Order of Operations Attempt

Ok. This will not work for EVERY problem you encounter but it will help I think.

1. Product/Quotient Rule (if it is not inside parenthesis)

2. Exponent Rule

3. Trig Rules/log rules, trig inverse rules, e rules

You work from the outside in. So if the quotient is inside a log you would skip to three because it would be inside parenthesis. Or if a fraction is raised to an exponent then you would default to step 2 because it is inside parenthesis. I hope this helps!

pzost # 2

This week in class i managed to pick up somewhat youseful information that i will be using in calculus. I learned arc trig functions. we learned a total of six functions. These functions include sin, cos, tan, sec, csc, and cot. The function that I understand the sin function the most. The sin trig function is one over the square root of 1 -U² times U prime. Even though i understand the concept of this function after i set the problem up and work it the first part ( plugging in and finding the derivitive, the first simple simplify) I get lost with the algebra. I have the very very very very very bad bad bad habbit of wanting to cancel out what i cant. I also get very very confused when i have the big massive fraction with in like two other fractions that i am suppose to multiply with some beast derivitive. And yes i need help wwith this.

Post #2

Alright, so...this week was extremely tough. I think most people probably agree on that, haha. We learned about finding equations for a tangent line, normal line, the points where y has horizontal tangents, and we worked on arc trig derivative formulas. I'm having so much trouble with solving the problem after I get the derivative, and I don't know why.

At least I get one thing we learned. It was finding the equation of the tangent line.
For example: (x^3-3x-1)(x+2) and given the point (1,-3)
First you just follow the product rule for the derivative: (x^3-3x-1)(1)+(x+2)(3x^2-3)
= (x^3-3x-1)+(3x^4-3x+6x^2-6) = 3x^4+x^3+6x^2-6x-7
Next plug in your x, which is 1: 3(1)^4+(1)^3+6(1)^2-6(1)-7 = 3+6-6-7 = -4
Finally, plug in to the point slope form: y+3=-4(x-1)

Something I am confused on is the horizontal tangents.
For example: y=3x^3+4x^2+5 at (1,12) So I understand how to get 9x^2+8x=0
But then what's next? How do I get the points?

Also, can someone help me on how to solve: arcsin3x/x ?
That would be the quotient rule. In the second part of the formula I'm suppose to copy the top, but I'm not suppose to have trig in the problem. So, do I take the derivative of it anyway? Or is there something else I'm suppose to do?

Overall, my confusion is mainly on simplifying these arc trig formulas once I have the derivative. They just seem massive and complicated. And then I get aggravated with it and give up. Hopefully, I get it by the test ha.

Post #2

I struggled this week. I’m not completely comfortable with anything we did this week but I think I understood how to find the equation of a tangent line over everything else. To find an equation of a tangent line, you can be giving a point or only giving an x value. If you are only giving an x value, you have to plug in x to find your y and make a point. Next you have to find the derivative of the equation. Once you found the derivative, you plug in for x and get your slope. The last step is to put in point slope form which is y-y1=m(x-x1).
An example of this is:

3x^3+4x^2=5 at (1,12) – you are already giving the point so you can skip step one

Derivative = 9x^2+8x

Plug in x = 9(1)^2+8(1)= 17

Plug into point slope form: y-12=(x-1)

I understand logs from last year and the first column of our page three worksheet but I didn’t understand the derivatives of logs. I know the formula is u prime/ulna but I get confused on the really big problems when you have to do more than one step to solve it. Making a list sort of helps but I end up doing it in the wrong order and my answer usually turns out completely wrong. If I do somehow happen to get the derivative right, I get stuck on the algebra part of the equation. I think I just need more practice on derivatives all together, especially before the test Wednesday. Help if you can and good luck on the test.

