math 1131q - calculus 1. · a very important theorem: differentiability is a stronger condition...

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Our first Midterm Exam will be on Tuesday 9/30

Two times (check your time following these instructions)(1) 6-8 PM at AUST 108, or(2) 9-11 PM at TLS 154.

There is a practice exam and solutions in outline in the website:http://alozano.clas.uconn.edu/math1131f14

Our first Midterm Exam will be on Tuesday 9/30Two times (check your time following these instructions)(1) 6-8 PM at AUST 108, or(2) 9-11 PM at TLS 154.

Covers Sections 1.1 - 3.3 (including 3.3, derivatives oftrigonometric functions).

There will be a review during class on Tuesday 9/30. Bring yourquestions, and/or I will go over the practice exam.

My office hours are as usual: Tuesdays 10:15-11:15 andThursdays 11-12, at MSB 312. TA’s office hours also as usual(see website).

There will be a review session on Monday 9/29 (by Amit Savkar;time and room TBA; will also be available online).

MATH 1131Q - Calculus 1.

Álvaro Lozano-Robledo

Department of MathematicsUniversity of Connecticut

Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 5 / 30

Derivatives(Rates of Change)

The Derivative as a Function

DefinitionThe derivative of a function f at a number a, denoted by f ′(a), isdefined by

f ′(a) = limh→0

f (a + h)− f (a)

hif this limit exists and it is finite.

We define a new function, the derivative of f :

DefinitionLet f (x) be a function. We define the derivative of f , denoted by f ′, by

f ′(x) = limh→0

f (x + h)− f (x)

The domain of f ′(x) are those values x where the limit exists and it isfinite.

Differentiability

DefinitionA function f (x) is continuous at a number a if

1 f (x) is defined at x = a, i.e., f (a) is well-defined,

2 limx→a

f (x) exists, and

3 limx→a

f (x) = f (a).

DefinitionA function f (x) is differentiable at x = a if f ′(a) exists, i.e., if the limit

f ′(a) = limh→0

f (a + h)− f (a)

exists and it is finite. We say f (x) is differentiable in a set of points (e.g., anopen interval) if it is differentiable at every point in the set.

Continuity and Differentiability

A very important theorem: differentiability is a stronger condition thancontinuity.

TheoremIf f (x) is differentiable at x = a, then f (x) is continuous at x = a.

Warning! The converse is not true. The function f (x) = |x | iscontinuous at x = 0, but not differentiable at x = 0.

Differentiation Rules

The Derivative of Elementary Functions

TheoremLet f (x) be a function.

If f (x) = c is constant for all x, then f ′(x) = 0.

If f (x) = x for all x, then f ′(x) = 1.

If f (x) = xn where n is a real number, then f ′(x) = nxn−1.

If f (x) = ax where a is a positive real number, then

f ′(x) = f ′(0)f (x) = f ′(0)ax where f ′(0) = limh→0

ah − 1

DefinitionWe define the number e as the unique real number a with the property

limh→0

ah − 1

Corollaryd

dx(ex) = ex .

DefinitionWe define the number e as the unique real number a with the property

limh→0

ah − 1

Corollaryd

dx(ex) = ex .

New Derivatives from Old

Some helpful rules of differentiation:

TheoremLet c be a constant, and let f and g be differentiable functions. Then:

1 ddx (cf (x)) = c d

dx f (x), i.e., (cf (x))′ = cf ′(x).

2 ddx (f (x) + g(x)) = d

dx f (x) + ddx g(x), i.e., (f + g)′ = f ′ + g′.

3 ddx (f (x)− g(x)) = d

dx f (x)− ddx g(x), i.e., (f − g)′ = f ′ − g′.

Example

What is the derivative of f (x) = x2ex?

We need a “product rule” for differentiation.

Example

We need a “product rule” for differentiation.

The Product Rule

Let f (x) and g(x) be two differentiable functions, and letp(x) = f (x)g(x). Then:

p′(x) = limh→0

p(x + h)− p(x)

= limh→0

f (x + h)g(x + h)− f (x)g(x)

= limh→0

f (x + h)g(x + h)− f (x)g(x + h) + f (x)g(x + h)− f (x)g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + f (x)

g(x + h)− g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + lim

h→0f (x)

g(x + h)− g(x)

h= f ′(x)g(x) + f (x)g′(x).

