This week in calculus, we went over a lot. We had to remember finite limits, infinite limits, and points of discontinuity from last year. We also went over some new material. The new things we went over were average speed (or slope), instantaneous speed, and derivatives. We were given derivative formulas to help us and give us shortcuts for the lengthy formulas we learned for average speed and instantaneous speed.
I learned that when working average speed, you will be given a word problem, and in that word problem there will be either two numbers or a hint (in the first two seconds) to two numbers. You will also have an equation at the end of the word problem (y=16t^2). These numbers in the word problem will become the x variables in your formula. To get the y variables, you plug both the x variables into the equation to get two separate ys. Once that is done, you plug your xs and ys into the midpoint formula (because midpoint finds slope/average speed). The midpoint formula is y2-y1/x2-x1. Once you get your number, you are to write the units next to it (unit length/unit time).
I also understand the product rule formula in our derivative formulas. If you are asked to take the derivative of a product you are to copy down the first term being multiplied times the second term's derivative plus copy down the second term and times that by the first term's derivative.
For the most part, I believe I understand all the concepts of derivatives so far. I just get mixed up when a formula has a fraction, such as the quotient rule, or if a problem ends up having a negative exponent. When I look at the problem, I always want to put the variable with the negative exponent under everything, instead of just what it's supposed to be over. I also confuse myself when I see a number minus a number times something else. Such as 5-3sinx. When I see this, I automatically think I can combine these two numbers into one.
Sunday, August 23, 2009
Week #1
First week of Calculus finally done and with little blood shed. After hours of doing Calculus homework and helping people with Calculus, I notice that some people did not understand some questions on the first packet. An example question: "Supposed w'(-1)=18 and w(-1)=0. Find g(-1) assuming (wg)'(-1)=32." After staring at it for a good 10 minutes I finally understood it.
First off, you should recognize that they are using the product rule (uv)'=u(v') + v(u'). (w=u and g=v)
Secondly, if they -1 is confusing anyone just ignore it because they all plug -1 in for x, meaning that they are all the same.
To start this problem, write out the fully expanded product rule. Then, just start plugging in numbers where they belong. You should get (0)(g') + (g)(18)=36. Simplify the problem, which leaves you with 18g=36. Just divide 36 by 18 you find that g(-1)=2.
This also applies to problems that use the quotient rule also, so do not be scared because it turns out to be relatively simple.
One thing I really did not grasp until recently, though, was being able to manipulate the problem algebraically before taking the derivate. A good example of this is -2/(x^3). Besides the fact that it is still a simple problem, you can really save time by just moving the x^3 up, giving you -2(x^(-3)), which makes it much easier that doing the quotient rule. This little trick can save you a lot of time on any test, including an ACT or Mu Alpha Theta competition test.
Another little side note...If you understand derivatives and want to start trying to gain speed while doing them, stop and make sure you do the math correctly first because if your going too fast to realize that you did not subtract your fractional exponents correctly, you will have to go back and redo half of the problem or just not notice and get all of the problems wrong.
First off, you should recognize that they are using the product rule (uv)'=u(v') + v(u'). (w=u and g=v)
Secondly, if they -1 is confusing anyone just ignore it because they all plug -1 in for x, meaning that they are all the same.
To start this problem, write out the fully expanded product rule. Then, just start plugging in numbers where they belong. You should get (0)(g') + (g)(18)=36. Simplify the problem, which leaves you with 18g=36. Just divide 36 by 18 you find that g(-1)=2.
This also applies to problems that use the quotient rule also, so do not be scared because it turns out to be relatively simple.
One thing I really did not grasp until recently, though, was being able to manipulate the problem algebraically before taking the derivate. A good example of this is -2/(x^3). Besides the fact that it is still a simple problem, you can really save time by just moving the x^3 up, giving you -2(x^(-3)), which makes it much easier that doing the quotient rule. This little trick can save you a lot of time on any test, including an ACT or Mu Alpha Theta competition test.
