Derivatives, Antiderivatives, and Integrals

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Calculus III

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CHAPTER 7 IMPLEMENTING STRATEGY

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Derivatives, Antiderivatives, and Integrals

Maxima and Minima

Mean Value Theorem

Then there's a number c in [a,b] such that

f'(c)=f(b)-f(a)/b-a

f(b)-f(a)=f'(c)(b-a)

CANNOT GO INTO THE NEGATIVES

Note: Rolle Theorem is a special case of MVT where fc) =f(b)-f(a)/b-a=0. It is all of MVT with added conditions that f|b|=f|a|

Let f be a function such that:

2. f is differentiable on [a,b]

1. f is continuous on a closed interval [a.b]

NO SHARP POINTS OR HOLES

Do not forget to test your x values

Critical Points

Critical Points does NOT GURANTEE max or min

Thus, we can rephrase Fermat's Theorem

if f has a local max or min at c, then c is called a critical point of f.

An exterior point c of the domain D of f such that, f'(c)=0 or f'(c) fails to exists

Locating Absolute Extreme Values on a closed interval

3. The largest value will be the absolute Max and the smallest will be the absolute min

2. Evaluate f at all critical numbers AND endpoints

1. Find all critical numbers (where f'(c)=0 or f'(c)=DNE)

And f(c) is the absolute minimum value of f on D. If f(c) is greater than or equal to f(x) for all x in D.

Fermat's Theorem

If f has a local max or min at c, f'(c) exists, then f'(c)=0

max and min are often referred as EXTREMA

Let f(c) be a number in the domain D of function. Then f(c) is the absolute maximum value of f on D if y is greater than or equal to y and f(c) is greater then or equal to f(x) for all x in D.

Extreme Value Theorem

If f is continuous on a closed interval [a,b], then f has an absolute maximum f(c) and minimum f(d) in [a,b].

c and d can be ENDPOINTS

Let c be a number in the domain D of a function f(c) is a local maximum value of f on D if f(c) is greater than or equal to f(x) for all values of x neighborhood around c.

Curve Sketching

Graphing Guidelines

7. Choose the appropriate window and graph

Make sure that you know what you are graphing

6. Determine concavity and inflection points

Derivative has to exist for it to change

IP is when the signs changes(negative and positive)

5. Identify extrema values.

Where the first derivative changes signs

4.Find the intervals of increasing and decreasing

Create number line to know where it is positive or negative

Critical points, y'=0

First Derivative

3. Find the Asymptotes

b. Vertical

denominator is 0

a. Horizontal

use limits to find

2. Find the intercepts

y-intercepts set x=0

x-intercepts set y=0

1. Identify the Domain/Interval of interest

log(x) is greater than 0

Even radicals is greater than or equal to 0

Denominator can not equal 0

Affect the Shape of a Graph

Concavity

Test for Concavity

If c is a point at which f''(x) changes sign at c, then f has an inflection point at c.

If f''(x) less than 0 on I, then f is concave down

If f''(x) greater than 0 on I, then f is concave up

if f is ontinuous at c and f changes concavity at c (from up to down or from down to up), then f has an inflection point at c.

if the graph of f lies below all of its tangents on an interval I, then it is said to be concave down.

Definition: If the graph of f lies above all of its tangents on interval I, then it is said to be concave up

Second Derivative Test

If f''(c)=0, then the second derivative test is inconclusive

If f''(c) is less than 0, then f has a local maximum.

Suppose f'' is continuous on an open interval containing c with f'(c)=0

First Derivative Test

Supposed f is continuous on an interval that contains a critical point c and is differentiable on the interval, except possibly c.

If f' positive on both sides of c or negative on both sides of c, then f has no extreme value at c.

if f' changes from negative to positive as x increases through c (from left to right), then f has a local minimum at c.

If f' changes sign from positive to negative as x increases through c (from left to right), then f has a local maximum at c.

There are infinite ways to make something true

A function f is decreasing on an interval I if f(x1)≥f(x2) whenever x1 ≥x2

A function f is increasing on an interval I if f(x1)⩽f(x2) whenever x1 ⩽ x2

How are they used?

An antiderivative is a function that accomplishes the opposite of what a derivative does. There are several antiderivatives of one function, but they always have the form of a function plus an arbitrary constant. Indefinite integrals rely heavily on antiderivatives. Antiderivatives and the Fundamental Theorem of Calculus are important for calculating the sum of items and how much they increased over time.

An integral is a function whose derivative is a specified function. Integration is mostly used to compute the areas of two-dimensional regions and the volumes of three-dimensional objects. Finding the integral of a function with respect to x hence implies calculating the area to the X-axis from the curve.