The Product Rule

p′(x) = limh→0

p(x + h)− p(x)

= limh→0

f (x + h)g(x + h)− f (x)g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + f (x)

g(x + h)− g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + lim

h→0f (x)

g(x + h)− g(x)

h= f ′(x)g(x) + f (x)g′(x).

The Product Rule

p′(x) = limh→0

p(x + h)− p(x)

= limh→0

f (x + h)g(x + h)− f (x)g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + f (x)

g(x + h)− g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + lim

h→0f (x)

g(x + h)− g(x)

h= f ′(x)g(x) + f (x)g′(x).

The Product Rule

p′(x) = limh→0

p(x + h)− p(x)

= limh→0

f (x + h)g(x + h)− f (x)g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + f (x)

g(x + h)− g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + lim

h→0f (x)

g(x + h)− g(x)

h= f ′(x)g(x) + f (x)g′(x).

The Product Rule

p′(x) = limh→0

p(x + h)− p(x)

= limh→0

f (x + h)g(x + h)− f (x)g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + f (x)

g(x + h)− g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + lim

h→0f (x)

g(x + h)− g(x)

h= f ′(x)g(x) + f (x)g′(x).

The Product Rule

p′(x) = limh→0

p(x + h)− p(x)

= limh→0

f (x + h)g(x + h)− f (x)g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + f (x)

g(x + h)− g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + lim

h→0f (x)

g(x + h)− g(x)

= f ′(x)g(x) + f (x)g′(x).

The Product Rule

p′(x) = limh→0

p(x + h)− p(x)

= limh→0

f (x + h)g(x + h)− f (x)g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + f (x)

g(x + h)− g(x)

= limh→0

f (x + h)− f (x)

hg(x + h) + lim

h→0f (x)

g(x + h)− g(x)

h= f ′(x)g(x) + f (x)g′(x).

The Product Rule

TheoremIf f (x) and g(x) are both differentiable, then

dx(f (x)g(x)) =

dxg(x) + f (x)

Example

The Product Rule

dx(f (x)g(x)) =

dxg(x) + f (x)

Example

The Product Rule

dx(f (x)g(x)) =

dxg(x) + f (x)

Example

If R(t) = v(t)p

t , where v(4) = 2 and v ′(4) = 3, find f ′(4).

The Product Rule

dx(f (x)g(x)) =

dxg(x) + f (x)

Example

Find the first derivative of f (x) =x2

We need a “quotient rule” of differentiation.

The Product Rule

dx(f (x)g(x)) =

dxg(x) + f (x)

Example

We need a “quotient rule” of differentiation.

The Quotient Rule

Let g(x) be a differentiable function, and let Q(x) = 1g(x) . Then:

Q′(x) = limh→0

Q(x + h)−Q(x)

= limh→0

1g(x+h) −

= limh→0

g(x)−g(x+h)g(x)g(x+h)

= limh→0

g(x)−g(x+h)h

g(x)g(x + h)

= limh→0

− g(x+h)−g(x)h

g(x)g(x + h)

=−g′(x)

(g(x))2

The Quotient Rule

Q′(x) = limh→0

Q(x + h)−Q(x)

= limh→0

1g(x+h) −

= limh→0

g(x)−g(x+h)h

g(x)g(x + h)

= limh→0

− g(x+h)−g(x)h

g(x)g(x + h)

=−g′(x)

(g(x))2

The Quotient Rule

Q′(x) = limh→0

Q(x + h)−Q(x)

= limh→0

1g(x+h) −

= limh→0

g(x)−g(x+h)h

g(x)g(x + h)

= limh→0

− g(x+h)−g(x)h

g(x)g(x + h)

=−g′(x)

(g(x))2

The Quotient Rule

Q′(x) = limh→0

Q(x + h)−Q(x)

= limh→0

1g(x+h) −

= limh→0

g(x)−g(x+h)h

g(x)g(x + h)

= limh→0

− g(x+h)−g(x)h

g(x)g(x + h)

=−g′(x)

(g(x))2

The Quotient Rule

Q′(x) = limh→0

Q(x + h)−Q(x)

= limh→0

1g(x+h) −

= limh→0

g(x)−g(x+h)h

g(x)g(x + h)

= limh→0

− g(x+h)−g(x)h

g(x)g(x + h)

=−g′(x)

(g(x))2

The Quotient Rule

Q′(x) = limh→0

Q(x + h)−Q(x)