Another little side note...If you understand derivatives and want to start trying to gain speed while doing them, stop and make sure you do the math correctly first because if your going too fast to realize that you did not subtract your fractional exponents correctly, you will have to go back and redo half of the problem or just not notice and get all of the problems wrong.
we learned a bunch this week like instantaneous speed and average speed and taking the derivative of many equations. i prettty much know how to take derivatives of like easy stuff. this week wasn't as hard as i thought it would be. especially when we first got all those formulas. im getting almost everything but i forget some small part of a problem every once in a while. i sometimes forget to subtract the one when taking the derivative and that makes everything wrong. when product rule and quotient rule come into play i either mess up with my signs or i do something dumb and get the entire thing wrong. i always find myslef messing up foil or something easy and stupid. i can use some help with knowing when to use quotient rule and when not to. when there is like 1/4x^4 i think im doing it right but im not sure. average speed and instantaneous speed is pretty easy once i learned how to do it the fast way. that part of calculus is easy. quotient rule is the most confusing for me with all the algebra involved. i make small mistakes that cost me. if anybody can help me solve some of this i'd appreciate it.
Ok, let me just say that I was somewhat scared when I walked in on Monday not because I didn't think I could do the work, but jsut because it was CALCULUS. However, I stressed myself out over nothing as usual.
To me, calculating the average speed was the easiest thing I've learned so far because it's just basic algebra with the slope formula. First, you identify the time you start and the time you end [t1, t2]. Each t value is the equivalent of a x value in the basic slope formula (y2 - y1)/(x2 - x1). Then you plug each t value in as your x in the equation given to describe the fall. The outputs are your y values. After plugging everything into your equation for slope, you get your average speed.
B-Rob stressed to us that we had to remember the Long way to find instantaneous speed (rate of change, definition of derivative), and I feel confident that I can do this becaus plugging into a formula is my thing. I like things thatare set in stone, so there's no question in my mind that I'll have a problem with this. You just take the limit as h goest to 0 of the function plus h minues the function all over h. So, this is it:
lim f(x + h) - f(x)
h-0 h
Then you just manipulate algebra to get the instantaneous speed or the derivative.
I really have a true understanding of everything, even after thirty-six formulas for derivatives were thrown on the board. For instance, the product rule states that when multiplying two things, you take one original times the derivative of the other and add that to the other switched up (this is better: (uv1 * vu1). I also understood how the quotient rule works and among other things (except, I keep forgetting the bottom of the quotient rule--my problem!).
The only things I'm unsure about do not actually involve doing the steps and formulas, just how to end the problem.
1. Like, after I take a derivative using the quotient rule and get a fraction, do I leave the bottom squared or foil it out?
2. Also, if I have a fractional exponent in the denominator, do I have to change it to a root and multiply by the conjugate to get rid of the radical on the bottom?
3. On the Section 2.3 sheet # 35, I didn't know what to do with three, so if anyone could help me, that would be great!
4. On the big packet (one with like everything), I didn't know how to do the ones where it started at a certain height. Could anyone explain to me an example (like #34?)? Thanks!
So overall, I think I had a good first week in Calculus. I learned a lot and am so psyched about the possiblity of getting college credit by taking the AP exam.
--Malerie
To me, calculating the average speed was the easiest thing I've learned so far because it's just basic algebra with the slope formula. First, you identify the time you start and the time you end [t1, t2]. Each t value is the equivalent of a x value in the basic slope formula (y2 - y1)/(x2 - x1). Then you plug each t value in as your x in the equation given to describe the fall. The outputs are your y values. After plugging everything into your equation for slope, you get your average speed.
B-Rob stressed to us that we had to remember the Long way to find instantaneous speed (rate of change, definition of derivative), and I feel confident that I can do this becaus plugging into a formula is my thing. I like things thatare set in stone, so there's no question in my mind that I'll have a problem with this. You just take the limit as h goest to 0 of the function plus h minues the function all over h. So, this is it:
lim f(x + h) - f(x)
h-0 h
Then you just manipulate algebra to get the instantaneous speed or the derivative.
I really have a true understanding of everything, even after thirty-six formulas for derivatives were thrown on the board. For instance, the product rule states that when multiplying two things, you take one original times the derivative of the other and add that to the other switched up (this is better: (uv1 * vu1). I also understood how the quotient rule works and among other things (except, I keep forgetting the bottom of the quotient rule--my problem!).
The only things I'm unsure about do not actually involve doing the steps and formulas, just how to end the problem.
1. Like, after I take a derivative using the quotient rule and get a fraction, do I leave the bottom squared or foil it out?
2. Also, if I have a fractional exponent in the denominator, do I have to change it to a root and multiply by the conjugate to get rid of the radical on the bottom?
3. On the Section 2.3 sheet # 35, I didn't know what to do with three, so if anyone could help me, that would be great!
4. On the big packet (one with like everything), I didn't know how to do the ones where it started at a certain height. Could anyone explain to me an example (like #34?)? Thanks!