Derivatives are used to calculate the rate of change of one item in relation to another. The derivatives can be used to determine the equation of tangent and normal line to a function's curve. A function's derivative can be used to get the linear approximation of a function at a given value.

Derivatives

mtan=0

mtan<0

Zero Slope

mtan>0

Negative Slope

example:x=0

Slope

Positive Slope

Interpretations of Derivative

Physical interpretation, if the signed distance s from the orgin at time t of an object in linear motion is given by the position s=f(t), the derivative f'(t0) is the velocity of the object at time t0

Geometric interpretation if y=f(x), the derivative f'(c) is the slope of the tangent line to the graph of f at point (c,f(c))

4. The function does not exist at a.

Discontinuous means DNE

3. There is a vertical tangent at a.

Optimization

Example with steps

A rancher has 400 ft of fence for constructing three adjacent rectangular corrals. One side will be formed by the barn and requires no fence. What dimensions of the corrals will maximize the enclosed area?

400=8x

400/8=x

x=50

4(50)+y=400

y=200

Step 5: Use the First Derivative Test to find the absolute max or min. If necessary, check end points.

A'=400-8x

Step 4: Write the optimization equation expressed with a single variable

A=400x-4x^2

Step 3: Use the constraints to eliminate all but one independent variable.

A=(400-4x)x

Step 2: Identify the Optimization equation. Write it in terms of the variables from the constraints.

Step 1:Read the problem, draw a picture of diagram to repents all quantities and identify the constraints

4x+y=400

Generally involves a condition called the constraints and on optimization functions.

Optimization typically ask to find a max or min value of an object function

opposite reciprical

Linear Approximations and Differentials

dy=f'(x)dx

If y=f(x), where f is differentiable function, then the differential dx is an independent variable . That is, dx can be given the value of any number. The differential dy is then defined in terms of dx by equation.

Implicit Differentiation

Find dy/dx of x^2+y^2=1

x^2(d/dx)+y^2(d/dx)=1(d/dx)

2x+2y=0

2x+2y(dy/dx)=0

2y(dy/dx)=-2x

dy/dx=-2x/2y

Answer: dy/dx=-x/y

this is why the dx/dy notation is important?

The derivative of y with respect to x.

Equations that are not written in terms of one variable are called implicit equations. Some implicit equations cannot be written explicitly.

x^2+y^2=1

y^2+x^2+1

y=+/- square root (x^2+1)

half a circle

f(x)=f(a)+f'(a)(x-a)

f'(x)=f(x)-f(a)/x-a

is called the linear approximation or tangent line approximation of f at a. The linear function whose graph is this tangent line, that is L(x)=f(a)+f'(a)(x-a) is called the linearization of f at a.

a function will fail to be differentiable at a point a if

2. The function is discontinuous at a.

1. There is a casp or kink at a (bounce)

Derivative of a function

Higher Order Derivatives

f'(x) f''(x) f'''(x) f^4(x)

y' y'' y''' y^4

Dx D^2x D^3x D^nx

dy/dx d^2y/dx^2 d^3y/dx^3 d^ny/dx^n

The derivative of f is f(x) prime=dy/dx=limf(a+h)+f(a)/h

Provided the limit exists and a is in the domain of f. If f(x) prime exists, we say f is differentiable at a. If f is differentiable on every point of an interval I, we say that f is differentiable on I

mtan=limf(a+h)+f(a)/h

h→0

Derivative at a point

Find the tangent line to the function f(x)=-3x^2-5x+1 at (1,-7), Then find the normal line

First thing you are looking for is f(a+h)

-3(a+h)^2-5(a+h)+1

New Notation

Alternate definition

f on the interval [a,a+h]

From this we can get the tangent line passing through point (a,f(a))

Using the point-slope formula

mtan=point

msec=(f(x)-f(a))/x-a and mtan=limf(x)-f(a)/x-a

mtan=f(a) prime

x→a

Remember that a function f on the interval [a,x]

Hyperbolic Funtioncs

Derivatives of Hyperbolic Functions

d/dx(sechx)=-sech x tanh x

d/dx(coth x)= -csch^2 x

d/dx(csch x)= -cschxcoth x

d/dx(sinhx)=cosh x

d/dx(cosh x)=sinh x

d/dx(tanhx)= sech^2 x

Hyperbolic Identities

cosh^2x-sinh^2x=1

cosh(x+y)=coshxcoshy+sinhxsinhy

1-tanh^2x=sech^2x

sin^2x+cos^2x=1

cosh(-x)=-cosh(x)

sinh(x+y)=sinhxcoshy+coshysinhx

sinh(-x)=-sinh x

Hyperbolic functions arose from the comparison of semicircular regions.

