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anonymous
 4 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.
anonymous
 4 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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mathslover
 4 years ago
Best ResponseYou've already chosen the best response.1give MEDALS as much as possible :P just joking

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here's a FEW of the things that are ESSENTIAL to differential calculus: \[\frac{d}{dx} (x^n) = nx^{n1}\] \[\frac{d}{dx} (e^x) = e^x\] \[\frac{d}{dx} (\ln x) = \frac 1x\] \[\frac{d}{dx} (\sin x) = \cos x\] \[\frac{d}{dx} (\cos x) = \sin x\] 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.

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.1that will be beneficial for me .. thanks lgba

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Chain Rule , ........ etc ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes.. i forgot chain rule.. \[\frac{d}{dx} (u) = u'du\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it goes like that right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Note in everything i wrote, u and v stand for FUNCTIONS of X

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0however, like i said...the things i wrote above are the most essential

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh. ill keep those points in mind. thanks!

goformit100
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks from me too... they are really benifitiel ... i'll keep it in my mind... thanks again

across
 4 years ago
Best ResponseYou've already chosen the best response.2To 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah...that's probably the least possible lol

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0When I hadn't started with Calculus, that \(\Delta x\) actually intimidated me. Then I learnt that you could just call that \(h\). haha

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0\[f'(x) = {\lim_{h \to 0} \;\;{f(x + h)  f(x) \over h}} \]

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0@lgbasallote Dude, you're a hero of OpenStudy. I look forward to seeing more tips from ya :) and believe me, I really needed them!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f \prime(x)=\lim_{x \rightarrow x _{0}}\frac{ f(x)f(x_{0}) }{ xx _{0} }\]

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0LGB, can you please write some tips for integration as well? It really gets messy when you start Calculus II and Integration by Parts!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea...for integration by parts...its all good as long as u remember those formulae! we need tips for when u dont!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{a}^{b}f(x)dx=F(a)F(b)=\int\limits_{b}^{a}f(x)dx\] \[\frac{ d }{ dx }(\int\limits_{a}^{f(x)}g(x)dx)=\frac{ d }{ dx }(G(f(x)G(a))=f \prime(x) \times g(f(x))\] \[=\frac{ d }{ dx }G(f(x))\] \[\int\limits_{}^{}\frac{ 1 }{ x }dx=\ln(x)\] \[\int\limits_{}^{}\frac{ f \prime(x) }{ f(x) }dx=\ln(f(x))\] \[\int\limits_{}^{}e^{x}dx=e^{x}\] \[\int\limits_{}^{}f \prime(x)\times e ^{f(x)}dx=e^{f(x)}\] \[\int\limits_{}^{}\sin(x)dx=\cos(x)\] \[\int\limits_{}^{}\cos(x)dx=\sin(x)\] \[\int\limits_{}^{}f \prime(x) \sin(f(x))dx=\cos(f(x))\] \[\int\limits_{}^{}f \prime(x) \cos(f(x))dx=\sin(f(x))\]

across
 4 years ago
Best ResponseYou've already chosen the best response.2@ParthKohli, pop quiz! Evaluate\[\int\frac{2x}{x^2+6}\,dx.\]:)

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0O Lord, please help me figuring out the technique I should use here. Trig substitution?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0No... not trig substitution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0jst substitute x^2 = u

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, right!\[du = 2xdx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's ln(x^2+6). You can use the f'(x)/f(x)case.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Woohoo!\[\int {1 \over u + 6}du\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1x^2+6=u will be more beneficial.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int (u + 6)^{1}du\]

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Well, okay... yes, I did learn in calculus II about the root technique thingy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then it'll just be integration of 1/u

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \int {1 \over u}du\]\[\lnu\]\[\lnx^2 +6\]?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I turned \(dx\) to \(du\).

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0\[u = x^2 + 6\]\[du = 2xdx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nope. that was the ryt answer. that + C.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Protect me from Calculus, O Heaven!

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1your answer was correct ln x^2+6 + c

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int {1 \over u}du\]\[\lnx^2 + 6 + C\]I always slip the + c off.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I need more practice. Ugh!

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1parth u learning limits also ??

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Already learnt, yes.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1ok, i was gonna give a tutorial on that....and definite integration...which u want first ?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I'd like a tutorial on the applications of definite integrals!

Kainui
 4 years ago
Best ResponseYou've already chosen the best response.0I 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

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0That's @TuringTest's intuition too!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but remember @Kainui i did not put in chain rule in what i wrote

Kainui
 4 years ago
Best ResponseYou've already chosen the best response.0The 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. \[d/dx[2y^0(x)^3]=6y^0x^2(d/dx(x))+0=6x^2\]

Kainui
 4 years ago
Best ResponseYou've already chosen the best response.0Also 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is differential calculus....how the heck did integration by parts come in o.O

Kainui
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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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