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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
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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mathsloverBest ResponseYou've already chosen the best response.1
give MEDALS as much as possible :P just joking
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.17
Here'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.
 one year ago

mathsloverBest ResponseYou've already chosen the best response.1
that will be beneficial for me .. thanks lgba
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.1
Chain Rule , ........ etc ??
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.17
yes.. i forgot chain rule.. \[\frac{d}{dx} (u) = u'du\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.17
it goes like that right?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.17
Note in everything i wrote, u and v stand for FUNCTIONS of X
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.17
however, like i said...the things i wrote above are the most essential
 one year ago

Vaidehi09Best ResponseYou've already chosen the best response.0
oh. ill keep those points in mind. thanks!
 one year ago

goformit100Best ResponseYou've already chosen the best response.0
Thanks from me too... they are really benifitiel ... i'll keep it in my mind... thanks again
 one year ago

acrossBest ResponseYou've already chosen the best response.2
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

lgbasalloteBest ResponseYou've already chosen the best response.17
yeah...that's probably the least possible lol
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
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

ParthKohliBest ResponseYou've already chosen the best response.0
\[f'(x) = {\lim_{h \to 0} \;\;{f(x + h)  f(x) \over h}} \]
 one year ago

ParthKohliBest 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!
 one year ago

Kathi26Best ResponseYou've already chosen the best response.0
\[f \prime(x)=\lim_{x \rightarrow x _{0}}\frac{ f(x)f(x_{0}) }{ xx _{0} }\]
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
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

Vaidehi09Best ResponseYou've already chosen the best response.0
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

Kathi26Best ResponseYou've already chosen the best response.0
\[\int\limits_{a}^{b}f(x)dx=F(a)F(b)=\int\limits_{b}^{a}f(x)dx\] \[\frac{ d }{ dx }(\int\limits_{a}^{f(x)}g(x)dx)=\frac{ d }{ dx }(G(f(x)G(a))=f \prime(x) \times g(f(x))\] \[=\frac{ d }{ dx }G(f(x))\] \[\int\limits_{}^{}\frac{ 1 }{ x }dx=\ln(x)\] \[\int\limits_{}^{}\frac{ f \prime(x) }{ f(x) }dx=\ln(f(x))\] \[\int\limits_{}^{}e^{x}dx=e^{x}\] \[\int\limits_{}^{}f \prime(x)\times e ^{f(x)}dx=e^{f(x)}\] \[\int\limits_{}^{}\sin(x)dx=\cos(x)\] \[\int\limits_{}^{}\cos(x)dx=\sin(x)\] \[\int\limits_{}^{}f \prime(x) \sin(f(x))dx=\cos(f(x))\] \[\int\limits_{}^{}f \prime(x) \cos(f(x))dx=\sin(f(x))\]
 one year ago

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

ParthKohliBest ResponseYou've already chosen the best response.0
O Lord, please help me figuring out the technique I should use here. Trig substitution?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
No... not trig substitution
 one year ago

Vaidehi09Best ResponseYou've already chosen the best response.0
jst substitute x^2 = u
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Yeah, right!\[du = 2xdx\]
 one year ago

Kathi26Best ResponseYou've already chosen the best response.0
It's ln(x^2+6). You can use the f'(x)/f(x)case.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Woohoo!\[\int {1 \over u + 6}du\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
x^2+6=u will be more beneficial.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
\[\int (u + 6)^{1}du\]
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Well, okay... yes, I did learn in calculus II about the root technique thingy
 one year ago

Vaidehi09Best ResponseYou've already chosen the best response.0
then it'll just be integration of 1/u
 one year ago

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

ParthKohliBest ResponseYou've already chosen the best response.0
I turned \(dx\) to \(du\).
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
\[u = x^2 + 6\]\[du = 2xdx\]
 one year ago

Vaidehi09Best ResponseYou've already chosen the best response.0
nope. that was the ryt answer. that + C.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Protect me from Calculus, O Heaven!
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
your answer was correct ln x^2+6 + c
 one year ago

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

ParthKohliBest ResponseYou've already chosen the best response.0
I need more practice. Ugh!
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
parth u learning limits also ??
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Already learnt, yes.
 one year ago

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

ParthKohliBest ResponseYou've already chosen the best response.0
I'd like a tutorial on the applications of definite integrals!
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
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

ParthKohliBest ResponseYou've already chosen the best response.0
That's @TuringTest's intuition too!
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.17
but remember @Kainui i did not put in chain rule in what i wrote
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
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

KainuiBest ResponseYou've already chosen the best response.0
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

lgbasalloteBest ResponseYou've already chosen the best response.17
this is differential calculus....how the heck did integration by parts come in o.O
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
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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