= limh→0

1g(x+h) −

= limh→0

g(x)−g(x+h)h

g(x)g(x + h)

= limh→0

− g(x+h)−g(x)h

g(x)g(x + h)

=−g′(x)

(g(x))2

The Quotient Rule

Q′(x) = limh→0

Q(x + h)−Q(x)

= limh→0

1g(x+h) −

= limh→0

g(x)−g(x+h)h

g(x)g(x + h)

= limh→0

− g(x+h)−g(x)h

g(x)g(x + h)

=−g′(x)

(g(x))2

The Quotient Rule

Let g(x) be a differentiable function. Then:

�′

=−g′(x)

(g(x))2

Now, let f (x) be a diff. function, and consider f (x)g(x) = f (x) · 1

g(x) . Then:

f (x) ·1

�′

f ′(x)1

g(x)+ f (x)

�′

= f ′(x)1

g(x)+ f (x)

−g′(x)

(g(x))2

=f ′(x)g(x)− f (x)g′(x)

(g(x))2

The Quotient Rule

�′

=−g′(x)

(g(x))2

g(x) .

f (x) ·1

�′

f ′(x)1

g(x)+ f (x)

�′

= f ′(x)1

g(x)+ f (x)

−g′(x)

(g(x))2

=f ′(x)g(x)− f (x)g′(x)

(g(x))2

The Quotient Rule

�′

=−g′(x)

(g(x))2

g(x) . Then:

f (x) ·1

�′

f ′(x)1

g(x)+ f (x)

�′

= f ′(x)1

g(x)+ f (x)

−g′(x)

(g(x))2

=f ′(x)g(x)− f (x)g′(x)

(g(x))2

The Quotient Rule

�′

=−g′(x)

(g(x))2

g(x) . Then:

f (x) ·1

�′

= f ′(x)1

g(x)+ f (x)

�′

= f ′(x)1

g(x)+ f (x)

−g′(x)

(g(x))2

=f ′(x)g(x)− f (x)g′(x)

(g(x))2

The Quotient Rule

�′

=−g′(x)

(g(x))2

g(x) . Then:

f (x) ·1

�′

= f ′(x)1

g(x)+ f (x)

�′

= f ′(x)1

g(x)+ f (x)

−g′(x)

(g(x))2

=f ′(x)g(x)− f (x)g′(x)

(g(x))2

The Quotient Rule

�′

=−g′(x)

(g(x))2

g(x) . Then:

f (x) ·1

�′

= f ′(x)1

g(x)+ f (x)

�′

= f ′(x)1

g(x)+ f (x)

−g′(x)

(g(x))2

=f ′(x)g(x)− f (x)g′(x)

(g(x))2

The Quotient Rule

TheoremLet f (x) and g(x) be two differentiable functions. Then:

dfdx · g(x)− f (x) · dg

(g(x))2.

Example

The Quotient Rule

TheoremLet f (x) and g(x) be two differentiable functions. Then:

dfdx · g(x)− f (x) · dg

(g(x))2.

Example

The Quotient Rule

How to remember the formula?�

�′

=g(x)f ′(x)− f (x)g′(x)

The Quotient Rule Song

If the quotient rule you wish to know,it’s low d-high less high d-low.Draw the line and down belowdenominator squared will go!

Example

Find the equation of the tangent line to the curve y =ex

1 + x2at x = 0.

Example

Find the derivative of f (x) = x3 sin(x).

We need the derivative of sin(x) first!

Example

We need the derivative of sin(x) first!

Derivatives of trigonometric functions

f (x) = sin(x)

−3 −2 −1 1 2 3 4 5 6 7

f ′(x) = (sin(x))′

−3 −2 −1 1 2 3 4 5 6 7

f (x) = cos(x)

−3 −2 −1 1 2 3 4 5 6 7

f ′(x) = (cos(x))′

−3 −2 −1 1 2 3 4 5 6 7

Theorem

(sin(x))′ = cos(x), and (cos(x))′ = − sin(x).

Example

Theorem

(sin(x))′ = cos(x), and (cos(x))′ = − sin(x).

Example

ExampleFind the derivative of f (x) = tan x .

Two important limits

Recall:

limx→0

The proof is in the book.

Let us calculate: limx→0

cos x − 1

Two important limits

Recall:

limx→0

The proof is in the book. Let us calculate: limx→0

cos x − 1

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math 1131q - calculus 1. · a very important theorem: differentiability is a stronger condition...

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