Differential Calculus: A refresher · GWU Math Camp 2014 Differential Calculus I –Rafael López-Monti Differential Calculus: A refresher (Part 1) Math Camp, August 2014 Draft for

Differential Calculus I – Rafael López-MontiGWU Math Camp 2014 Differential Calculus I – Rafael López-MontiGWU Math Camp 2014

Differential Calculus: A refresher

(Part 1)

Math Camp, August 2014

Draft for teaching only, do not cite. Please contact me If you find a typo

Rafael López-MontiDepartment of Economics – PhD Program

George Washington University

[email protected]

1

mailto:[email protected]


Let 𝒇:𝑿 → ℝ, where 𝑿 ⊆ ℝ𝒏. Then, we say that 𝒇 is continuous at 𝒙 ∈ 𝑿 if for all 𝜺 > 𝟎, exits 𝜹 > 𝟎 such that ∀𝒙 ∈ 𝑩𝜹( 𝒙) ∩ 𝑿 we havethat 𝒇 𝒙 − 𝒇( 𝒙) < 𝜺.

Alternatively, we can say that 𝒇:𝑿 → ℝ is continuous at a point 𝒙 ∈ 𝑿if ∀ {𝒙𝒌} such that 𝒙𝒌 ∈ 𝑿 for all k, and 𝒍𝒊𝒎

𝒌→∞𝒙𝒌 = 𝒙, it is the case that

𝒍𝒊𝒎𝒌→∞

𝒇(𝒙𝒌) = 𝒇( 𝒙)

A function is said to be continuous if it is continuous at every point in itsdomain.

CONTINUOUS FUNCTIONS:

Note: Roughly speaking, a function is continuous “if you can draw it withoutlifting the pencil”

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Examples:

Note: In the first one, the function equals the limit value and this so acontinuous function at x=a. In the 2nd and 3rd examples, the limits andfunction values do not equal each other.

x x x

f(x) f(x) f(x)

In order to check if 𝒇 is continuous at 𝒙 = 𝒂, we need:i. 𝒇 𝒂 exitsii. 𝒍𝒊𝒎

𝒙→𝒂𝒇(𝒙) exits

iii. and 𝒍𝒊𝒎𝒙→𝒂

𝒇(𝒙) = 𝒇(𝒂)

3


Examples of discontinuous functions:

i. 𝒇 𝒙 =𝟏

𝒙−𝟑at 𝒙 = 𝟑

ii. 𝒇 𝒙 = 𝟏

𝒙𝟐, 𝒙 ≠ 𝟎

𝟐, 𝒙 = 𝟎at 𝒙 = 𝟎

iii. 𝒇 𝒙 = 𝒙𝟐−𝒙−𝟐

𝒙−𝟐, 𝒙 ≠ 𝟐

𝟓 , 𝒙 = 𝟐at 𝒙 =2 ⟹ 𝒍𝒊𝒎

𝒙→𝟐𝒇(𝒙) ≠ 𝒇(𝟐)

4


Right and Left Continuity:

A simple example:

f(x)

x0

1

i. 𝒇 𝟎 = 𝟏ii. 𝒍𝒊𝒎

𝒙⟶𝟎+𝒇(𝒙) = 𝟏

iii. However, 𝒍𝒊𝒎𝒙⟶𝟎−

𝒇(𝒙) ≠ 𝟏

⇒ 𝑪𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔 𝒇𝒓𝒐𝒎 𝒕𝒉𝒆 𝒓𝒊𝒈𝒉𝒕

1. We say 𝒇:𝑿 → ℝ is left-continuous at 𝒂 ∈ 𝑿 if:

𝒍𝒊𝒎𝒙⟶𝟎−

𝒇(𝒙) = 𝒇(𝒂)

2. We say 𝒇:𝑿 → ℝ is right-continuous at 𝒂 ∈ 𝑿 if:

𝒍𝒊𝒎𝒙⟶𝟎+

𝒇(𝒙) = 𝒇(𝒂)

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Graphically:

If 𝒇: [𝒂, 𝒃] → ℝ is continuous, and 𝒇 𝒂 > 𝒇 𝒃 , then ∀𝑨 ∈ 𝒇 𝒃 , 𝒇 𝒂 there isat least 𝜽 ∈ [𝒂, 𝒃] such that 𝑨 = 𝒇(𝜽).

