Derivatives,
Antiderivatives,
and Integrals

Definite Integrals

Properties of Definite Integrals

b∫ak(f(x)dx=kb∫af(x)dx

b∫ak(f(x)+g(x))dx=b∫af(x)dx+b∫ag(x)dx

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

a∫af(x)=0

a∫bf(x)dx=-b∫af(x)dx

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

going from a to b

Fundamentals Theorem of Calculus

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.

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

\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].

Subtopic

Part 1

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)

Derivatives and Integrals are inverses of one another

Part 2

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

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.

Even and Odd Functions

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

If f is even,a∫-af(x)dx=2a∫0f(x)dx

if f is odd, a∫-af(x)dx=0

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

Indefinite Integrals
and Net Change

Indefinite Integrals of Trigonometric Functions

Derivative of trigonometric functions

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

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

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

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

d/dx(sin^-1(x/a))=-1/√ a^2-x^2

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

d/dx(sec^-1(x/a))-1/x√ a^2-x^2

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

d/dx(sinh(ax)=acosh ax

d/dx(cosh(ax)=asinh ax

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

Integral

∫cos(ax)dx=1/asin(ax)

∫sin(ax)dx=1/acos(ax)+C

∫sec^2xax)dx=1/atan(ax)+C

∫csc^2(ax)dx=-1/acot(ax)+C

∫sec(ax)dx=1/asec(ax)+C

∫csc(ax)cot(ax)dx=-1/acot(ax)dx

∫1/√ a^2-x^2dx=sin^-1(x/a)

∫1/√a^2+x^2dx=tan^-1(x/a)

∫1/x√ a^2-x^2dx=sec^-1(x/a)

∫sinh(ax)dx=1/acosh(ax)+C

∫cosh(ax)dx=1/asinh(ax)+C

Example: Evaluate the following

∫cos(x/2)dx

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

Answer:
2sin(x/2)+C

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

Integrals

∫cf(x)dx=c∫f(x)dx

∫1/xdx=ln|x|+C

b∫ak(f(x)+g(x))dx=b∫af(x)dx+b∫ag(x)dx

∫sindx=-cos+C

∫cosdx=sinx+C

∫secxtandx=secx+C

∫sec^2dx=tan x +C

∫csc^2dx=-cotx+C

∫cscxcotxdx=-cscx+C

∫x^ndx=x^n+1/n+1+C

∫e^x=e^+C

∫b^xdx=b^x/ln b +C

REMEMBER THE +C

Basic Integration Rules(Where k is a constant)

∫kdx=kx+C

∫kf(x)dx=k∫f(x)+C

∫x^ndx=1/n+1x^n+1 +C

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

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

∫e^axdx=1/ae^ax+C

∫1/xdx=ln|x|+C

∫b^xdx=1/ln b(b^x)

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

simplify

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

Antiderivative

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

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

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

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

Undoing chain rule

Do not forget to substitute u into final equation

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

Procedures for indefinite integrals

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.

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.

3. Evaluate the new integral with respect to u.

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

SUBSTITUTION RULE DOES NOT APPLY TO ALL INTEGRALS

Example

∫2(2x+1)^3dx

u=2x+1

du=2dx

f(u)=2u^3

1/4u^4+C

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

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

For trig functions, you must use DOUBLE ANGLE FORMULAS

Derivative of Polynomials/
Exponential Functions

lim(f(a+h)+f(a))/h
h→0

Constant Rule

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

Process

lim(f(a+h)+f(a))/h
h→0

lim (c-c)/h
h→0

lim 0=0
h→0

d/dx(c)=0

Exponential Functions

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).

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

Product and Quotient Rule

Product Rule

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

(fg)' does not equal f'g'

Example

3x^3e^x

f=3x^3

f'(x)=9x^2

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

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

g=e^x

g'(x)=e^x

Before starting Product Rule

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

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

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

Quotient Rule

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.

Example

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

T=x^2-3x-1

T'=2x-3

B=x^2-1

B'=2x

((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

Derivative of e

d/dxe^x=e^kx

K=real numbers

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

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

e=2.71828

Example

x^2e^2x

f'=2x

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

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

g'=2e^x

Limits and Derivatives of
Trig Functions

Two limits are easy to evaluate

lim sin x =sin x
h→0

lim sin h/h=1
h→0

lim cos x =cos x
h→0

lim cos h/h =0
h→0

Trig Functions

d/dx(sin x)=cos x

d/dx(cos x)=-sin x

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

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

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

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

Chain Rule

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)

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

The derivative of the inside times the derivative of the outside

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

Example

Sin^3(x)

(sinx)^3

u=sin x

u'=cos x

f(u)=x^3

f'(u)=3x^2

(cos x)^3 sin^2x

Answer:
3(cos x)sin^2 x

2 versions

u' times f'(u)

g'(x) times f'(u)

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

Derivatives of Log and inverse

Power, Product, Quotient, and Chain Rule still applies

Derivative of Natural Log

e^ln x=x

e^ln cancels

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

Domain of ln x is (0, positive infinity)

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

Hint: Use properties of Logarithmics

Example

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

Derivatives of Exponential Function(Other than e^x)

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

use this variation to differentiate b^x

e^xlnb

ln b e ^xln b

b is greater than 0

b cannot equal 1

d/dx(b^x)

ln b(b^x)

Example

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

f=x^3

f'=3x^2

s= 3^x

s'=(ln3)3^x

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

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

Logarithmic Differentiation

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.

