P1 2017 1 / 35

P1 Calculus II

Partial Differentiation & Multiple Integration

Prof David Murray

david.murray@eng.ox.ac.ukwww.robots.ox.ac.uk/∼dwm/Courses/1PD

4 lectures, MT 2017

P1 2017 2 / 35

2 Partial derivatives & function character

P1 2017 3 / 35

Lecture contents

In this lecture we develop a number of relationships which dependon the nature of the function of several variables.There is a danger that the various cases will become an unconnectedjumble.The key thing is to always ask the question

what is f a function of, exactlyand then to write only expressions that you know to be true in thatcase.

The topics are2.1 A function of a function.2.2 Composite functions I,II,III, including the Chain Rule for Partials.2.3 Implicit functions in two variables.2.4 ** Implicit functions in more than two variables.

P1 2017 4 / 35

2.1 A function of a function.f = f (u) where u = u(x , y).

P1 2017 5 / 35

Function of a function

Supposef (x , y) = xy sin(xy) .

We could find the partial derivatives in the usual way as(∂f∂x

)= y sin(xy) + xy2 cos(xy)

(∂f∂x

)= x sin(xy) + x2y cos(xy)

BUT!Notice that f is a function f = f (u) = u sin u of a single variable u = xy .So total df /du must exist.

When f = f (x , y) and u = u(x , y) is found such that f = f (u)

∂f∂x

=dfdu∂u∂x

and∂f∂y

=dfdu∂u∂y

P1 2017 6 / 35

Proof

We know that the total differentials of f (x , y) and u(x , y) are:

df =∂f∂x

dx +∂f∂y

dy , and du =∂u∂x

dx +∂u∂y

dy .

We are allowed to take ratios of total differentials, so that

dfdu

=

∂f∂x dx + ∂f

∂y dy∂u∂x dx + ∂u

∂y dy.

But x and y are independent ⇒ dx and dy are independent.Taking y to be fixed: dy = 0 Taking x to be fixed: dx = 0

dfdu

=∂f∂x dx∂u∂x dx

=∂f∂x

/∂u∂x

,dfdu

=

∂f∂y dy∂u∂y dy

=∂f∂y

/∂u∂y

.

The result quoted follows immediately.

P1 2017 7 / 35

♣ ExampleReminder:

∂f∂x

=dfdu∂u∂x

,∂f∂y

=dfdu∂u∂y

BTW don’t think that means that you dividing out the partial ∂x ’s and ∂y ’s.

Q: Find ∂f /∂x and ∂f /∂y when f = tan−1(y/x).

A: Set f = tan−1(u) with u = y/x . Then

dfdu

=1

(1+ u2)=

x2

x2 + y2

and∂u∂x

= −yx2 ;

∂u∂y

=1x

⇒∂f∂x

=dfdu∂u∂x

=−y

x2 + y2 ;∂f∂y

=dfdu∂u∂y

=x

x2 + y2

P1 2017 8 / 35

♣ Example (harder)Q: A quantity z(x , t) is given by z = f (x − ct) where c is a constant.Show that, no matter what f is exactly, z satisfies

∂2z∂x2 =

1c2∂2z∂t2 ,

Write u = x − ct, so that z = f (u), and use the result just proved.

∂z∂x

=dfdu

(∂u∂x

)=

dfdu

(1) =dfdu

But df /du is also just a function of u — let’s say df /du = g(u). So ifwe differentiate wrt x again

∂2z∂x2 =

dgdu

(∂u∂x

)=

dgdu

(1) =d2fdu2

P1 2017 9 / 35

Example ctd ...

Reminder: u = x − ct and z = f (u).Now repeat the exercise, but differentiating wrt t.

∂z∂t

=dfdu

(∂u∂t

)=

dfdu

(−c) = −cg(u) .

So if we differentiate wrt t again

∂2z∂t2 = −c

dgdu

(∂u∂t

)= −c

dgdu

(−c) = c2 d2f

du2 .

Hence∂2z∂x2 =

d2fdu2 =

1c2∂2z∂t2

Note that we have not had to say anything about the function f at all,and you might care to check the result for any arbitrary function.

P1 2017 10 / 35

2.2 Composite functions and the Chain Rule

There are various cases of composite functions, and we deal with them inturn.

This section will also introduce the chain rule, which you will use overand over again.