Functions

sinh x = e^x-e^-x/2

Range: (-∞, ∞)

Domain:(-∞, ∞)

tanh x = sinh x/cosh x=e^x-e^-x/e^x+e^-x

Range: (-1,1)

Domain: (-∞,∞)

sech x = 1/cosh x = 2/e^x+e^-x

Range: (0,1]

Domain: (-∞, ∞)

coth x=cosh x/sinh x=e^x+e^-x/e^x-e^-x

Range: (-∞, -1)u(1,∞)

csch x = 1/sinh x = 2/e^x-e^-x

Range: (-∞, 0)u(0,∞)

Domain: (-∞, 0)u(0,∞)

cosh x = e^x+e^-x/2

Range: [1,∞)

Domain: (-∞, ∞)

Derivatives of Log and inverse

Derivative of Logarithms

Example:

y=1/log6x^2

rewrite as 1/lnx^2/ln6

Do KCF(Keep Change and Flip)

ln6/2ln x

1/2(ln6)(ln x)^-1

-ln6/2x(ln x)^2

Using b^ylnb and implicit differentiation of y=logbX

we can find

logbx=ln x/lnb

d/dx(log b x=1/xlnb

Table of Derivatives of Inverse Trigonometric Functions

d/dx(cot^-1 x) = -1/1+x^2

d/dx(sec^-1 x) = 1/x√1-x^2

d/dx(sin^-1 x) = 1/√1-x^2

d/dx(cos^-1 x) = -1/√1-x^2

d/dx(csc^-1 x) = -1/x√1-x^2

d/dx(tan^-1 x) = 1/1+x^2

Logarithmic Differentiation

You should distinguish carefully between the Power Rule, where the base is variable and the exponent is constant, and the rule for differentiating exponential functions, where the base is constant and the exponent is variable.

b^x=b^xln b

Power Rule

x^n=nx^n-1

Steps

3. Solve the resulting equation for y' and replace y by f(x)

2. Differentiate implicitly with respect to x

!. Take natural logarithms of both sides of an equation y= f(x) and use the Laws of Logarithms to expand the expression.

The calculation of derivatives of complicated functions involving products, or powers can be simplified by taking logarithms

If we don't do logarithmic differentiation, we have to use Quotient Rule or Product Rule.

Derivatives of Exponential Function(Other than e^x)

Find the derivative of y=x^3(3^x)

3x^2(3^x)+(ln3)3^x(x^3)

x^2(3^x)(3+xln3)

s= 3^x

s'=(ln3)3^x

f=x^3

f'=3x^2

e^xlnb

ln b e ^xln b

d/dx(b^x)

ln b(b^x)

b cannot equal 1

b is greater than 0

Rules of exponents can redefine b^x=e^ln bx=e^xlnb

use this variation to differentiate b^x

Derivative of Natural Log

Use definition and implicit differentiation to find d/dx(ln x)

y=ln x

(dy/dx)e^y=x(dy/dx)

dydx(y^y)=x

dy/dx(1/e^y)=1/e^lnx

Answer: 1/x

Hint: Use properties of Logarithmics

Domain of ln x is (0, positive infinity)

because ln(1/1-x) is greater than 0

By definition y= ln x if and only if x=e^y

e^ln x=x

e^ln cancels

Power, Product, Quotient, and Chain Rule still applies

Chain Rule

2 versions

Given y=f(g(x)) then u=g(x) and y=f(u)

g'(x) times f'(u)

u' times f'(u)

Sin^3(x)

(sinx)^3

(cos x)^3 sin^2x

Answer: 3(cos x)sin^2 x

f(u)=x^3

f'(u)=3x^2

u=sin x

u'=cos x

Reminder: Substitute your u for g(x) when done finding derivative

The derivative of the inside times the derivative of the outside

Steps: 1. find the derivatives 2. substitute into formula 3. put g(x) back into f'(x) and multiply by g'(x). 4. Combine like terms

If g is differentiable at x and f is differentiable at g(x), then the composite function F= f o g defined by F(x)=f(g(x)) is differentiable at x and F' is given by the product

F'(x)=f'(g(x)) times g'(x)

Limits and Derivatives of Trig Functions

Trig Functions

d/dx(cot x) = -csc^2 x

d/dx(sec x)= sec x tan x

d/dx(csc x)= - csc x cot x

d/dx(tan x)=sec^2(x)

d/dx(cos x)=-sin x

d/dx(sin x)=cos x

Two limits are easy to evaluate

lim cos x =cos x h→0

lim cos h/h =0 h→0

lim sin x =sin x h→0

lim sin h/h=1 h→0

Product and Quotient Rule

Derivative of e

x^2e^2x

g'=2e^x

2xe^2x+2e^2x(x^2)