THE INTERMEDIATE VALUE THEOREM:

𝜃

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Suppose 𝒇:𝑿 → ℝ, where 𝑿 ⊆ ℝ is open, fix 𝒙 ∈ 𝑿 and define:

𝑯𝒙 = 𝒉 ∈ ℝ\{𝟎 : (𝒙 + 𝒉) ∈ 𝑿}

Now, ∀𝒉 ∈ 𝑯𝒙, evaluate the following expression:

𝒇 𝒙 + 𝒉 − 𝒇(𝒙)

𝒉

Since 𝒙 is fixed, this expression depends on 𝒉 , on the nonempty domain 𝑯𝒙.Note that 0 is the limit point of 𝑯𝒙, so we want to study:

𝒍𝒊𝒎𝒉→𝟎

𝒇 𝒙 + 𝒉 − 𝒇(𝒙)

𝒉

A function, 𝒇:𝑿 → ℝ is said to be differentiable at 𝒙 ∈ 𝑿, if ∃𝒃 ∈ ℝ such that:


𝒇 𝒙 + 𝒉 − 𝒇(𝒙)

𝒉= 𝒃

Therefore, a function is said to be differentiable if it is differentiable at all 𝒙 ∈ 𝑿

DIFFERENTIABLE FUNCTIONS ON ℝ:

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𝒙𝟎

y

x

f(𝒙𝟎+h)

f(𝒙𝟎)

𝒙𝟎 + 𝒉

f(𝒙)

𝒉

𝚫𝒚

𝜷

Note: Graphically, the derivative is the slope of the tangent line to the

function at the point 𝒙𝟎 : 𝒕𝒂𝒏𝜷 = 𝒍𝒊𝒎𝒉→𝟎

𝜟𝒚

𝒉= 𝒇′ 𝒙𝟎

Let 𝒇:𝑿 → ℝ be differentiable at the point 𝒙𝟎 ∈ 𝑿, the DERIVATIVE of 𝒇 at𝒙𝟎 can be defined as:

ⅆ𝒚

ⅆ𝒙𝒙𝟎

= 𝒇′ 𝒙𝟎 = 𝒍𝒊𝒎𝒉→𝟎

𝒇 𝒙𝟎 + 𝒉 − 𝒇(𝒙𝟎)

𝒉= 𝒍𝒊𝒎

𝒉→𝟎

𝜟𝒚

𝒉

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Proof: Note that if ℎ ∈ 𝐻 𝑥, then ∃ 𝑥 ∈ 𝑋: 𝑥 = 𝑥 + ℎ, then we can rewrite:

limℎ→0

𝑓 𝑥 + ℎ − 𝑓( 𝑥)

ℎ= lim

𝑥→ 𝑥

𝑓 𝑥 − 𝑓( 𝑥)

𝑥 − 𝑥

If a function is differentiable at 𝑥 ∈ 𝑋, then exits b ∈ ℝ such that:

lim𝑥→ 𝑥

𝑓 𝑥 − 𝑓( 𝑥)

𝑥 − 𝑥= 𝑏

Notice that this is true for any 𝑥 ∈ 𝑋\{ 𝑥}, rewriting 𝑓 𝑥 :

𝑓 𝑥 = 𝑓 𝑥 +𝑓 𝑥 − 𝑓 𝑥

𝑥 − 𝑥𝑥 − 𝑥

Since lim𝑥→ 𝑥

𝑓 𝑥 = 𝑓 𝑥 ∈ ℝ, lim𝑥→ 𝑥

𝑓 𝑥 −𝑓 𝑥

𝑥− 𝑥= 𝑏 ∈ ℝ, and lim

𝑥→ 𝑥𝑥 − 𝑥 = 0. By the properties of

limits:

lim𝑥→ 𝑥

𝑓 𝑥 = lim𝑥→ 𝑥

𝑓 𝑥 +𝑓 𝑥 − 𝑓 𝑥

𝑥 − 𝑥𝑥 − 𝑥 = 𝑓 𝑥 + 𝑏. 0 = 𝑓 𝑥

then 𝑓 is continuous at 𝑥 ∎

If a function 𝒇:𝑿 → ℝ is differentiable at 𝒙 ∈ 𝑿, then it is continuous 𝒙

THEOREM DIFFERENTIABILITY AND CONTINUITY:

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Let 𝑓, 𝑔: 𝑋 → ℝ be differentiable functions. Then ∀𝑥 𝜖 𝑋 and ∀𝑘 𝜖 ℝ:

Refreshing some derivative rules:

𝑓 ± 𝑔 ′ 𝑥 = 𝑓′ 𝑥 ± 𝑔′ 𝑥

𝑘. 𝑓 ′ 𝑥 = 𝑘. 𝑓′ 𝑥

𝑓. 𝑔 ′ 𝑥 = 𝑔 𝑥 . 𝑓′ 𝑥 + 𝑓 𝑥 . 𝑔′ 𝑥

𝑓

𝑔

′

𝑥 =𝑔 𝑥 . 𝑓′ 𝑥 − 𝑓 𝑥 . 𝑔′ 𝑥

𝑔(𝑥)2

(𝑓 𝑥 )𝑘)′= 𝑘. (𝑓 𝑥 )𝑘−1. 𝑓′ 𝑥

(𝑓(𝑥)𝑔(𝑥))′ = (𝑒𝑔 𝑥 .𝑙𝑛𝑓 𝑥 )′ = 𝑓(𝑥)𝑔(𝑥). 𝑓′ 𝑥 .𝑔(𝑥)

𝑓(𝑥)+ 𝑔′ 𝑥 .ln𝑓(𝑥)

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Now, derivatives of logarithmic and exponential functions :

𝑑

𝑑𝑥𝑎𝑏𝑥 = 𝑎𝑏𝑥 . 𝑏. ln 𝑎 ,for 𝑎 > 0

𝑑

𝑑𝑥𝑒𝑥 = 𝑒𝑥

(Let’s prove it….!)

𝑑

𝑑𝑥ln 𝑥 =

1

𝑥

𝑑

𝑑𝑥𝑥𝑥 = 𝑥𝑥(1 + ln 𝑥)

Derivatives of trigonometric functions:

𝑑

𝑑𝑥sin 𝑥 = cos 𝑥

𝑑

𝑑𝑥cos 𝑥 = −sin 𝑥

𝑑

𝑑𝑥tan 𝑥 =

𝑑

𝑑𝑥

sin 𝑥

cos 𝑥=

1

𝑐𝑜𝑠2(𝑥)= 1 + (tan 𝑥)2

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Graphically:

If 𝒇 is continuous on [𝒂, 𝒃] and differentiable (𝒂, 𝒃), then there exits at least onepoint 𝜽 ∈ (𝒂, 𝒃) such that:

𝒇′ 𝜽 =𝒇 𝒃 − 𝒇(𝒂)

(𝒃 − 𝒂)

THE MEAN VALUE THEOREM:

𝜃

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DERIVATIVES AND LIMITS: L’ H 𝐨pital’s rule

Suppose that 𝑓: (𝑎, 𝑏) → ℝ and 𝑔: (𝑎, 𝑏) → ℝ\{0}, where −∞ < 𝑎 < 𝑏 <

∞, are differentiable, 𝑥 ∈ (𝑎, 𝑏), and for 𝑙 ∈ ℝ⋃ −∞,+∞ :

𝒍𝒊𝒎𝒙→ 𝒙

𝒇′ 𝒙

𝒈′(𝒙)= 𝒍

If 𝑙𝑖𝑚𝑥→ 𝑥

𝑓(𝑥) =0 and 𝑙𝑖𝑚𝑥→ 𝑥

𝑔(𝑥) =0, or if they tend to ±∞ then:

𝒍𝒊𝒎𝒙→ 𝒙

𝒇

𝒈𝒙 = 𝒍

Examples: Evaluate the following limits,

i. 𝑙𝑖𝑚𝑥→4

𝑥2−16

𝑥−4

ii. 𝑙𝑖𝑚𝑥→∞

𝑒𝑥

𝑥2

iii.𝑙𝑖𝑚𝑥→0

𝑠𝑖𝑛 𝑥

𝑥

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The chain rule: Suppose that 𝑓: 𝑋 → 𝑌 is differentiable at 𝑥 ∈ 𝑋, and𝑔: 𝑌 → ℝ is differentiable at 𝑓(𝑥). Then, 𝑔 ∘ 𝑓 : 𝑋 → ℝ is differentiableat 𝑥, and:

𝒈 ∘ 𝒇 ′ 𝒙 = 𝒈′ 𝒇(𝒙) . 𝒇′ 𝒙 =ⅆ𝒈(𝒇 𝒙 )

ⅆ 𝒇(𝒙).ⅆ𝒇(𝒙)