P1 2017 11 / 35

Composite Functions I:What if f = f (x , y) and x = x(t) and y = y(t)?

Function f is said to be a composite function.

Notice that f is effectively a function of t alone

⇒ totaldfdt

exists.

We can write the following chain rule:

Chain Rule: When f = f (x , y) and x = x(t) and y = y(t):

dfdt

=∂f∂x

dxdt

+∂f∂y

dydt

.

P1 2017 12 / 35

Proof

To prove this, write the total differential:

df =∂f∂x

dx +∂f∂y

dy

Now we know that f , x and y are functions of t alone so the df /dt,dx/dt, and dy/dt all exist.

So, dividing by dt (which we are allowed to do as it is a total differential)

dfdt

=∂f∂x

dxdt

+∂f∂y

dydt

P1 2017 13 / 35

Composite functions I /ctd:What if f is f (x1, x2, x3, . . . , xn) with xi = xi(t)?

Again f must be a function of t alone, so df /dt exists.

The Chain Rule is

dfdt

=∂f∂x1

dx1

dt+∂f∂x2

dx2

dt+ . . . =

n∑i=1

∂f∂xi

dxi

dt.

The proof follows the previous pattern:

df =∂f∂x1

dx1 +∂f∂x2

dx2 + . . .

then divide both sides by dt.

P1 2017 14 / 35

♣ ExampleQ: Find df /dt when

f = f (x , y , z) = xy + yz + zx ; and x = t, y = e−t , z = cos t

A: From the chain rule for composites

dfdt

=∂f∂x

dxdt

+∂f∂y

dydt

+∂f∂z

dzdt

= (y + z)dxdt

+ (x + z)dydt

+ (x + y)dzdt

= (e−t + cos t)(1) + (t + cos t)(−e−t) + (t + e−t)(− sin t)= e−t(1− t − cos t − sin t) + cos t − t sin t.

You could & should check the result by explicit substitution.

(Note that you might find the direct method easier. Making things easieris not the point here — the point is finding general relationships thatapply between derivatives, both total and partial.)

P1 2017 15 / 35

Composite functions II:If z = f (x , y) and x = x(t1, t2, ...) & y = y(t1, t2, ...)

Similar to previous, but now f is effectively a function of several variablest1, t2, . . ..

Let’s fix all but one of the ti .

We would end up with a composite function as above ...... but now instead of total derivatives df /dt and dxi/dt we must havepartial derivatives ∂f /∂tj and ∂xi/∂tj

Thus

∂f∂t1

=∂f∂x∂x∂t1

+∂f∂y∂y∂t1

∂f∂t2

=∂f∂x∂x∂t2

+∂f∂y∂y∂t2

and so on.

P1 2017 16 / 35

Composite functions II: chain rule for partials

We can now easily generalize this result to a functionf (x1, x2, x3, ..., xn) with xi = xi (t1, t2, ..., tm).

After chomping your way through the indices you should find that

The Chain Rule for Partials

∂f∂tj

=∂f∂x1

∂x1

∂tj+∂f∂x2

∂x2

∂tj+∂f∂x3

∂x3

∂tj+ . . . : j = 1, . . . ,m.

It turns out that the chain rule for partials is very commonly used intransforming from one set of coordinates to another.

Here is one example, but we shall return to it again (and again).

P1 2017 17 / 35

♣ Example: Chain rule for partials

Q: x = r cosφ and y = r sinφ defines the transformation betweenCartesian and plane polar coordinates.Find ∂f /∂r and ∂f /∂φ when f (x , y) = x2 + y2.

A: Using the chain rule for partials:

∂f∂r

=∂f∂x∂x∂r

+∂f∂y∂y∂r

= 2x cosφ+ 2y sinφ

= 2r cos2φ+ 2r sin2φ = 2r∂f∂φ

=∂f∂x∂x∂φ

+∂f∂y∂y∂φ

= 2x(−r sinφ) + 2y(r cosφ)

= −2r2 cosφ sinφ+ 2r2 sinφ cosφ = 0 .

P1 2017 18 / 35

Direct checking

For x = r cosφ and y = r sinφ and f = x2 + y2 we just used the chainrule for partials to show that

∂f∂r

= 2r∂f∂φ

= 0 .

This can be checked directly.f = x2 + y2, ⇒ f = r2 cos2φ+ r2 sin2φ = r2.Hence

∂f∂r

= 2r .