Post 2

the second week of calc is over and i feel like i know a good bit even though its eof my hardest classes.it's getting better. I am understanding everything a lot better.

last year i slacked during the log chaper and i didnt know it because i played block dude everyday. I get everything from last year with the normal logs too. john taught me all the logs from last year. Log4=x which equals x=2 thats prettty easyyy. i understand the natural log properties from this past week when you plug everthing into the formula and simplify. Its pretty easy most times. we have ln(x^5). You plug into the ln formula, 1/u * u', and you get 1/x^5 * 5x. When you multiply you get 5x/x^5 which simplifies to 5/x (you cancel the x's). I understand that you may have to deal with multiple formulas in one problem. An example would be ln(x/(x^5 + 1)). In this problem you can see that you have to deal with natural log and quotient rule.

Something I don't understand is the difference from u^n and a^u. and i dont understand the arc trig functions with all the algebra. all the algebra in these problems mess me up and i end up messing it up. i hear sarah is pretty good at those btw hahahaha

everyone is at my house at the moment studying the test we have on wednesday. is that a test or a quiz because we are not too sure. can someone answer that question as well?

Post Number Two

So everyone, the second week of calculus is over. Of course it's gotten harder, but not everything in life comes easily right? I felt good at the beginning of the week because we were reviewing quotient rule and product rule. Though at the time of learning them I thought those were impossible, I'm now wishing that we were back to that again. I'm confused about everything overall, but there are some things that I did understand this week (Thank God).

For example, I caught on to horizontal tangent points real quick. All you do is find the derivative of the problem given then set equal to zero and solve for x. We were drilled on solving for x last year, so I felt extremely comfortable doing that. An example problem of this is Find the points where y has horizontal tangents y=3x^3 + 4x^2 + 5. Therefore, y'=9x^2 + 8x because you just take the simple derivative of a polynomial. Once finding the derivative, set equal to zero giving you 9x^2 + 8x = 0. Factor out an x giving you x(9x + 8) = 0. You then find that your x points are 0 and -8/9. Next, you must plug in your x's to the original equation in order to find your ys. 3(0)^3 + 4(0)^2 + 5 and 3(-8/9)^3 + 4(-8/9)^2 + 5. These come out to give you the points (0,5) and (-8/9, 6.053). Box it off and that's your answer :)

Also, i understood the arc problems until we got to problems like #53 y=1/2(1/2ln x+1/x-1 + arctanx). I'm not exactly sure what you do first because there's a natural log and an arc formula. If anyone can direct me on this it'd be greatly appreciated.

I still remember the log properties from advanced math so I do okay on those problems. I also do fine on natural log problems unless they have more than one function in it. Like I said I get extremely lost then and have no idea where to start. Basically, I'm doing fine in this class except when it gets to the simplifying parts. I know it's just algebra but I just can't seem to grasp how to do all the right cancellations. I'm sure it'll take a lot of practice which I'm working on but it's just so frustrating not getting it time after time.

Finally, I want to extend the offer to arrange a study group before Wednesday to prepare for the test. I can always use a lot of help so if anyone is interested please let me know because I believe we can all help each other if we work together.

Overall, i'm looking forward to another stressful week full of work and headaches in Calculus--SIKEEEEEEE.

Week #2

Week #2 has finally come to a close and things just got harder. With the introduction of e^u and more complex derivatives, I was actually starting to struggle a little. A good example problem would be y=ln(e^x)

First off one should should identify the steps of your problem. In this case they would be
1. Natural Log
2. e^u

you problem should be 1
--- . (e^x)'
e^x

then you find the derivative of e^x which is e^x . x' (x'=1)

so your final problem should be 1
--- . e^x . 1
e^x

After this you have to simplify algebraically, giving you e^x ,which equals 1.
----
e^x

I still mess up while simplifying because once I get a huge derivative i end up missing a sign or just plain messing up with different rules, for example the log rules. I kept thinking that log₁₀4⁴ cancelled leaving log₁₀1. Does anyone have any advice on remembering rules and such. It would be greatly appreciated.

