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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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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give MEDALS as much as possible :P just joking
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.
that will be beneficial for me .. thanks lgba

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welcome
Chain Rule , ........ etc ??
POINT NOTED.
yes.. i forgot chain rule.. \[\frac{d}{dx} (u) = u'du\]
it goes like that right?
Note in everything i wrote, u and v stand for FUNCTIONS of X
however, like i said...the things i wrote above are the most essential
GooD work.
oh. ill keep those points in mind. thanks!
Thanks from me too... they are really benifitiel ... i'll keep it in my mind... thanks again
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
yeah...that's probably the least possible lol
When I hadn't started with Calculus, that \(\Delta x\) actually intimidated me. Then I learnt that you could just call that \(h\). haha
\[f'(x) = {\lim_{h \to 0} \;\;{f(x + h) - f(x) \over h}} \]
@lgbasallote Dude, you're a hero of OpenStudy. I look forward to seeing more tips from ya :) and believe me, I really needed them!
\[f \prime(x)=\lim_{x \rightarrow x _{0}}\frac{ f(x)-f(x_{0}) }{ x-x _{0} }\]
LGB, can you please write some tips for integration as well? It really gets messy when you start Calculus II and Integration by Parts!
yea...for integration by parts...its all good as long as u remember those formulae! we need tips for when u dont!
\[\int\limits_{a}^{b}f(x)dx=F(a)-F(b)=-\int\limits_{b}^{a}f(x)dx\] \[\frac{ d }{ dx }(\int\limits_{a}^{f(x)}g(x)dx)=\frac{ d }{ dx }(G(f(x)-G(a))=f \prime(x) \times g(f(x))\] \[=\frac{ d }{ dx }G(f(x))\] \[\int\limits_{}^{}\frac{ 1 }{ x }dx=\ln(x)\] \[\int\limits_{}^{}\frac{ f \prime(x) }{ f(x) }dx=\ln(f(x))\] \[\int\limits_{}^{}e^{x}dx=e^{x}\] \[\int\limits_{}^{}f \prime(x)\times e ^{f(x)}dx=e^{f(x)}\] \[\int\limits_{}^{}\sin(x)dx=-\cos(x)\] \[\int\limits_{}^{}\cos(x)dx=\sin(x)\] \[\int\limits_{}^{}f \prime(x) \sin(f(x))dx=-\cos(f(x))\] \[\int\limits_{}^{}f \prime(x) \cos(f(x))dx=\sin(f(x))\]
@ParthKohli, pop quiz! Evaluate\[\int\frac{2x}{x^2+6}\,dx.\]:)
O Lord, please help me figuring out the technique I should use here. Trig substitution?
No... not trig substitution
jst substitute x^2 = u
Yeah, right!\[du = 2xdx\]
It's ln(x^2+6). You can use the f'(x)/f(x)case.
Woohoo!\[\int {1 \over u + 6}du\]
x^2+6=u will be more beneficial.
\[\int (u + 6)^{-1}du\]
ah yes...@hartnn
Well, okay... yes, I did learn in calculus II about the root technique thingy
then it'll just be integration of 1/u
\[ \int {1 \over u}du\]\[\ln|u|\]\[\ln|x^2 +6|\]?
Oh well, nope.
I turned \(dx\) to \(du\).
oops
\[u = x^2 + 6\]\[du = 2xdx\]
nope. that was the ryt answer. that + C.
Protect me from Calculus, O Heaven!
your answer was correct ln |x^2+6| + c
\[\int {1 \over u}du\]\[\ln|x^2 + 6| + C\]I always slip the + c off.
I need more practice. Ugh!
parth u learning limits also ??
Already learnt, yes.
ok, i was gonna give a tutorial on that....and definite integration...which u want first ?
I'd like a tutorial on the applications of definite integrals!
*noted
....wow
... wow * 2
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
That's @TuringTest's intuition too!
but remember @Kainui i did not put in chain rule in what i wrote
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. \[d/dx[2y^0(x)^3]=6y^0x^2(d/dx(x))+0=6x^2\]
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.
this is differential calculus....how the heck did integration by parts come in o.O
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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