1. ## Desperate Student

So....Coming up on the end of the semester and I haven't exactly done so well in my Calc. 1A class, however I still have a chance of passing if I pass the final with a C or better. Right now I don't have any specific questions (though I'm sure before the semester is over I will ), but I was wondering if anyone had some priceless knowledge about studying/memorizing that I might want to keep in mind. Any opinions would be helpful!

2. Originally Posted by CALsunshine
So....Coming up on the end of the semester and I haven't exactly done so well in my Calc. 1A class, however I still have a chance of passing if I pass the final with a C or better. Right now I don't have any specific questions (though I'm sure before the semester is over I will ), but I was wondering if anyone had some priceless knowledge about studying/memorizing that I might want to keep in mind. Any opinions would be helpful!
$\int{u}dv=uv-\int{v\cdot{du}}$

remember taht $\int{\frac{u'(x)}{u(x)}dx}=ln|u(x)|+C$
O and three VERY helpful identities $sec^2(x)=1+tan^2(x)$,

$cos^2(x)=\frac{1+cos(2x)}{2}$,

$sin^2(x)=\frac{1-cos(2x)}{2}$

another thing is you always want to get integrals into the form $\int{f(g(x))\cdot{g'(x)}dx}$

I will be back later for more

3. Hi!! I've been in your situation before and it is NO FUN!

So I'll share the things I have learned with you:

First, I am already studying for my Calc II final even though we have two unit tests to take, then the final. My final is on May 12th.

I have been to class everyday but two. I have notes that I have kept in order from the beginning. I go through and highlight really important information that I have not memorized as well as I would like, and I highlight things that I know that are important, even if I am pretty familiar with them.

If you do not have notes because you've missed several days, get a syllabus or notes from a friend and make copies. Go through and look at the important stuff and make a list of the sections that you need to refresh yourself with. You do not even have to make it a long, labourous session.

For example, I am planning on putting in 30 minutes per night with past material until my final. Let's say that I start on Monday, April 21. If I do 30 minutes per night until my final, that is 10.5 hours of studying. That's a lot!

Also, when I do my 30 minutes per night, I already have a game plan. I pick out specific problems from each section of my book that we have done. Sometimes it takes a little more than 30 minutes to get through the problems, but I usually do about 4 problems/night plus or minus a few depending on the length. So if I do about 4 problems/night until my final, that's give or take, 84 problems!

I guess my point is to pick a time each night where you are just going to look at calc until your final. It does you so much better to look at it for short bursts over a period of a couple weeks than to cram for 8 hours the night before and take the test on no sleep. Believe me, I've tried...my 94.1% in calc I moved to an 89.3% after the final! Let's just say I was not too thrilled.

Good luck!

4. You should know all basic algebra, this also includes trig. identities and other stuff, it's very important before learnin' calculus.

If you have problems, post them and all here we're gonna help ya with.

5. ## :)

Thanks guys! I'll probably be back with more specific derivative/integral problems soon enough

6. I don't know how you learn, but I learn best by doing. Meaning don't bother with "studying" in the traditional sense, it is not as effective. Instead, I would suggest going back through your old sections, under the assumption that your book has odd numbered questions answered in the back, just sit down and start working out the odd problems. Don't waste your energy on presentation or tidiness, just focus on making sure you can get the right answer (ideally you can do this quickly). If you cannot figure out the answer, then look back through the chapter at the formulas provided, and the examples that are worked out. Work on doing this quickly and effectively, focus on the material itself only so much as you need in order to get the correct solution. Then verify your solutions in the back of the book. If you cannot get the correct solution, and you cannot figure out what you are doing wrong, post what you have done on here for some assistance.

For me, this approach works very effectively, my memory of formulas, techniques, methods, material is much much greater when I learn it by applying it rather than by taking notes or reading through dull chapters.

When you are finished, go back and do the same problems over again, this is a memorization technique, it will help solidify the concepts in your mind.

I would probably suggest do all the odd problems in one section, then go back and do them over again, before going onto the next section, but feel free to experiment and see what works best for you, you may find that you prefer to do all the sections before going over them again.

In the end, Calculus is deceptively simple, there are very few topics covered, mostly they are all variations of the same theme. Calc1 I felt was particularly like this, they teach you what a derivative is, then throw you 20 different ways to apply it. This is very beneficial, it means that the complexity is much lower than it appears, you just need to put in enough dedication to understand the nuances, and become familiar with the material.