Post #2

This week in Calculus AB, things got a little harder. I do not understand how to solve a problem with many parts. I am slowly catching on, but still confused. The thing I understand most in Calculus this week was the arc trig identities.
For example, I understand how to do a problem like arccos (1/x2). First, you must know the formula for arccos, which is (1/ √1-u²) (u’). Take this problem step by step to solve it efficiently. First, find your “u,” which, in this problem, is (1/x²). Now, plug it into the first part of your equation that asks for the “u” to receive the beginning of the equation which should be (1/ (√1-(1/(x²)²). Now you must find u’, or the derivative of u. You can find this by doing the quotient rule, which is: [ (copy the bottom) (derivative of the top) – (copy the top) (derivative of the bottom) ] / (bottom)². The equation should look like ( (x²)(0) – [ (2)(1) ] ) / (x²)² = (-2x)/ (x^4) = (-2) / (x^3). You should use brackets to remind yourself when you are simplifying that you have a negative to distribute. So far, your problem should look like 1/ ( √1- (1/x^4) ) ( (-2) / (x^3) ). The rest of the problem can be solved by algebra to obtain an answer of (2/x^3√1-(1/x^4)). These types of problems are the simplest to solve for me because it only involves plugging in formulas to solve.
If anyone can help me understand order of operations for derivatives it would help me a lot. I’m not sure if you are supposed to take derivatives of everything or if you should work from inside out, then take derivative. I am just very confused on how you should work problems that include more than one rule.

Post #2

The second week of calculus is over! Although I think Calculus is still one of my hardest classes, it's getting better. I am understanding everything a lot better, and I've learned that it's good to ask questions.

Something I understoood this week was the natural log properties. You simply plug everthing into the formula and simplify. Let's say we have ln(x^2). You plug into the ln formula, 1/u * u', and you get 1/x^2 * 2x. When you multiply you get 2x/x^2 which simplifies to 2/x (you cancel the x's). I also understand that you may have to deal with multiple formulas in one problem. An example would be ln(x/(x^2 + 1)). In this problem you can see that you have to deal with natural log and quotient rule. You would plug it into 1/u * u' first and then into the u prime you would do the quotient rule, (vu' - uv') / v^2.

I also still get everything from last year with just normal logs. Take log 1000 = x. You would take away the log and swap 1000 and x. You then have 10^x = 1000. So from algebra you would know that 10^3 = 1000. So x = 3 would be your answer.

Something I don't understand is the difference from u^n and a^u. Take the example (ln x)^4. Would you plug it into u^n = nu^(n-1) or a^u = a^ulna * u'? I am still unsure about the whole concept.

Hope this helped somebody.
Ryan.

Post #2

Well, to begin, week two in my opinion was much more difficult that week one. At first we reviewed the stuff from the week before, and I understood everything like the quotient and product rules. Then the next day we did the arcsin stuff, and I understood how to do that. So I'm going to show and example of it. For the problem 2arcsin(x-1), you would have to use the formula arcsinu = (1/(sqrt 1-u^2)) times u^1. To start off, you just move the 2 into the next step without doing anything, then you put the rest of the problem into the formula, but do not forget to multiply it by the derivative of u, I've done that before. So, then you get (2)(1/(sqrt1-(x-1)^2)) (1), which you further simplify to make it 2/sqrt1-(x-1)^2, then in the next step you multiply out the (x-1)^2 to get 2/(sqrt(1-(x^2)-2x+1)), then your final answer become 2/ (sqrt( -(x^2)-2x-2)). I hope you could all understand that example, I tried my best to type it, but I'm not sure if I put all the parentheses and other stuff the right way. So I understood that, and the simple log problems, like how log27 9 = 27^x = 9, then you solve for x and you find that x=2/3. I also understand how you are supposed to expand log problems like, ln2/3 = ln2-ln3. The only problems I struggle with are the extremely long problems that you have to do loads of algebra and use like 7 different rules on different sections of the problem, for example,
-[sqrt((x^2)+4)/2x^2] -1/4ln[(2+sqrt(x^2+4))/x] one thing that helped me was writing down what you had to do for each part of the problem before you solve, like how for the first part of it, you need to use quotient rule, exponent rule, and work inside exponent; in the second part, you need to take care of the ln, then use quotient rule, then exponent rule, then work the inside exponent. But I suck at doing log problems like 10log4t/t because I suck at that kind of log stuff. I think you use quotient rule, and do all kinds of simplifying after that, but I just hate it.

mher---out.