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lgbasallote Group Title

A tip on how to become good in Differential Calculus: MEMORIZE as LITTLE as possible, and learn to MANIPULATE as MUCH as possible.

  • one year ago
  • one year ago

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  1. mathslover Group Title
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    give MEDALS as much as possible :P just joking

    • one year ago
  2. lgbasallote Group Title
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    Here's a FEW of the things that are ESSENTIAL to differential calculus: \[\frac{d}{dx} (x^n) = nx^{n-1}\] \[\frac{d}{dx} (e^x) = e^x\] \[\frac{d}{dx} (\ln x) = \frac 1x\] \[\frac{d}{dx} (\sin x) = \cos x\] \[\frac{d}{dx} (\cos x) = -\sin x\] Then of course, the product and quotient rules: product rule: \[\frac{d}{dx}(uv) = udv + vdu\] quotient rule: \[\frac{d}{dx} (\frac uv) = \frac{vdu - udv}{v^2}\] Then, five of the trigonometric formulas: \[\sin (\alpha \pm \beta) = \sin \alpha \cos \beta \pm \sin \beta \cos \alpha\] \[\cos (\alpha \pm \beta) = \cos \alpha \cos \beta \pm \sin \alpha \sin \beta\] \[\sin^2 + \cos^2 = 1\] \[\tan^2 + 1 = \sec^2\] \[1 + \cot ^2 = \csc^2\] With these, and a little manipulation, you can survive Differential Calculus. However, of course, as you go on with the course, you use other formulas as well and you'll just remember it. For example \[\frac{d}{dx} \sin h x = \cosh x\] However, if you can't remember it, you can always figure it out using the basic information I wrote above. This tip was taken from a true Calculus master (who happened to be a summa cum laude), and has been proven effective by hundreds of students whom he have taught.

    • one year ago
  3. mathslover Group Title
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    that will be beneficial for me .. thanks lgba

    • one year ago
  4. lgbasallote Group Title
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    welcome

    • one year ago
  5. Yahoo! Group Title
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    Chain Rule , ........ etc ??

    • one year ago
  6. hartnn Group Title
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    POINT NOTED.

    • one year ago
  7. lgbasallote Group Title
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    yes.. i forgot chain rule.. \[\frac{d}{dx} (u) = u'du\]

    • one year ago
  8. lgbasallote Group Title
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    it goes like that right?

    • one year ago
  9. lgbasallote Group Title
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    Note in everything i wrote, u and v stand for FUNCTIONS of X

    • one year ago
  10. lgbasallote Group Title
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    however, like i said...the things i wrote above are the most essential

    • one year ago
  11. Eyad Group Title
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    GooD work.

    • one year ago
  12. Vaidehi09 Group Title
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    oh. ill keep those points in mind. thanks!

    • one year ago
  13. goformit100 Group Title
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    Thanks from me too... they are really benifitiel ... i'll keep it in my mind... thanks again

    • one year ago
  14. across Group Title
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    To be totally honest with you, remembering that\[\frac{df}{dx}=f'(x)=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\]is much better than memorizing all of that up there, but I'm a masochist, so, yeah. :P

    • one year ago
  15. lgbasallote Group Title
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    yeah...that's probably the least possible lol

    • one year ago
  16. ParthKohli Group Title
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    When I hadn't started with Calculus, that \(\Delta x\) actually intimidated me. Then I learnt that you could just call that \(h\). haha

    • one year ago
  17. ParthKohli Group Title
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    \[f'(x) = {\lim_{h \to 0} \;\;{f(x + h) - f(x) \over h}} \]

    • one year ago
  18. ParthKohli Group Title
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    @lgbasallote Dude, you're a hero of OpenStudy. I look forward to seeing more tips from ya :) and believe me, I really needed them!

    • one year ago
  19. Kathi26 Group Title
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    \[f \prime(x)=\lim_{x \rightarrow x _{0}}\frac{ f(x)-f(x_{0}) }{ x-x _{0} }\]

    • one year ago
  20. ParthKohli Group Title
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    LGB, can you please write some tips for integration as well? It really gets messy when you start Calculus II and Integration by Parts!

    • one year ago
  21. Vaidehi09 Group Title
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    yea...for integration by parts...its all good as long as u remember those formulae! we need tips for when u dont!

    • one year ago
  22. Kathi26 Group Title
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    \[\int\limits_{a}^{b}f(x)dx=F(a)-F(b)=-\int\limits_{b}^{a}f(x)dx\] \[\frac{ d }{ dx }(\int\limits_{a}^{f(x)}g(x)dx)=\frac{ d }{ dx }(G(f(x)-G(a))=f \prime(x) \times g(f(x))\] \[=\frac{ d }{ dx }G(f(x))\] \[\int\limits_{}^{}\frac{ 1 }{ x }dx=\ln(x)\] \[\int\limits_{}^{}\frac{ f \prime(x) }{ f(x) }dx=\ln(f(x))\] \[\int\limits_{}^{}e^{x}dx=e^{x}\] \[\int\limits_{}^{}f \prime(x)\times e ^{f(x)}dx=e^{f(x)}\] \[\int\limits_{}^{}\sin(x)dx=-\cos(x)\] \[\int\limits_{}^{}\cos(x)dx=\sin(x)\] \[\int\limits_{}^{}f \prime(x) \sin(f(x))dx=-\cos(f(x))\] \[\int\limits_{}^{}f \prime(x) \cos(f(x))dx=\sin(f(x))\]

    • one year ago
  23. across Group Title
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    @ParthKohli, pop quiz! Evaluate\[\int\frac{2x}{x^2+6}\,dx.\]:)

    • one year ago
  24. ParthKohli Group Title
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    O Lord, please help me figuring out the technique I should use here. Trig substitution?