Answer 2xe^2x+2x^2e^2x

f'=2x

e=2.71828

d/dxa^kx=klna(a^kx)

d/dxe^x=e^kx

K=real numbers

lime^h-1/h=1 h to 0

Quotient Rule

(x^2+3x-1)/x^2-1

((2x-3)(x^2-1)-2x(x^2-3x-1)/(x^2-1)^2

Make sure to combine all like terms: 2x^3-3x^2-2x+3-2x^3+6x+2x/(x^2-1)^2

Answer: 4x^2-3/(x^2-1)^2

B=x^2-1

B'=2x

T=x^2-3x-1

T'=2x-3

If f and g are differentiable at x and g(x) does not equal 0, then

d/dx f/g= (f'g-g'f)/g^2 or T'B-B'T/B^2

The derivative of the top times bottom minus the derivative of the bottom times the top all over the bottom squared.

Product Rule

Before starting Product Rule

Assume u=f(x) and v=g(x)

change in v=f(x+change in x)-g(x)

change in u=f(x+change in x)-f(x)

3x^3e^x

g=e^x

g'(x)=e^x

9x^2e^x+e^x(3x^3)

Answer: ex^3e^x+9x^2e^x

f=3x^3

f'(x)=9x^2

(fg)' does not equal f'g'

If f and g are differentiable,

d/dxfg=f'g+g'f

The derivative of the first times the second plus the derivative of the second times the first

Derivative of Polynomials/ Exponential Functions

Exponential Functions

f'(x)=lim(f(a+h)+f(a))/h h→0

f'(x)=lim b^x+h-b^x/h h→0

f'(x)=lim b^xb^h-b^x/h h→0

f'(x)=lim b^x(b^h-1)/h h→0

f'(0)f'(x)=limb^h-1 h→0

f(x)=b^x

f'(x)=f'(0)b^x

The rate of change of any exponential function is proportional to the function itself (slope is proportional to height).

Constant Rule

if c is a constant then f(x)=C

Process

lim (c-c)/h h→0

lim 0=0 h→0

d/dx(c)=0

lim(f(a+h)+f(a))/h h→0

Substitution Rule

For trig functions, you must use DOUBLE ANGLE FORMULAS

Substitution Rule for Definite Integrals

If g' is continuous on [a,b] and f is continuous on the range of u=g(x) then

b∫af(g(x)g'(x)dx=g(b)∫g(a) f(u)du

Example

∫2(2x+1)^3dx

1/4u^4+C

1/4(2x+1)^4+C

f(u)=2u^3

u=2x+1

du=2dx

SUBSTITUTION RULE DOES NOT APPLY TO ALL INTEGRALS

Procedures for indefinite integrals

4. Substitute g(x) into the result and add C

3. Evaluate the new integral with respect to u.

2. Substitute du=g'(x)dx and f(u) in the integral.

Note: you will need to solve for dx before you can substitute it in.

1. Given an indefinite integral involving a composite f(g(x)), identify an inner function g(x) such that a constant multiple of g'(x) appears in the integrand.

constants can be anywhere, does not affect the solving process.

Do not forget to substitute u into final equation

Undoing chain rule

if u=g(c) is a differentiable function whose range is an interval I and f is continuous on I then,

∫f(g(c)g'(c)=∫f(u)du

Antiderivatives and Integrals

Antiderivative

Finding a Particular Solutions

The antiderivative F(x)+C is the general solution. If we want to find a particular solution, we need to have a point (x,y) of F(x).

Example: F(x)=4x+2 with F(1)=8

∫(4x+2)dx

4(1/1+1)x^1+1+2x+C

2x^2+2x+C

2(1)^2+2(1)+C=8

c=4

F(x)=2x^2+2x+4

The process of finding the antiderivative is called integration. We use a symbol called the integral sign and write

dy=∫f(x)dx=F(x)+C

Tells you what you are looking for

Definition: a function F is an antiderivative of a function f when for every x in the domain of f F'(x)=f(x)

So the antiderivative of f(x)=3x^2 is F(x)=x^3+C, where C is a constant

Indefinite Integrals and Net Change

Example: ∫(3x^3+3x^2-5x-2e^2x)dx

3(1/3+1)x^3+1+3(1/2+1)x^2+1-5(1/1+1)x^1+1-2(1/2)e^2x+C

3(1/4)x^4+3(1/3)x^3-5(1/2)x^2-e^2x+C

Answer: 3/4x^4+x^3-5/2x^2-e^2x+C

simplify

Basic Integration Rules(Where k is a constant)