So overall, I think I had a good first week in Calculus. I learned a lot and am so psyched about the possiblity of getting college credit by taking the AP exam.
--Malerie
Post 1
In the first class in calculus that we learned something new, I was completely lost. I was mad because I thought "I'm going to fail because I'll be lost the whole year." And Mrs. Robinson telling us we needed to know EVERYTHING from Advanced Math didn't help either. But once I got the hang of it I realized that it's not really all that bad. I guess I was getting nervous about the whole AP college credit thing. Now I'm kind of anxious to see how the rest of the school year plays out in first hour.
Something that I understood well this week was most things to do with the Derivative Formulas. I guess it just clicked in my head? Anyway I think it is very easy because one you establish what your u and v are then you can simply plug in to one (or more) of the formulas. Although I was a little confused about how to solve a derivative with multiple formulas in it in the beginning, but I'm pretty sure I got it now. And of course probably the easiest thing is that the derivative of any number is 0.
Something I am still a little confused about is average speed and instantaneous speeed. I get it for the most part but I get confused about the [f(x+h) - f(x)]/h (or instantaneous speed) formula. I always get confused on how to plug in a number into a formula that you have to plug into another formula.
One problem that I didn't quite know how to solve was a problem on the worksheet we had to do for homework this weekend (pg. 126 - chapter 2).
Number 37:
(x^2 + c^2) / (x^2 - c^2), c is a constant.
I get that you have to plug into the quotient rule. But I get stuck whenever you have to take the prime of (x^2 + c^2). I get that the derivative of x^2 is 2x. I do not know what to do with the c^2 however. Do you treat is like a number or like x? If treated like a number the derivative of it would be 0, but if treated like x the derivative would be 2c. It's racking my brain on what to do.
In conclusion to the first week, I think that as a class we are on our way to success on the AP. We can only achieve this goal if we keep up the hard work all year though. So everybody be ready to learn tomorrow!
-Ryan B.
Something that I understood well this week was most things to do with the Derivative Formulas. I guess it just clicked in my head? Anyway I think it is very easy because one you establish what your u and v are then you can simply plug in to one (or more) of the formulas. Although I was a little confused about how to solve a derivative with multiple formulas in it in the beginning, but I'm pretty sure I got it now. And of course probably the easiest thing is that the derivative of any number is 0.
Something I am still a little confused about is average speed and instantaneous speeed. I get it for the most part but I get confused about the [f(x+h) - f(x)]/h (or instantaneous speed) formula. I always get confused on how to plug in a number into a formula that you have to plug into another formula.
One problem that I didn't quite know how to solve was a problem on the worksheet we had to do for homework this weekend (pg. 126 - chapter 2).
Number 37:
(x^2 + c^2) / (x^2 - c^2), c is a constant.
I get that you have to plug into the quotient rule. But I get stuck whenever you have to take the prime of (x^2 + c^2). I get that the derivative of x^2 is 2x. I do not know what to do with the c^2 however. Do you treat is like a number or like x? If treated like a number the derivative of it would be 0, but if treated like x the derivative would be 2c. It's racking my brain on what to do.
In conclusion to the first week, I think that as a class we are on our way to success on the AP. We can only achieve this goal if we keep up the hard work all year though. So everybody be ready to learn tomorrow!
-Ryan B.
post 1
We have learned many things this past week. We learned from average speed to instantaneous speed, also we learned derivatives and their formulas. And within these things we have learned I understande some where on the other side I am having some problems with some.
The thing I understand the best is how to take derivatives of some different things. The reason is that the formulas on the derivatives are not hard to understand. One of the formulas is the product rule. This rule is one of the easiest to me. It is because you just follow the formula and its easy algebra. All you do is write the first take the derivative of the second and add that to where you write the second and take the derivative of the first. Also the derivatives any number. This is easy because the derivative of any number is equal to 0. Another thing is to take a derivative of anything with an x raised to a number. All that has to be done is multiply the coefficient by the exponent and then subtract the exponent by one. An example of this is 3x^3+4x^2-5x-5. The derivative is 9x^2+8x-5.
Also I understand how to do the average speed and instantaeous speed even though I did not really understand this at all at the beginning. The reason I started to understand this is because i realize that average has to do with over a time period and slope and instantaneous is has to do with now. This helps me do these problems.
One thing i do not understand fully is how to simplify a derivative of something like 1/3 x -3 . The reason is that i dont know how fix it when you have to make a fraction to get rid of the negative exponent.