And because f has no dependence on the other variable φ, it is obviousthat

∂f∂φ

= 0 .

P1 2017 19 / 35

Composite functions III:What if f = f (x , y) and y = y(x)?

Clearly f is a composite function of x alone. So this is like t being x .

Effectively, x = x(x) — but one wouldn’t bother to write this down!

So,dfdt

=∂f∂x

dxdt

+∂f∂y

dydt

morphs intodfdx

=∂f∂x

[dxdx

]+∂f∂y

dydx

Ie,dfdx

=∂f∂x

+∂f∂y

dydx

.

Note that this expression contains both total and partial derivates of fwrt x . They are different, and they can co-exist!

P1 2017 20 / 35

♣ Example: Composite Functions III

Reminder:dfdx

=∂f∂x

+∂f∂y

dydx

.

Q: Suppose z = xy + x/y and y =√

x . Find dz/dx .

A: Using the results just obtained,

dzdx

= y +1y+ x(1−

1y2 )

12x−1/2

= x1/2 + x−1/2 +12

(x1/2 − x−1/2

)=

32x1/2 +

12x−1/2

a result which can be checked directly in this case, using z = x3/2 + x1/2.

P1 2017 21 / 35

2.3 Implicit Functions in 2 variables

P1 2017 22 / 35

Implicit functionsIn calculus and analytic geometry it is common to see the equations ofplane curves written as

f (x , y) = 0.

For example the equation of a circle of radius√2 with centre at (0, 1) is

x2 + (y − 1)2 − 2 = 0

In this case it is easy to write an explicit expression for y

y = y(x) = 1± (2− x2)1/2

but often one cannot rearrange in this way.

But if we were to solve for y numerically we might trace out a curvey = y(x) which was single valued and differentiable.

In other words, f (x , y) = 0 may define function y = y(x) implicitly.

P1 2017 23 / 35

Implicit functions /ctd

Another way of seeing this is to think about independence andconstraints.

Writing f (x , y) might seem to involve two independent variables. But wehave written one constraint f (x , y) = 0, and this reduces the degrees offreedom from 2 to 1.

It turns out that for many purposes we don’t need an explicit expressionfor the function. It is enough to characterize the function by itsderivative.

So how do we compute the derivative dy/dx for a function y definedimplicitly by f (x , y) = 0?

P1 2017 24 / 35

Derivative of an implicit function

We saw that for f (x , y) with y = y(x),

dfdx

=∂f∂x

+∂f∂y

dydx

.

But f (x , y) = 0 defines y = y(x) implicitly — so this result still holds!

In addition, we know that f = 0 always ⇒df /dx = 0 always.Hence

0 =∂f∂x

+∂f∂y

dydx

⇒dydx

= −∂f /∂x∂f /∂y

.

P1 2017 25 / 35

♣ Example: Implicit functionsFirst consider an example where the implicit function can be derivedexplicitly. This gives us a check on our results.

Q: Find dy/dx when f (x , y) = x − x2y3 = 0.

A: Use the result for implicit functions just derived:

dydx

= −∂f /∂x∂f /∂y

= −

(1− 2xy3

−3x2y2

).

As a check, here we can find y explicitly as y = x−1/3. Hencedy/dx = −x−4/3/3.

Now substitute for y in the earlier result

dydx

= −

(1− 2xy3

−3x2y2

)= −

(1− 2xx−1

−3x2x−2/3

)= −

(−1

−3x4/3

)= −x−4/3/3 .

P1 2017 26 / 35

♣ Another Example: Implicit functionsBut remember! The whole point about implicit functions is that you donot need an explicit y = y(x) to get information about the derivatives.This is made clear in the next example.

Q: Find dy/dx when f (x , y) = ax2 + 2hxy + by2 + 2gx + 2fy + c = 0.

A: You may recognize f as a conic section — an ellipse or hyperbola,depending on the constants.

Now we can’t find y as a function of x , but using the result for implicitfunctions we have

dydx

= −∂f /∂x∂f /∂y

= −2ax + 2hy + 2g2by + 2hx + 2f

.

P1 2017 27 / 35

2.4 Implicit Functions in more than 2 variables

This section goes a bit beyond the syllabus, but is helpful because it testsyour ability to think about functions and partial derivatives.

