

w=t Hence

dw dt 3. w = 2yex

dw =1 dt =1

t=3

ln z, x = ln t2 + 1 , y = tan

dw dt

= =

1

t, z = et , t = 1.

@w dx @w dy @w dz + + @x dt @y dt @z dt 4t tan 1 t + 1

252

CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

w = 2 tan So

dw dt

1

t t2 + 1

dw = 4t tan dt 1

= 4 tan

1+1=

1

t

t+1

+1

t=1

For the problems below, do the following: Find

@z @z and using the chain rule. @u @v

Find

@z @z and by …rst expressing z as a function of u and v. @u @v

Evaluate

@z @z and at the given point (u; v). @u @v

4. z = 4ex ln y, x = ln (u cos v), y = u sin v, (u; v) = 2;

@z @u

= =

4

.

@z @x @z @y + @x @u @y @u 4 cos v (ln (u sin v) + 1)

and @z @v

= =

@z @x @z @y + @x @v @y @v 4u sin v ln (u sin v) +

4u cos2 v sin v

z = 4u cos v ln (u sin v) So

@z = 4 cos v (ln (u sin v) + 1) @u

and @z = @v

4u sin v ln (u sin v) +

4u cos2 v sin v

p @z 2; = 2 (ln 2 + 2) @u 4 p @z 2; = 2 2 (ln 2 2) @v 4

3.4. THE CHAIN RULE

253

For the problems below, do the following: Find

@w @w and using the chain rule. @u @v

Find

@w @w and by …rst expressing w as a function of u and v. @u @v

Evaluate

@w @w and at the given point (u; v). @u @v

5. w = xy + yz + xz, x = u + v, y = u

@w @u

= =

v, z = uv, (u; v) =

1 ;1 . 2

@w @x @w @y @w @z + + @x @u @y @u @z @u 4uv + 2u

and @w @v

= =

@w @x @w @y @w @z + + @x @v @y @v @z @v 2 2v + 2u w = u2

So

and

v 2 + 2u2 v

@w = 2u + 4uv @u @w = @v @w

and @w

2v + 2u2

1 ;1 2 @u

1 ;1 2 @v

=3

=

3 2

For the problems below, do the following: Find

@u @u @u , and using the chain rule. @x @y @z

Find

@u @u @u , and by …rst expressing u as a function of u and v. @x @y @z

254

CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES Evaluate

6. u =

p q

@u @u @u , and at the given point (x; y; z). @x @y @z

q , p = x + y + z, q = x r

@u @x

= =

y + z, r = x + y

z, (x; y; z) =

@u @p @u @q @u @r + + @p @x @q @x @r @x 0

and @u @y

= =

@u @p @u @q @u @r + + @p @y @q @y @r @y z 2

(z

y)

and @u @z

= =

@u @p @u @q @u @r + + @p @z @q @z @r @z y 2

(z

y) u=

So

and

and

y z

y

@u =0 @x @u z = 2 @y (z y) @u = @z (z @u

p

@u

p

and

and @u

y 2

y)

3; 2; 1 =0 @x 3; 2; 1 =1 @y

p

3; 2; 1 = @z

2

p

3; 2; 1 .

3.4. THE CHAIN RULE

255

Draw a tree diagram and write a formula for the each derivative below. 7.

8.

dz for z = f (x; y), x = g (t), y = h (t). dt dz @z dx @z dy = + dt @x dt @y dt @w @w and for w = h (x; y; z), x = f (u; v), y = g (u; v) and z = k (u; v). @u @v @w @x @w @y @w @z @w = + + @u @x @u @y @u @z @u and

9.

@w @w and for w = g (x; y), x = g (u; v) and y = k (u; v). @u @v @w @w @x @w @y = + @u @x @u @y @u and

10.

12.

@w @x @w @y @w = + @v @x @v @y @v

@z @z and for z = f (x; y), x = g (t; s) and y = h (t; s). @t @s @z @z @x @z @y = + @t @x @t @y @t and

11.