Do not hand in your test until the instructor demands it. Even if every other kid in your class has already left, you are allotted a given amount of time. Use every bit of it, when you finish, go back and recheck your work, do not feel pressured to turn it in until you have no choice.

Anyway, best of luck, keep in mind that you've made it this far, and if you are willing to dedicate yourself in this last month, you can make it through this. It won't always be easy, but a month of dedication now will save you the many months (as well as hundreds of dollars) that would be lost if you had to retake the course. Keep in mind that even though it may be tough and require a lot of energy and dedication, the sacrifice is worth it. When you get discouraged in one area switch focuses to another. Keep your eye on the prize, don't let demotivation prevent you from making progress. With the right attitude and mindset, you can pull it off.
Originally Posted by Mathstud28
$cos^2(x)=\frac{1+cos(2x)}{2}$,

$sin^2(x)=\frac{1-cos(2x)}{2}$
Invaluable formulas, I use them very frequently.

7. Originally Posted by CALsunshine
So....Coming up on the end of the semester and I haven't exactly done so well in my Calc. 1A class, however I still have a chance of passing if I pass the final with a C or better. Right now I don't have any specific questions (though I'm sure before the semester is over I will ), but I was wondering if anyone had some priceless knowledge about studying/memorizing that I might want to keep in mind. Any opinions would be helpful!
Another thing I will point out to you is this...when it comes to derivatives...unlike integrals they are cut-and-dry....if you learn all the following techniques you will be able to tackle all but the most difficult integrals you will learn...I suggest looking at an integration/differentiation table for your test...the following are the best for me

Basic rules
$\frac{D[f_1(x)\pm{f_2(x)}]}{dx}=f_1'(x)\pm{f_2'(x)}$

$\frac{D[cf(x)]}{dx}=cf'(x)$ where c is a constant

$\frac{D[x^{n}]}{dx}=nx^{n-1}$

$\frac{D\bigg[e^{u(x)}\bigg]}{dx}=u'(x)\cdot{e^{u(x)}}$

$\frac{D[log_a(u(x))]}{dx}=\frac{D\bigg[\frac{ln(u(x))}{ln(a)}\bigg]}{dx}=\frac{u'(x)}{ln(a)u(x)}$...which implies that if the base is e then the derivative is $\frac{u'(x)}{u(x)}$

now we get to some rules

Product rule
$\frac{D[u\cdot{v}]}{dx}=uv'+vu'$ where u and v are functions of x

Quotient rule
$\frac{D\bigg[\frac{u}{v}\bigg]}{dx}=\frac{u'v-vu'}{v^2}$

Chain rule
$\frac{D[f(g(x))]}{dx}=f'(g(x))\cdot{g'(x)}$

Now that I have discussed both I will talk about the fact that sometimes easier than the quotient rule is applying the chain rule to quotients by using the fact taht $\frac{u}{v}=u\cdot{v^{-1}}$...therefore $\frac{D\bigg[\frac{u}{v}\bigg]}{dx}=\frac{D[u\cdot{v^{-1}}]}{dx}=u'\cdot{v^{-1}}+u\cdot{-v^{-2}}\cdot{v'}$ by the chain rule

Another very useful technique is logarithmic differentiatiow which takes full advantage of logarithims ability to simplify complex functions...this one comes in two flavors...simplification of large quotients...or differentiating when there is an x in the exponent and in the base

I will give you a general case for the second one $y=f(x)^{g(x)}$..now applying our logarithims(prefably always ln(x) since it is the nicest to work with) we get $\ln(y)=\ln\bigg(f(x)^{g(x)}\bigg)$...now using the amazing simplification techniques of logarithim this turns into $\ln(y)=g(x)\cdot{\ln(f(x))}$...now differentiating we get $\frac{y'}{y}=g'(x)\cdot{\ln(f(x))}+g(x)\cdot\frac{ f'(x)}{f(x)}$...the right side was gotten using the fact that $y=u(x)$ and the left side was gotten using the product rule...and we multiply both sides by y to get the final answer remember that $y=f(x)^{g(x)}$

for the first use I willshow you an example $y=\frac{\sqrt[3]{x^2-1}(x+1)}{\sqrt{x-1}(x+7)^3}$..now using that tecnhique we have $\ln(y)=\ln\bigg(\frac{\sqrt[3]{x^2-1}(x+1)}{\sqrt{x-1}(x+7)^3}\bigg)$...simplifying we get $\ln(y)=\frac{1}{3}\ln(x^2-1)+ln(x+1)-\frac{1}{2}\ln(x-1)-3\ln(x+7)$...differentiating we get $\frac{y'}{y}=\frac{2x}{3(x^2-1)}+\frac{1}{x+1}-\frac{1}{2(x-1)}-\frac{3}{x+7}$...now just multiply both sides by y(the original function) and you have your answer...I will post some practice questions you can work and we can critique you

