Derived that the derivative for the natural log is one over the argument

So…

To find the derivative of an exponential function take the natural logs of both sides. Don’t forget that y is a function of x so the chain rule applies.

A composite function uses the output of one function for the input of the other function: f(g(x)).

Trigonometric applications of the Chain Rule

Simple and Composite uses of the chain rule

Trig stuff, and note the use of the SQUEEZE THEOREM at the end

WE GOT OUR TESTS BACK
THEY WERE SCALED

QUESTIONS THAT WE WENT OVER:
PART 1: 2,4,6,7
QUESTION 2: REMEMBER TO FACTOR OUT AS MUCH AS POSSIBLE-SYNTHETIC DIVISION ON CUBIC. ANSWER B

QUESTION 4, WHAT DOES THIS FUNCTION LOOK LIKE?
ANSWER: C
THE LIMIT OF THE FUNCTION APPROACHING INFINITY IS ZERO, SO THE X AXIS IS THE VERTICAL ASYMPTOTE.
THE DENOMINATOR IS THE DIFFERENCE OF SQUARES OF ONE, MEANING THERE ARE TWO VERTICAL ASYMPTOTES, AT 1 AND -1.

QUESTION  6: WHAT MUST BE TRUE ABOUT THE FUNCTION?
ANSWER: E
ALL OF THESE OPTIONS CAN BE TRUE, BUT NONE OF THEM HAVE TO BE.

QUESTION 7: SOLVE FOR K
ANSWER: -1/2
MAKE SURE TO WRITE
OR ELSE IT MAKES NO SENSE

PART 2

QUESTION 3: FIND f ‘(2)
ANSWER: f ‘(2)=42
THERE WERE MULTIPLE WAYS TO DO THIS, ALL USING SOME FORM OF THE DIFFERENCE QUOTIENT.

QUESTION 5: FIND DISCONTINOUS POINTS AND LABEL
THIS CAME DOWN TO FACTORING CORRECTLY AND KNOWING YOUR DIFFERENT TYPES OF DISCONTINUITY

THEN WE PLAYED A GAME PLOTTING DERIVATIVE FUNCTIONS
FRONT TEAM WON!!!!!

DISPLACEMENT, VELOCITY, AND ACCELERATION
s(t)= DISPLACEMENT
v(t)=VELOCITY
a(t)=ACCLERATION

THIS LED TO THE DISCOVERY OF SECOND DERIVATIVES:

s’(t)=v(t)
s”(t)=v’(t)=a(t)

AKA: THE DERIVATIVE EQUATION OF A DISPLACEMENT EQUATION IS VELOCITY, AND THE DERIVATIVE EQUATION OF A VELOCITY EQUATION IS ACCELERATION, MAKING THE SECOND DERIVATIVE OF DISPLACEMENT ACCELERATION.

Example:

Definition of Derivative: f ‘ (x)

Forward Difference (quotient difference)=

Backward Difference=

Symmetric Difference (often used with table)=

At a Point=

Derivative Worksheet:

1.

2.

f’(x)=(-2)

POWER FUNCTION RULE (also on page 92)

Power function continued…

Example 1

Example 2

Calculator:

On the AP exam, a calculator can be used for finding x intercepts, numerical derivatives and numerical integration.

To find the derivative at x=2 when f(x)=x^5:

Plug in x^5 as Y1 –> Math –> Option #8 –>and then plug in nDeriv(Y1,x,2)–> 2nd calc gets a value.

This will get you an answer of 80.00004.  Don’t forget that .00004 is a small margin of error.

PROPERTIES OF DIFFERENTIATION (pg 93)

NOTATION OF DERIVATIVE

Derivative with respect to x of function y:

REMINDER:

HW: pg 95- 10 Qs, #1-25 odd

-Challenge problem

-Major Quiz Friday

-Always show exactly how you found an extreme (maxima or minima) by physically setting f’(x)=0

We also began work with slope fields.  Slope fields help to determine what the family of curves will look like if the anti-derivative of a differential function is graphed. (f’(x)=2x is the differential equation of f(x)=x^2+c)

-The tangent lines represent the slope at the same point on the original function (the slope of the tangent line at a point is equal to the slope of the original function at that same point)

We also received a packet on slope fields that has problems for which we have to determine and graph the slope field.

