The Leibniz rule: ECON3120/4120 curriculum - UiO

University of Oslo / Department of Economics / NCF

version 20180404

The Leibniz rule: ECON3120/4120 curriculum

About this document:

The Leibniz rule for dierentiating integrals is now curriculum.

This note claries what to expect (this semester!).

What you need to know:

Z b(x)

f (x, t) dt

If

F (x) =

Let

a = a(x), b = b(x)

f = f (x, t)

and

t),

(note: integration wrt.

?

be given functions .

then

a(x)

0

0

Z

0

b(x)

?

f (x, t) dt

?x

F (x) = f (x, b(x)) ¡¤ b (x) ? f (x, a(x)) ¡¤ a (x) +

a(x)

(?)

?

In this Wikipedia permalink the curriculum part stops before the Contents table.

What you need to be able to do:

Calculate derivatives when you are asked and when

you need it, just like any other dierentiation rule.

Why it works? An application of the chain rule: A function H(x, y, z) has dierential

dH = Hx0 dx + Hy0 dy + Hz0 dz , and if y = a(x) and z = b(x), then dy = a0 (x) dx and





dz = b0 (x)dx, so that F (x) = H(x, a(x), b(x)) has dierential dF = Hx0 +Hy0 a0 +Hz0 b0 dx .

Rz

Let H(x, y, z) =

f (x, t) dt and calculate partial derivatives.

y

?

Hz0 , note that Rx is treated as constant when we integrate with respect to t. If

z

G (t) = g(t), then y g(t) = G(z) ? G(y), and the derivative wrt. z is G0 (z) = g(z).

If we now introduce a constant in the notation and write f (x, t) for this  g(t) with x

as parameter, then it is the t variable that should be put equal to z . So

For

0

Hz0 (x, y, z) = f (x, z)

Hy0 (x, y, z) = ?f (x, y)

?

Analogously,

?

From the denition of partial derivatives,

?

?x

Z

Rz

z

f (x, t) dt = lim

y

y

because

f (x + , t) dt ?

¡ú0

Rz



y

y

is the lower limit (recall

Hx0 (x, y, z)

f (x, t) dt

Rz

y

=?

Ry

z

).

equals

Z

= lim

¡ú0

y

z

f (x + , t) ? f (x, t)

dt



?

?

and by moving the lim inside the integral sign , the integrand becomes

f (x, t).

?x

Inserting in

?

?

F 0 = Hx0 + Hy0 a0 + Hz0 b0

yields the three terms on the right-hand side of (?).

in Mathematics 2, they will be assumed suciently nice, and no precise conditions will be given



for

mobile-view version or for copy/paste. Note the permalink, immune to Wikipedia edits.

?

yes, you are allowed to do that in Mathematics 2, as all functions are nice enough; no, it is not obvious

1

Examples/exercises:

This used to be Mathematics 3 curriculum, so problems can be

?

found in the Mathematics 3 compendium, The following are just examples:

(1) Let

F (x) =

Rx

x

x2

tete dt.

Answer: By the Leibniz rule,

But

b(0) = a(0) = 0,

F 0 (0) = 0, with and without using Leibniz's rule.



R b(0)  x

0

0

b(0)eb(0)e b0 (0) ? a(0)ea(0)e a0 (0) + a(0) tete ¡¤ tex x=0 dt.

R0

Show that

so all the terms are zero (the integral because it is

0

). Solving

without: exercise. Take note that it is a bit more work, maybe?

F 00 (0).

R b(x)

x

x

0

b(x)ex 0

Answer: The formula yields F (x) = b(x)e

b (x)?a(x)ea(x)e a0 (x)+ a(x) [tete ¡¤tex ]dt,

Rx

2 x

x

2

0

xex

with b(x) = x and a(x) = x ; so, F (x) = xe

? 2x3 ex e + x2 [tete ¡¤ tex ] dt. The

x

xexe

xex

derivative at zero of xe

is limx¡ú0

= 1 (exercise (i): what did I just do?) and

x

3 x2 ex

2 x2 ex

the derivative zero of 2x e

is limx¡ú0 2x e

= 0 (exercise (ii): as (i)  what did I

(2) Let

F (x)

as in Example 1. Find

just do?) To dierentiate the integral term, use the Leibniz rule again. Exercise (iii):

?

show that you get zero from that term, so the answer is 1.

R e ?1 (1+x2 )t

d

(3) This you need the Leibniz rule for: Find

t e

dt.

dx

1

Re

R e ?1 ? (1+x2 )t

(1+x2 )t

dt. Notice that now the bothersome

t ?x e

dt = 1 2xe

Answer: We get

1

t=e

 (1+x2 )e

2

2

2x

2x

t?1 is gone! The rest is routine: 1+x

e(1+x )t = 1+x

? e1+x

2

2 e

t=1

(4) More generally: Given continuously dierentiable functions a, b and w , which do never

R b(x) ?1 tw(x)

d

t e

dt.

hit zero. Explain how to calculate

dx a(x)

R b(x)

b(x)w(x) 0

Answer: We get e

b (x)/b(x) ? ea(x)w(x) a0 (x)/a(x) + w0 (x) a(x) etw(x) dt. The latter

integral is solved by the substitution u = tw .

(5) Let

p(t) = p?(t) +

R ¡Ìp(t)

0

dierential equation for

2

eq ?p(t) dq . Deduce (but

p without evaluating the

do not try to solve!) a second-order

integral.

q ! We have p? =

R ¡Ìp q2 ?p

¡¤ (?p?) dq . The last term is ?p? ¡¤ (p ? p?). Gathering terms, we

p? + e

¡¤

+ 0 e




1

2

get 1 + p ? ¡Ì p? ? (p?) = p?.

2 p

R ¡Ìp q 2

R ¡Ìp q 2

R ¡Ìp q 2

?p

?p

?p d

Alternatively: rewrite as p?+e

e

dq

. Then p? = p??p?e

e

dq+e

e dq .

dt 0 p

0

0

?p p d

The second term is ?p?(p ? p?) again. For the third, use Leibniz's rule: e e ¡¤

p(t).

dt

Answer: Beware that now the variable to be integrated is called

p?p

?

p?

¡Ì

2 p



prob-

lems 4-01/02/10/11 (and, possibly with some hints, -03 and -04, which involve unknown functions).

?

An alternative, and somewhat more sophisticated approach on the integral, using the intermediate value

x

x2 and x (make the area

under graph argument, where the interval has width x ? x). Divide by x (as we seek the limit of

x

F 0 (x)/x as per exercises (i) and (ii)) to get T 2 ex+T e ¡¤ [1 ? x]. T is not known when x > 0, but is

2

squeezed between x and x, and must ¡ú 0 as x ¡ú 0 (like an argument in the last part of Term

theorem: for xed

x,

the integral is

T 2 ex+T e ¡¤ [x ? x2 ]

for some

T

between

2

Paper Problem 2!). So we get 0 from the integral and again, we are left with only the contribution

x

limx¡ú0 b(x)eb(x)e b0 (x) = 1.

2

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