Also I am having some problems with taking the derivative of a number to the square root, cube root, fourth root, etc. I can start it off but can't ever finish it correctly.
The thing I understand the best is how to take derivatives of some different things. The reason is that the formulas on the derivatives are not hard to understand. One of the formulas is the product rule. This rule is one of the easiest to me. It is because you just follow the formula and its easy algebra. All you do is write the first take the derivative of the second and add that to where you write the second and take the derivative of the first. Also the derivatives any number. This is easy because the derivative of any number is equal to 0. Another thing is to take a derivative of anything with an x raised to a number. All that has to be done is multiply the coefficient by the exponent and then subtract the exponent by one. An example of this is 3x^3+4x^2-5x-5. The derivative is 9x^2+8x-5.
Also I understand how to do the average speed and instantaeous speed even though I did not really understand this at all at the beginning. The reason I started to understand this is because i realize that average has to do with over a time period and slope and instantaneous is has to do with now. This helps me do these problems.
One thing i do not understand fully is how to simplify a derivative of something like 1/3 x -3 . The reason is that i dont know how fix it when you have to make a fraction to get rid of the negative exponent.
Also I am having some problems with taking the derivative of a number to the square root, cube root, fourth root, etc. I can start it off but can't ever finish it correctly.
Post 1
When I first walked in calculus class I was like "ahhh, help" and sadly I still am. I do understand how to do the product and quotient rule. I think I get the concept of everything, but I always seem to get stuck in the middle of each problem.
The only thing I'm fully confident in doing is finding the derivatives of -4x^3+2x^2-3x+1. For each term, you take the exponent and multiply it by the constant and subtract one from the exponent. Also, the derivative of any number is 0. for -4x^3: 3(-4)= -12 3-1=2 for 2x^2: 2(2)=4 2-1=1 for -3x: -3(1) = -1 1-1=0 So, then you have -12x^2+4x-3.
Also, I think I'm getting average speed, but can anyone tell me if I'm doing this right: Given the position equation s(t)=3t^2-3t-4, find the anverage velocity from t=2 to t=4. So, I ended up getting (2,4). I then plugged in 2 and then 4 to the equation 3t^2-3t-4. When I plugged in I got 2 and 32. After that I plugged (2,4) and 2, 32 in to the slope formula. For my final answer I got 15.
I think I'm having trouble more with the algebra kind of stuff. for example: [(-3x^2+5x-5)(sinx)] I understand copy the first, derivative of second, minus, copy the second, derivative of first (-3x^2+5x-5)(cosx)-(sinx)(-6x+5) but then what's the next step to solving?
Also, I don't understand problems like x(1-4/x+3) Do I use the product rule or no? I just don't know where to start for this one.
Yeah I'm kind of confused on a lot, but hopefully someone can help me before the test. (pleaseeeee haha)
The only thing I'm fully confident in doing is finding the derivatives of -4x^3+2x^2-3x+1. For each term, you take the exponent and multiply it by the constant and subtract one from the exponent. Also, the derivative of any number is 0. for -4x^3: 3(-4)= -12 3-1=2 for 2x^2: 2(2)=4 2-1=1 for -3x: -3(1) = -1 1-1=0 So, then you have -12x^2+4x-3.
Also, I think I'm getting average speed, but can anyone tell me if I'm doing this right: Given the position equation s(t)=3t^2-3t-4, find the anverage velocity from t=2 to t=4. So, I ended up getting (2,4). I then plugged in 2 and then 4 to the equation 3t^2-3t-4. When I plugged in I got 2 and 32. After that I plugged (2,4) and 2, 32 in to the slope formula. For my final answer I got 15.
I think I'm having trouble more with the algebra kind of stuff. for example: [(-3x^2+5x-5)(sinx)] I understand copy the first, derivative of second, minus, copy the second, derivative of first (-3x^2+5x-5)(cosx)-(sinx)(-6x+5) but then what's the next step to solving?
Also, I don't understand problems like x(1-4/x+3) Do I use the product rule or no? I just don't know where to start for this one.
Yeah I'm kind of confused on a lot, but hopefully someone can help me before the test. (pleaseeeee haha)
week #1
In calculus last week, we learned about derivatives, instantaneous speed, and average speed. We learned thirty-six formulas for derivatives, and also learned some for instantaneous speed and average speed. Although we learned many formulas, we only used about ten last week for derivatives. The two most important formulas that we learned, in my opinion, were quotient rule and product rule.