Homework:

Do the packets and other handouts

Class of December 15, 2008

1)    We handed in the Bonus

2)    We started applications of derivatives (see below)

3)    We finished and handed in the worksheet we started last class

Derivative Applications

• Maximum and minimum points of a graph are where the derivative is zero
• Pg. 371 #2
• y= 4-x² from x=0 to x=2; the parabola is rotated, and there is a cone inscribed; we need to find the maximum volume of the cone
• VOCAB: paraboloid: 3D parabola rotated

• In order to find the maximum volume of the cone, we need to determine the height and radius of the cone.
• V=1/3πr²h; this is the equation of volume of the cone. In order to find the maximum volume, we need to find the maximum height, or where the derivative is equal to zero. However, this equation is in terms of two variables, and we need it to have only one.
• The radius of the cone is equal to the x value at the outer-most edge of the cone; the height of the cone is equal to the y value at the top of the cone.
• Therefore, V (r)= 1/3πr²(4-r²)= 4/3πr²-1/3πr⁴
• V’=8/3πr-4/3r³ then becomes the derivative of the above equation. In order to find the maximum point, we need to find where the derivative is equal to zero, so set this equation equal to zero. However, the answers this way aren’t necessary maximum points. In order to be maximum points, the derivative to the left has to be positive, and the derivative on the right has to be negative. REMEMBER TO WRITE THIS OUT, ESPECIALLY ON THE AP TEST!!
• V’= 4/3πr(2-r²)=0
  • r=0 or 2-r²= 0 or r= +/- √2, only +√2 (-√2 is not within the given domain)
• While r=√2 is the only possible maximum point, because maximum points cannot be at zero in this problem, it cannot be assumed that it will be the maximum. It is only the maximum if:
  • x<√2, V’>0
  • x> √2, V’<0
• In order to check to see if these two are right, we must plug a number less than √2 and greater than √2 into the derivative equation and see if they give us a number larger than zero and less than zero (respectively).
• REMEMBER: Always check the endpoints! The endpoints could be a maximum or minimum point. Plug the endpoints into the equation and see which volume is the highest.
• NOTE: The technique for analysis of maximum and minimum point problems is on page 372!

HOMEWORK: pg. 367 #39 and max./min. problems on pg. 372 # 1, 7, 11, 15; quiz corrections due next class, and the MC packet is due on Friday. No quiz on Friday!!!

Class notes for 11/12

We went over †he curved quiz.  See Mrs. Evrard for more personal questions.  The multiple choice packet is due on Friday, and the graded hw is due tuesday.  On the homework calender it also says that we have a quiz tomorrow.  Use the quizzes in my classes to prepare, and also use the derivative plotter on Mrs. Evrard’s homepage.  Enjoy.

Product Rule

d(uv)=u’v+uv’

This can also be written with g(x) and h(x)

Quotient Rule

d(u/v)=(u’v-uv’)/(v^2)

d((8x+1)^6/(5x+2)^9)=
(6(8x+1)^5 * (8) * (5x+2)^9 – (8x+1)^6 * (5x+2)^9 * 9) / ((5x+2)^9)^2

d((8x+1)^6/(5x+2)^9)=
(3) * (8x+1)^5 * (5x+2)^8 [(16) * (5x+2) - (15) * (8x+1)] / (5x+2)^18

d((8x+1)^6/(5x+2)^9)=
(3) * (8x+1)^5 * [(16) * (5x+2) - (15) * (8x+1)] / (5x+2)^10

When the numerator is constant, it’s easiest not to use the quotient rule, because
1000/(3-t)=1000(3-t)^-1

Using the quotient rule to solve for the derivative of tan x

sin x/ cos x

d(sin x/ cos x)= (cos x*cos x – sin x * sin x) / (cos^2 x)

d(sin x/ cos x)= (cos^2 x)^-1  = sec^2 x

Homework: The packet on IVT and be prepared to discuss it next class