ⅆ𝒙

COMPOSITE FUNCTIONS:

Given the functions 𝑓: 𝑋 → 𝑌 and 𝑔: 𝑌 → ℝ , we define the compositefunction 𝑔 ∘ 𝑓 : 𝑋 → ℝ as follows: ∀𝑥 ∈ 𝑋, 𝑔 ∘ 𝑓 𝑥 = 𝑔(𝑓 𝑥 )

Example: Let 𝒇 𝒙 = 𝟐𝒙 + 𝟑 and 𝒈(𝒙) = −𝒙𝟐 + 𝟓i. Find 𝒈 ∘ 𝒇 𝒙ii. Find 𝒈 ∘ 𝒇 ′ 𝒙

Suppose that 𝒇:𝑿 → 𝒀 is continuous at 𝒙 ∈ 𝑿, and 𝒈:𝒀 → ℝ is continuous at𝒇(𝒙). Then, 𝒈 ∘ 𝒇 :𝑿 → ℝ is continuous at 𝒙.

THEOREM:

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(i) Recall that 𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑓 𝑥

Therefore 𝑔 𝑓 𝑥 = 𝑔 2𝑥 + 3

= −(2𝑥 + 3)2 + 5 = − 4𝑥2 + 12𝑥 + 9 + 5= −4𝑥2 − 12𝑥 − 4

Then 𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑓 𝑥 = −4𝑥2 − 12𝑥 − 4

(ii) Indirect way 𝑔 ∘ 𝑓 ′ 𝑥 =𝑑𝑔(𝑓 𝑥 )

𝑑 𝑓(𝑥).𝑑𝑓(𝑥)

𝑑𝑥= 𝑔′ 𝑓 𝑥 . 𝑓′ 𝑥

𝑔′ 𝑓 𝑥 = −2 2𝑥 + 3 = −4𝑥 − 6

𝑓′ 𝑥 = 2𝑔 ∘ 𝑓 ′ 𝑥 = −4𝑥 − 6 . 2 = −𝟖𝒙 − 𝟏𝟐

Alternatively, by using the result (i): 𝑔 𝑓 𝑥 = −4𝑥2 − 12𝑥 − 4,

then 𝒈 ∘ 𝒇 ′ 𝒙 = −𝟖𝒙 − 𝟏𝟐

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1234

INVERSE FUNCTIONS:

Definition: Let 𝒇:𝑿 ⟶ 𝒀, be a function. We say 𝒇 is one-to-one (or injective) ,iff for all 𝒙𝟏, 𝒙𝟐 ∈ 𝑿, if 𝒇 𝒙𝟏 = 𝒇 𝒙𝟐 then 𝒙𝟏 = 𝒙𝟐, or equivalently if 𝒙𝟏 ≠ 𝒙𝟐then 𝒇 𝒙𝟏 ≠ 𝒇 𝒙𝟐

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Definition: Let 𝒇:𝑿 ⟶ 𝒀, be a function. We say 𝒇 is onto (or surjective) , iff forany 𝒚 ∈ 𝒀 there exits some 𝒙 ∈ 𝑿 such that 𝒚 = 𝒇 𝒙 .

Graphically:

123

𝑿

1234

𝒀𝒇

one-to-one but NOT onto

1234

𝑿

123

𝒀𝒇

ontobut NOT one-to-one

1234

𝑿 𝒀𝒇

one-to-one AND onto


Definition: Suppose 𝒇:𝑿 ⟶ 𝒀 is a one-to-one correspondence function(i.e. one-to-one and onto). Then there is a function 𝒇−𝟏: 𝒀 ⟶ 𝑿, calledTHE INVERSE of 𝒇 defined as follows:

𝒇−𝟏 𝒚 = 𝒙 ⇔ 𝒇 𝒙 = 𝒚

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Note: If a function 𝑓: 𝑋 ⟶ 𝑌 is a one-to-one correspondence, then itassociates one and only one value of 𝑦 to each value in 𝑥

Definition: A one-to one correspondence (or bijection) from set 𝑿 to 𝒀 is afunction 𝒇: 𝑿 ⟶ 𝒀 which is both one-to-one AND onto

However: The conditions for the existence of an inverse function can berelaxed to restrict to those of a one-to-one functions…..Why?