8. Hi !

$\frac{D[x^{n}]}{dx}=nx^{n-1}$
I think it's better giving this formula :

$\frac{D[(f(x))^{n}]}{dx}=n f'(x) (f(x))^{n-1}$

Once you've mastered the formulas of derivatives, you will be able to recognize more easily some integrals involving simple substitutions

9. ## Problems

These are NOT in order in which I covered them to make them more challenging

Find the derivatives of the following equations
1. $y=x^{\frac{1}{x}}$

2. $y=(x+3)^2-(x+7)^5$

3. $y=\frac{x+7}{2x^2-15x}$

4. $y=e^{2x}$

5. $y=\ln(x^3+7x-5)$

6. $y=(x+3)(x-7)^3$

7. $y=log_{10}(x^3-1)$

8. $y=(3x^2+1)^4$

You can do these if you wish and we will critique you if you post them...hope I have been some help!

Mathstud

10. Originally Posted by Moo
Hi !

I think it's better giving this formula :

$\frac{D[(f(x))^{n}]}{dx}=n f'(x) (f(x))^{n-1}$

Once you've mastered the formulas of derivatives, you will be able to recognize more easily some integrals involving simple substitutions
But why post a specific case of the chain rule? I guess that is helpful! Thanks!

11. There are chain rules uses you should know immediately, and I think this one may be known raw.
What I wanted to say is that this formula is better than giving the derivative for x^n
My opinion, not everybody's

12. ## Ok

Sorry I am posting up a storm...but another useful bit of info would be

Equation of a tangent line to the point $(x_0,f(x_0))$
$y-f(x_0)=f'(x_0)(x-x_0)$

Equation of normal line at point $(x_0,f(x_0))$
$y-x_0=\frac{-1}{f'(x_0)}(x-x_0)$

and I know everyone hates it but L'hopital's rule

which states that if the limit $\lim_{x \to c}\frac{f(x)}{g(x)}$ gives one of the 8 or so indeterminate forms(most commonly $\frac{0}{0}$ or $\frac{\infty}{\infty}$...then $\lim_{x \to c}\frac{f(x)}{g(x)}=\lim_{x \to c}\frac{f'(x)}{g'(x)}$

13. Ok last one I think

haha

Commonly thrown at you are the following

Rolle's Theorem "If f is continous on [a,b] and differentiable on (a,b), and f(a)=f(b) then there exists one value c between a and b such that $f'(c)=0$"...which basically says that if on an interval a function meets the requirements of the theorem there is a max or a min

Mean-Value theorem "If f is continuous on [a,b] and differentiable on (a,b). then there exists one value c between a and b such that $f'(c)=\frac{f(b)-f(a)}{b-a}$ which basically says that there is one point on an interval where the actual slope is equal to the average slope(i.e. $\frac{\Delta{y}}{\Delta{x}}$)

Extreme Value theorem
"If f is continous on [a,b] then there exists an absolute maximum point and an absolute minimum point"

that one is self-explanitory

14. ## Last one

Ok this is the last one

First derivative test:
Find the points that make the first derivative 0, the first derviative undefined, and the points that make the original function undefined...set up test intervals based on theses numbers....test an element of each interval in the first derivative...where the sign's of the evaluated points change from + to - it is a max...and where they change from - to+ is a min

Second derivative test:
Take ONLY the points that make the first derivative 0...plug those points into the second derivative if $f''(c)<0$ c is a max and if $f''(c)>0$ c is a min...if $f''(c)=0$ then the test is inconclusive...c is now a candidate for an inflection point

Test for inflection points:
Find the points where the second derivative is undefined, the points where the second dervative is equal to 0, and the points that make the original function undefined...Set up test intervals based on theses numbers...test an element of each interval in the second derviative...where there is a sign change there is an inflection point

15. Haha you better get an 100% on your test after all this work I have put into typing haha

When it comes to studying basic calc nothing beats this site

Visual Calculus

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