∫b^xdx=1/ln b(b^x)

∫e^axdx=1/ae^ax+C

∫(f(x)+/-g(x))dx=∫f(x)dx +/- ∫(g(x)dx+C

there is an invisible one, when its ∫dx=x

∫x^ndx=1/n+1x^n+1 +C

∫kf(x)dx=k∫f(x)+C

∫kdx=kx+C

REMEMBER THE +C

Integrals

∫b^xdx=b^x/ln b +C

∫e^x=e^+C

∫x^ndx=x^n+1/n+1+C

∫cscxcotxdx=-cscx+C

∫csc^2dx=-cotx+C

∫sec^2dx=tan x +C

∫secxtandx=secx+C

∫cosdx=sinx+C

∫sindx=-cos+C

∫1/xdx=ln|x|+C

∫cf(x)dx=c∫f(x)dx

Net Change Theorem

The integral of the rate is the net change

We know f'(x) is the rate of change of y=f(x) with respect to x. Then f(b)-f(a) is the change in y when x goes from a to b. Since y could go up, then down, the up again, between the intervals

Indefinite Integrals of Trigonometric Functions

Integral

Example: Evaluate the following

∫cos(x/2)dx

1/1/2sin(x/2)+C

Answer: 2sin(x/2)+C

∫cosh(ax)dx=1/asinh(ax)+C

∫sinh(ax)dx=1/acosh(ax)+C

∫1/x√ a^2-x^2dx=sec^-1(x/a)

∫1/√a^2+x^2dx=tan^-1(x/a)

∫1/√ a^2-x^2dx=sin^-1(x/a)

∫csc(ax)cot(ax)dx=-1/acot(ax)dx

∫sec(ax)dx=1/asec(ax)+C

∫csc^2(ax)dx=-1/acot(ax)+C

∫sec^2xax)dx=1/atan(ax)+C

∫sin(ax)dx=1/acos(ax)+C

∫cos(ax)dx=1/asin(ax)

Derivative of trigonometric functions

d/dx(csc ax)=-a csc(ax)cot(ax)

d/dx(cosh(ax)=asinh ax

d/dx(sinh(ax)=acosh ax

d/dx(tan^-1(x/a))=1/a^2+x^2

d/dx(sec^-1(x/a))-1/x√ a^2-x^2

d/dx(sec ax)=a sec(ax)tan(ax)

d/dx(sin^-1(x/a))=-1/√ a^2-x^2

d/dx(cot ax)=acsc^2(ax)

d/dx(tan ax)= asec^2(ax)

d/dx(cos ax)=-a sin(ax)

d/dx(sin ax)= a (cos(ax)

Definite Integrals

Example: Use geometry to find are of y=1+square root of 4-x^2

this is a half circle

r=4

A=pir^2(1/2)

1/2pi(2)

2pi

A=lw

(4)(1)

2pi+4

Fundamentals Theorem of Calculus

Even and Odd Functions

if f is odd, a∫-af(x)dx=0

If f is even,a∫-af(x)dx=2a∫0f(x)dx

Let a be a real positive number and let f be an integrable function on the interval[-a,a].

Part 2

b∫af(t)dt=f(b)-f(a)

where f is any antiderivative off, that is, a function such that f'=f

plug in a and b into the antiderivative and then subtract them, according to the formula.

Part 1

Derivatives and Integrals are inverses of one another

Let f be continuous for f on[a,b]

g(x)=x∫af(t)dt

is continuous on[a,b] and differentiable on (a,b) and g'(x)=f(x)

Let f be continuous for f to be greater than or equal to a. The area function with constant endpoint a is.

A(x)=x∫af(t)dt

Subtopic

\where x is greater than or equal to a. That gives the net area of the region bounded by the graph of f and the t-axis on the interval [a,x].

Area:b∫a f(x)dx)=limnEk=1(f(x)change of x)

The area function/integral above represents the Net area. Area above the axis is positive, while the area below the axis is negative. Thus, we can get the TOTAL net area of a function.

going from a to b

Limit of Integration/limit of Summation(a,b) (portion of the graph we are trying to find the area of).

Integrand-function under the integral

variable of integration- dx

Properties of Definite Integrals

a∫bf(x)dx=-b∫af(x)dx

a∫af(x)=0

with a less than c less then, b∫ak(f(x)+g(x))dx=b∫af(x)dx+b∫ag(x)dx

b∫ak(f(x)+g(x))dx=b∫af(x)dx+b∫ag(x)dx

b∫ak(f(x)dx=kb∫af(x)dx