Definition: Suppose 𝒇:𝑿 ⟶ 𝒀 is just a one-to-one function, and let 𝑪 ⊆ 𝒀 bethe image of 𝒇. Then there is a function 𝒇−𝟏: 𝑪 ⟶ 𝑿, called THE INVERSE of 𝒇defined as follows:

𝒇−𝟏 𝒚 = 𝒙 ⇔ 𝒇 𝒙 = 𝒚


y

x

f(𝒙)=𝒙𝟐

g(𝒙)=𝒇−𝟏= 𝒙

y=x

Note: The graphs of 𝑓 and 𝑔 = 𝑓−1 aresymmetric with respect to the 45 degreeline (y=x). This suggests that there must berelationship between their derivatives…..

Example:

Let 𝒇:𝑿 → ℝ be a continuous and differentiable function (𝑪𝟏) defined on theinterval 𝑿 ⊆ ℝ. If 𝒇′ 𝒙 ≠ 𝟎 ∀𝒙 ∈ 𝑿, then:

i. 𝒇 is invertible on Xii. Its inverse 𝒈 = 𝒇−𝟏 is 𝑪𝟏 on the interval 𝒇(𝑿)iii. For all z in the domain of the inverse function g:

𝒈′ 𝒛 =𝟏

𝒇′ 𝒈(𝒛)Proof: Left as an exercise (hint: you may want to use the Chain Rule!)

THE INVERSE FUNCTION THEOREM:

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HIGHER ORDER DERIVATIVES:

Suppose that 𝑓: 𝑋 → ℝ is differentiable, where X ⊆ ℝ. Then, ∀𝑥 ∈ 𝑋, ∃𝑓′(𝑥) ∈ ℝ.In other words, the derivative itself assigns a real number 𝑓′(𝑥) to each 𝑥 ∈ 𝑋.This means that 𝑓′𝑚𝑎𝑝𝑠 𝑋 → ℝ .

Consider 𝒇: 𝑿 → ℝ , and 𝒇 is 𝑪𝟏. If 𝒇′(𝒙) is differentiable at 𝒙 ∈ 𝑿, we say that 𝒇 istwice differentiable at 𝒙, and define the second-order derivative of 𝒇 at 𝒙,denoted by 𝒇′′ 𝒙 , to be the derivative of 𝒇′(𝒙) at 𝒙.

In other words, we define 𝒇′′ 𝒙 to be (𝒇′)′(𝒙), whenever 𝒇′ is differentiable at 𝒙.This means that ∃𝒃 ∈ ℝ such that:


𝒇′ 𝒙 + 𝒉 − 𝒇′(𝒙)

𝒉= 𝒃

and we define 𝒇′′ 𝒙 = 𝒃.

If 𝑓: 𝑋 → ℝ is twice differentiable (so it’s twice differentiable ∀𝑥 ∈ 𝑋) , then𝑓′′: 𝑋 → ℝ . if 𝑓′′ is continuous, we say that the function 𝒇 is twicecontinuously differentiable and 𝒇 ∈ 𝑪𝟐 , where 𝐶2 is the class of twicecontinuously differentiable functions. It follows then that 𝐶2 ⊆ 𝐶1.

19


Generalize to 𝒌𝒕𝒉 order derivative:

Consider 𝑓: 𝑋 → ℝ , and 𝑓 is 𝐶k−1. If 𝑓𝑘−1 is differentiable at 𝑥 ∈ 𝑋, we say

that 𝑓 is k times differentiable at 𝑥, and define the kth order derivative of 𝑓 at

𝑥, denoted 𝑓k 𝑥 , to be the derivative of 𝑓𝑘−1(𝑥) at 𝑥.

Once again, if 𝑓: 𝑋 → ℝ is k times differentiable (so it’s k times differentiable∀𝑥 ∈ 𝑋) , then 𝑓𝑘 𝑥 :𝑋 → ℝ . If 𝑓𝑘is continuous, we say that the function 𝒇 isk times continuously differentiable and 𝒇 ∈ 𝑪𝒌, where 𝐶𝑘 is the class of ktimes continuously differentiable functions. It follows then that :

𝐶𝑘 ⊆ 𝐶𝑘−1 ⊆ ⋯ ⊆ 𝐶2 ⊆ 𝐶1

Example: Consider the following function:

𝒇 𝒙 =

𝒙𝟑

𝟑, 𝒙 ≥ 𝟎

−𝒙𝟑

𝟑, 𝒙 < 𝟎

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A quick interpretation of the First and Second Derivative:Let consider a function 𝑓: 𝑋 → ℝ, , where X ⊆ ℝ, and evaluate the derivatives atthe point 𝑝 ∈ 𝑋.