@w @w @x @w @y @w @z = + + @v @x @v @y @v @z @v

@z @x @z @y @z = + @s @x @s @y @s

@w @w and for w = g (u), u = h (s; t). @s @t dw @u @w = @s du @s and @w dw @u = @t du @t @w @w and for w = f (x; y), x = g (r) and y = h (s). @r @s @w @w dx = @r @x dr and @w @w dy = @s @y ds

256

CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

dy Assuming the relations below de…ne y as an implicit function of x, …nd dx at the given point. 13. x3

2y 2 + xy = 0, (1; 1) dy = dx

3x2 + y 4y + x

Hence 4 dy (1; 1) = dx 3 14. x2 + xy + y 2

7 = 0, (1; 2) dy = dx

2x + y x + 2y

Hence dy (1; 2) = dx

4 5

Assuming the relations below de…ne z as an implicit function of x and y, @z @z and at the given point. …nd @x @y 15. z 3

xy + yz + y 3

2 = 0, (1; 1; 1). @z = @x

3z 2

y +y

So @z (1; 1; 1) 1 = @x 4 and @z = @y

x + z + 3y 2 3z 2 + y

So @z (1; 1; 1) = @y

3 4

16. sin (x + y) + sin (y + z) + sin (x + z) = 0, ( ; ; ). @z Fx = = @x Fz

1

@z Fy = = @y Fz

1

and

3.4. THE CHAIN RULE

257

@w when r = 1, s = @r z = sin (r + s).

2

17. Find @w @r

= =

1, if w = (x + y + z) , x = r s, y = cos (r + s),

@w @x @w @y @w @z + + @x @r @y @r @z @r 2 (r s + cos (r + s) + sin (r + s)) (1

So

sin (r + s) + cos (r + s))

@w (1; 1) = 12 @r

Additional Problems 18. If z = f (x; y) where f is di¤erentiable, x = g (t), y = h (t), g (3) = 2, dz g 0 (3) = 5, h (3) = 7, h0 (3) = 4, fx (2; 7) = 6, fy (2; 7) = 8, …nd dt when t = 3. dz @f (x; y) dx (t) @f (x; y) dy (t) = + dt @x dt @y dt When t = 3, we have dz dt

=

(6) (5) + ( 8) ( 4)

=

62

19. The pressure of 1 mole of perfect gas is increasing at a rate of 0.05 kPa/s and the temperature is increasing at a rate of 0.15 K/s. Knowing that dV P V = 8:31T …nd when P = 20kP a and T = 320K. dt 8:31T We have V = so P dV @V dT @V dP = + dt @T dt @P dt = 0:270 08L=s 20. If z = f (x; y) where x = r cos and y = r sin 1. (a) Find

@z @z and . @r @ @z @r

= =

@z @x @z @y + @x @r @y @r @z @z cos + sin @x @y

and @z @

= =

@z @x @z @y + @x @ @y @ @z @z r sin + r cos @x @y

258

CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES 2

@z @z + @x @y The right hand side is

2

(b) Show that

@z @r

2

+

1 r2

@z @

=

2

=

cos

=

@z @x

@z @r

2

+

1 r2

@z @z + sin @x @y 2

+

@z @y

2

@z @ 2

+

1 r2

r sin

@z @z + r cos @x @y

2

@z @z + = 0 (hint: let u = x y then use the @x @y chain rule to compute the partial derivatives). @z dz @u dz @z dz @u dz Let u = x y, then = = and = = . @x du @x du @y du @y du Therefore,

21. If z = f (x

y), show that

@z @z + @x @y

= =

fx = cos (x + ct)

dz du 0

dz du

2 sin (2x + 2ct)

so fxx

Hence the result.

=

sin (x + ct)

4 cos (2x + 2ct)

=

(sin (x + ct) + 4 cos (2x + 2ct))

2