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lgbasallote

  • 2 years ago

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

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  1. mathslover
    • 2 years ago
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    give MEDALS as much as possible :P just joking

  2. lgbasallote
    • 2 years ago
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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.

  3. mathslover
    • 2 years ago
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    that will be beneficial for me .. thanks lgba

  4. lgbasallote
    • 2 years ago
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    welcome

  5. Yahoo!
    • 2 years ago
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    Chain Rule , ........ etc ??

  6. hartnn
    • 2 years ago
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    POINT NOTED.

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

  8. lgbasallote
    • 2 years ago
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    it goes like that right?

  9. lgbasallote
    • 2 years ago
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    Note in everything i wrote, u and v stand for FUNCTIONS of X

  10. lgbasallote
    • 2 years ago
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    however, like i said...the things i wrote above are the most essential

  11. Eyad
    • 2 years ago
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    GooD work.

  12. Vaidehi09
    • 2 years ago
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    oh. ill keep those points in mind. thanks!

  13. goformit100
    • 2 years ago
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    Thanks from me too... they are really benifitiel ... i'll keep it in my mind... thanks again

  14. across
    • 2 years ago
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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

  15. lgbasallote
    • 2 years ago
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    yeah...that's probably the least possible lol

  16. ParthKohli
    • 2 years ago
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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

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

  18. ParthKohli
    • 2 years ago
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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!

  19. Kathi26
    • 2 years ago
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    \[f \prime(x)=\lim_{x \rightarrow x _{0}}\frac{ f(x)-f(x_{0}) }{ x-x _{0} }\]

  20. ParthKohli
    • 2 years ago
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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!

  21. Vaidehi09
    • 2 years ago
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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!

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

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

  24. ParthKohli
    • 2 years ago
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    O Lord, please help me figuring out the technique I should use here. Trig substitution?

  25. ParthKohli
    • 2 years ago
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    No... not trig substitution

  26. Vaidehi09
    • 2 years ago
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    jst substitute x^2 = u

  27. ParthKohli
    • 2 years ago
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    Yeah, right!\[du = 2xdx\]

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

  29. ParthKohli
    • 2 years ago
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    Woohoo!\[\int {1 \over u + 6}du\]

  30. hartnn
    • 2 years ago
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    x^2+6=u will be more beneficial.

  31. ParthKohli
    • 2 years ago
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    \[\int (u + 6)^{-1}du\]

  32. Vaidehi09
    • 2 years ago
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    ah yes...@hartnn

  33. ParthKohli
    • 2 years ago
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    Well, okay... yes, I did learn in calculus II about the root technique thingy

  34. Vaidehi09
    • 2 years ago
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    then it'll just be integration of 1/u

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

  36. ParthKohli
    • 2 years ago
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    Oh well, nope.

  37. ParthKohli
    • 2 years ago
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    I turned \(dx\) to \(du\).

  38. ParthKohli
    • 2 years ago
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    oops

  39. ParthKohli
    • 2 years ago
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    \[u = x^2 + 6\]\[du = 2xdx\]

  40. Vaidehi09
    • 2 years ago
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    nope. that was the ryt answer. that + C.

  41. ParthKohli
    • 2 years ago
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    Protect me from Calculus, O Heaven!

  42. hartnn
    • 2 years ago
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    your answer was correct ln |x^2+6| + c

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

  44. ParthKohli
    • 2 years ago
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    I need more practice. Ugh!

  45. hartnn
    • 2 years ago
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    parth u learning limits also ??

  46. ParthKohli
    • 2 years ago
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    Already learnt, yes.

  47. hartnn
    • 2 years ago
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    ok, i was gonna give a tutorial on that....and definite integration...which u want first ?

  48. ParthKohli
    • 2 years ago
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    I'd like a tutorial on the applications of definite integrals!

  49. hartnn
    • 2 years ago
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    *noted

  50. lgbasallote
    • 2 years ago
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    ....wow

  51. ParthKohli
    • 2 years ago
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    ... wow * 2

  52. Kainui
    • 2 years ago
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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

  53. ParthKohli
    • 2 years ago
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    That's @TuringTest's intuition too!

  54. lgbasallote
    • 2 years ago
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    but remember @Kainui i did not put in chain rule in what i wrote

  55. Kainui
    • 2 years ago
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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. \[d/dx[2y^0(x)^3]=6y^0x^2(d/dx(x))+0=6x^2\]

  56. Kainui
    • 2 years ago
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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.

  57. lgbasallote
    • 2 years ago
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    this is differential calculus....how the heck did integration by parts come in o.O

  58. Kainui
    • 2 years ago
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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.

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