First Derivative:The first derivative tells us whether a function is increasing or decreasing, and byhow much, thus:

i. If𝑑𝑓

𝑑𝑥𝑝 > 0, then 𝑓(𝑥) is an increasing function at x = p

ii. If𝑑𝑓

𝑑𝑥𝑝 < 0, then 𝑓(𝑥) is a decreasing function at x = p

iii. If𝑑𝑓

𝑑𝑥𝑝 = 0, then x = p is called a “critical point” of 𝑓(𝑥), a priori we don’t

know anything about the behavior of 𝑓(𝑥): it may be increasing, decreasing,or attends a local maximum or a local minimum at that point. [Next Section:Static Optimization]

Second Derivative:The second derivative tells us if the first derivative is increasing or decreasing.

i. If𝑑2𝑓

𝑑𝑥2𝑝 > 0, then 𝑓(𝑥) is convex at x = p

ii. If𝑑2𝑓

𝑑𝑥2𝑝 < 0, then 𝑓(𝑥) is concave at x = p

iii. If𝑑2𝑓

𝑑𝑥2𝑝 = 0, no new information.

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FUNCTIONS ON ℝ𝒌:

Let 𝑧: 𝑋 → ℝ, where 𝑋 ⊆ ℝ𝒌, in other words, 𝑧 = 𝑓(𝑥1, 𝑥2, . . 𝑥𝑘−1 , 𝑥𝑘 ) then:

𝜕𝑧

𝜕𝑥𝑖𝕩 =

𝜕𝑓

𝜕𝑥𝑖𝕩 = 𝐷𝑥𝑖𝑓, where 𝕩 ∈ 𝑋

all denote the derivative of 𝑓(𝑥1, 𝑥2, . . 𝑥𝑘−1 , 𝑥𝑘 ) with respect to 𝑥𝑖 , when allthe other variables are held constant. When 𝑓(𝑥1, 𝑥2, . . 𝑥𝑘−1 , 𝑥𝑘 ) ispartially differentiable with respect to 𝑥𝑖 for every 𝑖 ∈ {1, ..., k}, we say that𝒇 is partially differentiable.

Geometric interpretation for a function mapping from ℝ𝟐 to ℝ :

The partial derivatives of a function oftwo variables, 𝑧 = 𝑓(𝑥, 𝑦): 𝑓1

′(𝑥0, y0) isthe slope of the tangent line 𝑙𝑥, while𝑓2′(𝑥0, y0) is the slope of the tangent

line 𝑙𝑦

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Now, if the function𝜕𝑓

𝜕𝑥𝑖(𝕩) is differentiable w.r.t. 𝑥𝑗 at 𝕩, then we get the

second derivative of 𝑓 with respect to 𝑥𝑗 and 𝑥𝑖:

𝜕2𝑓

𝜕𝑥𝑖𝜕𝑥𝑗(𝕩) =

𝜕𝜕𝑓𝜕𝑥𝑖𝜕𝑥𝑗

(𝕩)

We define the Hessian matrix of 𝑓 at 𝕩 , denoted by 𝐻𝑓 𝕩 , as a squaredmatrix (k x k) of the second-order derivatives

𝑯𝒇 𝕩 =

𝝏𝟐𝒇

𝛛𝒙𝟏𝛛𝒙𝟏(𝕩) ⋯

𝝏𝟐𝒇

𝛛𝒙𝟏𝛛𝒙𝒌(𝕩)

⋮ ⋱ ⋮𝝏𝟐𝒇

𝛛𝒙𝒌𝛛𝒙𝟏(𝕩) ⋯

𝝏𝟐𝒇

𝛛𝒙𝒌𝛛𝒙𝒌(𝕩)

If 𝒇: 𝑿 → ℝ, where 𝑿 ⊆ ℝ𝒌, is twice-partially differentiable, then:𝝏𝟐𝒇

𝝏𝒙𝒊𝝏𝒙𝒋(𝕩) =

𝝏𝟐𝒇

𝝏𝒙𝒋𝝏𝒙𝒊(𝕩), for 𝒊, 𝒋 = 𝟏, 𝟐, … , 𝒌

Therefore, Hessian matrix is symmetric!

THE YOUNG’S THEOREM:

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