The thing that I am most comfortable with is average speed. The difference between instantaneous and average speed, is that instantaneous speed occurs only at one instant, while average speed occurs over a certain period of time. For average speed, you’re given a problem where an anvil falls off the roof onto Ryan G.’s head and then you are asked to find the average speed during the first 2 seconds of falling if y=16t^2 describes the fall. Your x’s in the problem would be (0, 2) because you are asked to find the average speed in the first two seconds. Then, to find your y’s, you would simply plug in your first x to the problem y=16t^2, which is 0, to find your first y. (y=16t^2, y=16(0)^2, y=0) After that, you plug in your second x to find your second y. (y=16t^2, y=16(2)^2, y=16(4), y=64) Therefore, your y’s in the problem would be (0, 64). After that, you use slope formula to find your answer. (y2-y1/x2-x1, 64-0/2-0, 64/2=32) So, your average speed would be 32 m/s.
The thing that we learned in calculus that I was most confused with was simplifying the derivatives. For example, after using the quotient rule or product rule to solve a derivative, after I plugged in everything I wouldn’t know what to do. I know that you take the derivative of the bottom times the top minus the derivative of the top times the bottom all over the bottom squared, but after I finish that first step I get really mixed up with all the cancellation rules and how to simplify it. Also, I do not know when the problem is completely simplified. I know that the easiest way to become comfortable with it is just by practicing, but if anyone can help that would also be nice. :)
The thing that I am most comfortable with is average speed. The difference between instantaneous and average speed, is that instantaneous speed occurs only at one instant, while average speed occurs over a certain period of time. For average speed, you’re given a problem where an anvil falls off the roof onto Ryan G.’s head and then you are asked to find the average speed during the first 2 seconds of falling if y=16t^2 describes the fall. Your x’s in the problem would be (0, 2) because you are asked to find the average speed in the first two seconds. Then, to find your y’s, you would simply plug in your first x to the problem y=16t^2, which is 0, to find your first y. (y=16t^2, y=16(0)^2, y=0) After that, you plug in your second x to find your second y. (y=16t^2, y=16(2)^2, y=16(4), y=64) Therefore, your y’s in the problem would be (0, 64). After that, you use slope formula to find your answer. (y2-y1/x2-x1, 64-0/2-0, 64/2=32) So, your average speed would be 32 m/s.
The thing that we learned in calculus that I was most confused with was simplifying the derivatives. For example, after using the quotient rule or product rule to solve a derivative, after I plugged in everything I wouldn’t know what to do. I know that you take the derivative of the bottom times the top minus the derivative of the top times the bottom all over the bottom squared, but after I finish that first step I get really mixed up with all the cancellation rules and how to simplify it. Also, I do not know when the problem is completely simplified. I know that the easiest way to become comfortable with it is just by practicing, but if anyone can help that would also be nice. :)
Post #1
I started this week off really confused but by Friday I was only slightly confused. The thing I understand the most is quotient rule. The formula for quotient rule is vu^1-uv^1/v^2 or the derivative of the top times the bottom – the derivative of the bottom times the top over the bottom squared. An example is sin x / x-1. Take the derivative of the top which is (cos x) times the bottom (x-1) – the derivative of the bottom (1) times the top (sin x) over the bottom squared (x-1) ^2. From there, it is just simple algebra. The answer comes out to (cos x) (x-1)-sin x/(x-1) ^2. You do not use quotient rule when there is only an x on the bottom. You just bring the x to the top and make the exponent negative then use the formula U^n.
The thing I am most confused about is when to or when not to use the product rule. I understand you use it when you’re multiplying obviously, but do you only use it when you’re multiplying quadratics and trig functions? I know you don’t use it when you are multiplying 8 cos x so I’m thinking if there is a number in the problem you do not use product rule but does that apply to all problems? I also get confused when there is a problem such as x (quadratic). I don’t know if I have to distribute the x or use product rule and take the derivative of the x. I also struggled with problems like x^2 x^-5^3+1/2x^4pi^3 or number 45 on our first packet, whether I can simplify an answer or not, and the wording on some problems, but I think I will eventually catch on with more practice. All in all this week has been overwhelming but I’m looking forward to spending the year with everyone.
