Calculus

Question #4

Question #5

Question #6

Question #7

Question #8 – Find the domain

Question #6a

Question #6b

Example #5 Evaluate

Example #6 Evaluate

Example #7 Evaluate

Quiz questions! No outside help allowed
Question 1a – Evaluate exponentially

Question 1b – Evaluate exponentially

Question 2 – Evaluate logarithmically

Question 3 – Evaluate Logarithmically

Logs are another way to write exponents.

This shows ‘The common log’. When a base is left out, it is assumed to be base ‘10’

My curiousity was sparked by a Chain Rule problem, which included ln in the answer. I did not know how to handle ln. ln is ‘the natural log’. It involves a relationship with an unknown variable ‘e’, which is its base.

Some practice problems

Example #1 evaluate exponentially

Example #2 evaluate exponentially

Example#3 Evaluate Logarithmically

Example #4 Evaluate Logarithmically

Ran out of time. Using problems from West Texas A&M University

Practicing problems from UCDavis.

Problem 1

Problem 2

Note to Self: Square roots. To find the derivative of a square root, change the square root into a fraction and proceed, using the exponent rule. Ie. The derivative of the square root of x is 1/2x^-1/2

Problem 3

According to the shown solution this answer is wrong. I should have included (-4+35x⁴) inside of the exponent “29”. I don’t know why I should do this. Is it because a simple version of this might look like 3(x⁵)²(x⁴)? I believe with that sort of problem I could write out 3(x⁵)(x⁴)² and if I solve either one of those two I would get the same answer.

Lets try that:

Let x = 2
3(x⁵)²(x⁴)
3(32)²(14)
3(1024)(14)
43008

3(2⁵)(2⁴)²
3(32)(14)²
1806336

That’s not right either.
I’m going to have to do some research on this. And in the mean time make a mental note, to move the values bracketing the (u) formula to the end! Solved; See Problem 6

Problem #4

OK… So I answered this one wrong as well. (The correct answer is shown though..) This time I should not have pulled g’(x) inside of the exponent, like I needed to in problem #3. So what is going on here? How to do I handle exponents? Solved; See Problem 6
As a side note. I’ve learned how to “Align along = sign” How nice!

Problem #5

Ok. This one immediately threw me for a loop with the division AND the exponent. First things first, I was able to reduce u, leaving the inner function division free but with negative exponents. After that it was smooth flowing, except by the end of my problem, I still have three pesky negative exponents which I need to simplify out of existence.

Problem #6

I got this one right! And it was EASY!
D_xy Sin(x) = Cos(x)
I believe I’ve also solved the exponent trouble I was having. The trick is to simplify first when u is still a variable, and then once simplified completely, replace u

To review my knowledge of limits, I need to think back. A Limit expresses a relationship between two changing but related variables, Epsilon eand Theta strong>d.To review my understanding of limits, the limit is a variable L, which has a relationship to a secondary variable a. a is the change along the x-axis as we approach the limit, which is represented by the variable L. To think back on what I understand, the variable L is bordered on the y-axis by Epsilone. As far as limits are concerned, we are in absolute space. We do not have negative limits. So to start out, I know the limit must be greater than Zero. Both variables, Epsilon e and Theta d must be larger than Zero, or else the limit does not exist. So I can write e > 0.

The purpose of this exercise is to describe the relationship between Epsilon and Theta. So what can I write about Theta? Theta d represents the border of variable a along the x-axis. Since both Epsilon and Theta represent both sides of the border, there are absolute values.So what exactly is the relationship between our variables? L, a,e and d, as well as our line, y = f(x)

L is the limit. It is reached by x – a. Variable a is found when you look at a limit and there is an equivalency x ->a. Epsilon is a real number, greater than 0. The limit is a point on the line f(x). Epsilon deals with the y vertice, which is described as a function of f(x). So the variable x describes the location along the x-axis, and f(x) describes the location along the y-axis. Lets break this problem off into x and y axis sections. To the x axis, we will give the variables, x,a and d To the y-axis we give the variables f(x), L and Epsilon. x – a should be less than d and f(x) – L should be less than e. The reason our x and y coordinates are less than variables Epsilon and Theta, is because, the Limit is contained inside of the variables Epsilon and Theta, because they form the boundrays. Because the limit is not negative, we need to use absolute values for both of the expressed changes, x-a and f(x)-L so we would say |x-a| < d and |f(x)-L| < e .

There is only one more thing I am leaving out, which is that the limit is in positive space, so the whole thing has to be greater than 0. Back at the beginning of this essay, I stated e > 0 I could say the same thing by writing 0 < e . We can include this note about the limit being in positive space, by sticking that Zero Less Than in front of the entire expression to write. 0< |x-a| < d and |f(x)-L| < e Can I use this relationship to find Theta with a known Epsilon.

>

The first step is to solve the f(x)-L side, because we know what f(x) is, we know what L is and we know what Epsilon is. Using our known variables we will be able to find the values of x.

So now we have found out an exact value of variable x, it is between 2.00083 and 1.99916. x – a should give us a value for Theta

*Incase you haven’t noticed, I’m having some trouble typing out the Greek letter variables. I’ll figure that out some other time.