P1 2017 28 / 35

[**] Implicit functions in more variablesWith a function of two variables, one equation f (x , y) = 0 was sufficientto determine y = y(x).

Suppose now that we have functions of 3 variables. Now we require twosuch functions, say

f (x , y , z) = 0 g(x , y , z) = 0

to define y = y(x) and z = z(x) implicitly.

Now if we write down the chain rule, we have

dfdx

=∂f∂x

+∂f∂y

dydx

+∂f∂z

dzdx

= 0

anddgdx

=∂g∂x

+∂g∂y

dydx

+∂g∂z

dzdx

= 0 .

P1 2017 29 / 35

/ctd

These last equations

∂f∂x

+∂f∂y

dydx

+∂f∂z

dzdx

= 0

and∂g∂x

+∂g∂y

dydx

+∂g∂z

dzdx

= 0 .

are two simultaneous equations in dy/dx and dz/dx which can be solved

dydx

=

(∂f∂z∂g∂x

−∂f∂x∂g∂z

)/

(∂f∂y∂g∂z

−∂f∂z∂g∂y

)and

dzdx

=

(∂f∂x∂g∂y

−∂f∂y∂g∂x

)/

(∂f∂y∂g∂z

−∂f∂z∂g∂y

).

P1 2017 30 / 35

[**] ♣ ExampleTry an example where the implicit functions can be derived explicitly.

Q: Find dy/dx and dz/dx when

f (x , y , z) = x + y + z = 0 g(x , y , z) = x − y + 2z = 0.

A: fx = fy = fz = 1 and gx = 1, gy = −1 and gz = 2. Thus using thesimultaneous equations

dydx

= −13

anddzdx

= −23

In this case we can solve explicitly and check the result.

f + g = 2x + 3z = 0 ⇒ dz/dx = −2/32f − g = x + 3y = 0 ⇒ dy/dx = −1/3

So the result is verified.

P1 2017 31 / 35

[**] Partial diff & implicit functions

Rather than having 2 functions of 3 variables, we have only one, vizf (x , y , z) = 0. This defines z = z(x , y) implicitly.

Instead of a second implicit function, suppose we know a point(x0, y0, z0) which satisfies f (x0, y0, z0) = 0. Also suppose that near thispoint, f and its first partial derivatives are continuous and ∂f /∂z = 0.

An existence theorem states that in the region around (x0, y0) there isprecisely one differentiable function z(x , y) which satisfies f (x , y , z) = 0and is such that z0 = z(x0, y0).

P1 2017 32 / 35

Partial Differentiation and implicit functions /ctd

Suppose now we fix y at y = y0; then f (x , y0, z) = 0.

This constraint implicitly defines z = z(x) in the neighbourhood of x0.

So, with y fixed at y0 we have f (x , y0, z) = 0 and z = z(x).

We’ve seen that for f (x , y) = 0, y = y(x) that:dydx

= −∂f /∂x∂f /∂y

.

Rewrite this for z instead of y :dzdx

= −∂f /∂x∂f /∂z

.

Nearly right — but we have fixed y at y0, so that dz/dx should be apartial not a total derivative. Thus

∂z∂x

= −∂f /∂x∂f /∂z

.

P1 2017 33 / 35

Differentiation of implicit functions /ctd

But our choice of x , y , z was arbitrary. So the general result is

For f = f (x1, x2, . . . , xn) = 0 provided certain conditions are met

∂xi

∂xj= −

∂f /∂xj

∂f /∂xi

P1 2017 34 / 35

♣ Example: Differentiation of implicit functions

Return to the perfect gas law — now written as f (p,V ,T ) = 0. Recallthat we asked the value of

∂p∂V

.∂V∂T

.∂T∂p

.

Using the result just obtained we could now write:

∂p∂V

.∂V∂T

.∂T∂p

=

[−∂f∂V

/∂f∂p

] [−∂f∂T

/∂f∂V

] [−∂f∂p

/∂f∂T

]= −1 .

Note this result is independent of the form the perfect gas law.

P1 2017 35 / 35

Summary

In this lecture we have considered partial differentiation in the context offunctions that have particular attributs.

We considered:2.1 A function of a function.2.2 Composite functions I,II,III, including the Chain Rule for Partials.2.3 Implicit functions in two variables.2.4 ** Implicit functions in more than two variables.

Next lecture: Changing variables ...... will use the Chain Rule for Partials over and over.