The thing I am most confused about is when to or when not to use the product rule. I understand you use it when you’re multiplying obviously, but do you only use it when you’re multiplying quadratics and trig functions? I know you don’t use it when you are multiplying 8 cos x so I’m thinking if there is a number in the problem you do not use product rule but does that apply to all problems? I also get confused when there is a problem such as x (quadratic). I don’t know if I have to distribute the x or use product rule and take the derivative of the x. I also struggled with problems like x^2 x^-5^3+1/2x^4pi^3 or number 45 on our first packet, whether I can simplify an answer or not, and the wording on some problems, but I think I will eventually catch on with more practice. All in all this week has been overwhelming but I’m looking forward to spending the year with everyone.
Post Number One
This week we learned many different things, including average speed, instantaneous speed, and derivatives. Who knew average speed actually meant slope. I have to admit i was a bit overwhelmed when we had to copy all of those derivative formulas down, but of course once we were taught what they were I felt much better.
The first thing we learned was average speed, and i'm pretty comfortable with that. What I am not comfortable with is instantaneous speed. I get the point and know the formula, but I get mixed up while plugging everything in..as in the example Find the instantaneous speed at t=2. y=16t2 as lim of h goes to 0. Since i have the notes on this problem, I can do it without a problem. But when I need to do a different problem without my notes, I get confused with what and where to plut into the formula as lim h goes to 0 f(x+h) - f(x) all over h. Everytime I work it i never get the right amount of h's in the end, and I get completely frustrated after that. I think I just need to practice more without getting frustrated, then I might get it. The other thing I don't always understand is the quotient rule. I know that the formula is vu1 - uv1 all over v2. All that means is copy bottom(derivative of top) - top(derivative of bottom) all over bottom squared. This is easy for me to explain but when simplifying I always try to cross out too much. I need someone to verify what I can and cannot cross out, or my whole answer is screwed up. One more thing I don't understand is when we have to find the derivative of a number raised to something then that is raised to something else. Example: x raised to pi squared. Do you just bring pi in front of x and subract one from pi like any other polynomial?
Well, atleast I understood something throughout the week in Calculus, better than nothing right? Now for what I AM comfortable doing. First, I finally get product rule. It's so simple, I don't understand why i never got it in the beginning. Simply put, it's copy top(derivative of bottom) + copy bottom(derivative of top). The only thing you need to remember is the different derivative formulas that will come up in the problems. Though the next thing I understood was completely easy and everyone understood it, it's still something so i'll say it anyway: derivatives of polynomials. If I had a test of just that, I may actually pass it.
Next week, I think i'll probably catch on a lot quicker. I was a little emotional and stressed out this week due to different reasons, but I'm better now so hopefully i'm ready even though we're moving on. We have a good class, and i think we're ready for whatever we are going to face.
Thank youuuu :)
The first thing we learned was average speed, and i'm pretty comfortable with that. What I am not comfortable with is instantaneous speed. I get the point and know the formula, but I get mixed up while plugging everything in..as in the example Find the instantaneous speed at t=2. y=16t2 as lim of h goes to 0. Since i have the notes on this problem, I can do it without a problem. But when I need to do a different problem without my notes, I get confused with what and where to plut into the formula as lim h goes to 0 f(x+h) - f(x) all over h. Everytime I work it i never get the right amount of h's in the end, and I get completely frustrated after that. I think I just need to practice more without getting frustrated, then I might get it. The other thing I don't always understand is the quotient rule. I know that the formula is vu1 - uv1 all over v2. All that means is copy bottom(derivative of top) - top(derivative of bottom) all over bottom squared. This is easy for me to explain but when simplifying I always try to cross out too much. I need someone to verify what I can and cannot cross out, or my whole answer is screwed up. One more thing I don't understand is when we have to find the derivative of a number raised to something then that is raised to something else. Example: x raised to pi squared. Do you just bring pi in front of x and subract one from pi like any other polynomial?
Well, atleast I understood something throughout the week in Calculus, better than nothing right? Now for what I AM comfortable doing. First, I finally get product rule. It's so simple, I don't understand why i never got it in the beginning. Simply put, it's copy top(derivative of bottom) + copy bottom(derivative of top). The only thing you need to remember is the different derivative formulas that will come up in the problems. Though the next thing I understood was completely easy and everyone understood it, it's still something so i'll say it anyway: derivatives of polynomials. If I had a test of just that, I may actually pass it.
Next week, I think i'll probably catch on a lot quicker. I was a little emotional and stressed out this week due to different reasons, but I'm better now so hopefully i'm ready even though we're moving on. We have a good class, and i think we're ready for whatever we are going to face.
Thank youuuu :)
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