    • one year ago
  25. ParthKohli Group Title
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    No... not trig substitution

    • one year ago
  26. Vaidehi09 Group Title
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    jst substitute x^2 = u

    • one year ago
  27. ParthKohli Group Title
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    Yeah, right!\[du = 2xdx\]

    • one year ago
  28. Kathi26 Group Title
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    It's ln(x^2+6). You can use the f'(x)/f(x)case.

    • one year ago
  29. ParthKohli Group Title
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    Woohoo!\[\int {1 \over u + 6}du\]

    • one year ago
  30. hartnn Group Title
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    x^2+6=u will be more beneficial.

    • one year ago
  31. ParthKohli Group Title
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    \[\int (u + 6)^{-1}du\]

    • one year ago
  32. Vaidehi09 Group Title
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    ah yes...@hartnn

    • one year ago
  33. ParthKohli Group Title
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    Well, okay... yes, I did learn in calculus II about the root technique thingy

    • one year ago
  34. Vaidehi09 Group Title
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    then it'll just be integration of 1/u

    • one year ago
  35. ParthKohli Group Title
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    \[ \int {1 \over u}du\]\[\ln|u|\]\[\ln|x^2 +6|\]?

    • one year ago
  36. ParthKohli Group Title
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    Oh well, nope.

    • one year ago
  37. ParthKohli Group Title
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    I turned \(dx\) to \(du\).

    • one year ago
  38. ParthKohli Group Title
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    oops

    • one year ago
  39. ParthKohli Group Title
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    \[u = x^2 + 6\]\[du = 2xdx\]

    • one year ago
  40. Vaidehi09 Group Title
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    nope. that was the ryt answer. that + C.

    • one year ago
  41. ParthKohli Group Title
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    Protect me from Calculus, O Heaven!

    • one year ago
  42. hartnn Group Title
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    your answer was correct ln |x^2+6| + c

    • one year ago
  43. ParthKohli Group Title
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    \[\int {1 \over u}du\]\[\ln|x^2 + 6| + C\]I always slip the + c off.

    • one year ago
  44. ParthKohli Group Title
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    I need more practice. Ugh!

    • one year ago
  45. hartnn Group Title
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    parth u learning limits also ??

    • one year ago
  46. ParthKohli Group Title
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    Already learnt, yes.

    • one year ago
  47. hartnn Group Title
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    ok, i was gonna give a tutorial on that....and definite integration...which u want first ?

    • one year ago
  48. ParthKohli Group Title
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    I'd like a tutorial on the applications of definite integrals!

    • one year ago
  49. hartnn Group Title
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    *noted

    • one year ago
  50. lgbasallote Group Title
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    ....wow

    • one year ago
  51. ParthKohli Group Title
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    ... wow * 2

    • one year ago
  52. Kainui Group Title
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    I think it should be noted that the quotient rule is kind of meaningless to remember when you have the product rule already. Division is the same thing as multiplying. For instance 5x/(3x^2+2x)=5x*(3x^2+2x)^-1

    • one year ago
  53. ParthKohli Group Title
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    That's @TuringTest's intuition too!

    • one year ago
  54. lgbasallote Group Title
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    but remember @Kainui i did not put in chain rule in what i wrote

    • one year ago
  55. Kainui Group Title
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    The less you memorize, the better. I pretty much consider the power, chain, and product rule all in one rule these days to condense it further, it just usually turns out to be 1 and not important. An example of this might be: \[2x^3=2y^0(x)^3\] Well the derivative of the outside is 6x and multiplied by the derivative of the inside, which is 1. The derivative of the coefficient is also 0 so you end up adding 0 as well. \[d/dx[2y^0(x)^3]=6y^0x^2(d/dx(x))+0=6x^2\]

    • one year ago
  56. Kainui Group Title
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    Also a note about the integration by parts is a great thing to include here along with chain rule lol. They are both essential in my mind.

    • one year ago
  57. lgbasallote Group Title
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    this is differential calculus....how the heck did integration by parts come in o.O

    • one year ago
  58. Kainui Group Title
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    Yeah, good point. Another helpful thing is to think about what the graph looks like, it will help you visualize what the derivative will be. This is particularly useful for optimization problems where you are making a graph where the highest point is the thing you want, and taking the derivative and setting it equal to 0 to find that peak.

    • one year ago
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