Steps

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

2. Differentiate implicitly with respect to x

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

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.

Power Rule

x^n=nx^n-1

b^x=b^xln b

Table of Derivatives of Inverse Trigonometric Functions

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

d/dx(csc^-1 x) = -1/x√1-x^2

d/dx(cos^-1 x) = -1/√1-x^2

d/dx(sin^-1 x) = 1/√1-x^2

d/dx(sec^-1 x) = 1/x√1-x^2

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

Derivative of Logarithms

Using b^ylnb and implicit differentiation of y=logbX

we can find

d/dx(log b x=1/xlnb

logbx=ln x/lnb

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

Hyperbolic Funtioncs

Hyperbolic functions arose from the comparison of semicircular regions.

Functions

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

Domain: (-∞, ∞)

Range: [1,∞)

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

Domain: (-∞, 0)u(0,∞)

Range: (-∞, 0)u(0,∞)

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

Domain: (-∞, 0)u(0,∞)

Range: (-∞, -1)u(1,∞)

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

Domain: (-∞, ∞)

Range: (0,1]

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

Domain: (-∞,∞)

Range: (-1,1)

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

Domain:(-∞, ∞)

Range: (-∞, ∞)

Hyperbolic Identities

sinh(-x)=-sinh x

sinh(x+y)=sinhxcoshy+coshysinhx

cosh(-x)=-cosh(x)

sin^2x+cos^2x=1

1-tanh^2x=sech^2x

cosh(x+y)=coshxcoshy+sinhxsinhy

cosh^2x-sinh^2x=1

Derivatives of Hyperbolic Functions

d/dx(tanhx)= sech^2 x

d/dx(cosh x)=sinh x

d/dx(sinhx)=cosh x

d/dx(csch x)= -cschxcoth x

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

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

Derivative at a point

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

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

x→a

mtan=f(a) prime

New Notation

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

Using the point-slope formula

mtan=point

Alternate definition

f on the interval [a,a+h]

Example

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

opposite reciprical

Derivative of a function

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

h→0

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

h→0

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

Higher Order Derivatives

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

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

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

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

Linear Approximations
and Differentials

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

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

2. The function is discontinuous at a.

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.

Implicit Differentiation

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

this is why the dx/dy notation is important?

The derivative of y with respect to x.

Example

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

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.

Optimization

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

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

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?

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

4x+y=400

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

A=lw

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

A=(400-4x)x

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

A=400x-4x^2

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

A'=400-8x

400=8x

400/8=x

x=50

4(50)+y=400

y=200

Discontinuous means DNE

3. There is a vertical tangent at a.

example:x=0

4. The function does not exist at a.

Interpretations of Derivative

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))

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

Slope

Positive Slope

mtan>0

Negative Slope

mtan<0

Zero Slope

mtan=0

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.

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.

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.

Affect the Shape of a Graph

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

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

There are infinite ways to make something true

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' changes sign from positive to negative as x increases through c (from left to right), then f has a local maximum 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' positive on both sides of c or negative on both sides of c, then f has no extreme value at c.

Second Derivative Test

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

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

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

Concavity

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

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

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.

Test for Concavity

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

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

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

Graphing Guidelines

1. Identify the Domain/Interval of interest

Denominator can not equal 0

Even radicals is greater than or equal to 0

log(x) is greater than 0

2. Find the intercepts

x-intercepts set y=0

y-intercepts set x=0

3. Find the Asymptotes

a. Horizontal

use limits to find

b. Vertical

denominator is 0

4.Find the intervals of increasing and decreasing

First Derivative

Critical points, y'=0

Create number line to know where it is positive or negative

5. Identify extrema values.

Where the first derivative changes signs

6. Determine concavity and inflection points

IP is when the signs changes(negative and positive)

Derivative has to exist for it to change

7. Choose the appropriate window and graph

Make sure that you know what you are graphing

Extreme Value Theorem

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.

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].

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.

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

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.

Locating Absolute Extreme Values on a closed interval

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

2. Evaluate f at all critical numbers AND endpoints

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

Critical Points

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

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.

Critical Points does NOT GURANTEE max or min

Do not forget to test your x values

NO SHARP POINTS OR HOLES

Mean Value Theorem

Let f be a function such that:

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

2. f is differentiable on [a,b]

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|

CANNOT GO INTO THE NEGATIVES

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)