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

- lgbasallote

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- katieb

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- mathslover

give MEDALS as much as possible :P just joking

- lgbasallote

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.

- mathslover

that will be beneficial for me .. thanks lgba

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## More answers

- lgbasallote

welcome

- anonymous

Chain Rule , ........ etc ??

- hartnn

POINT NOTED.

- lgbasallote

yes.. i forgot chain rule..
\[\frac{d}{dx} (u) = u'du\]

- lgbasallote

it goes like that right?

- lgbasallote

Note in everything i wrote, u and v stand for FUNCTIONS of X

- lgbasallote

however, like i said...the things i wrote above are the most essential

- anonymous

GooD work.

- anonymous

oh. ill keep those points in mind. thanks!

- goformit100

Thanks from me too... they are really benifitiel ... i'll keep it in my mind... thanks again

- across

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

- lgbasallote

yeah...that's probably the least possible lol

- ParthKohli

When I hadn't started with Calculus, that \(\Delta x\) actually intimidated me. Then I learnt that you could just call that \(h\). haha

- ParthKohli

\[f'(x) = {\lim_{h \to 0} \;\;{f(x + h) - f(x) \over h}} \]

- ParthKohli

@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

\[f \prime(x)=\lim_{x \rightarrow x _{0}}\frac{ f(x)-f(x_{0}) }{ x-x _{0} }\]

- ParthKohli

LGB, can you please write some tips for integration as well? It really gets messy when you start Calculus II and Integration by Parts!

- anonymous

yea...for integration by parts...its all good as long as u remember those formulae! we need tips for when u dont!

- anonymous

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

- across

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

- ParthKohli

O Lord, please help me figuring out the technique I should use here. Trig substitution?

- ParthKohli

No... not trig substitution

- anonymous

jst substitute x^2 = u

- ParthKohli

Yeah, right!\[du = 2xdx\]

- anonymous

It's ln(x^2+6). You can use the f'(x)/f(x)case.

- ParthKohli

Woohoo!\[\int {1 \over u + 6}du\]

- hartnn

x^2+6=u will be more beneficial.

- ParthKohli

\[\int (u + 6)^{-1}du\]

- anonymous

ah yes...@hartnn

- ParthKohli

Well, okay... yes, I did learn in calculus II about the root technique thingy

- anonymous

then it'll just be integration of 1/u

- ParthKohli

\[ \int {1 \over u}du\]\[\ln|u|\]\[\ln|x^2 +6|\]?

- ParthKohli

Oh well, nope.

- ParthKohli

I turned \(dx\) to \(du\).

- ParthKohli

oops

- ParthKohli

\[u = x^2 + 6\]\[du = 2xdx\]

- anonymous

nope. that was the ryt answer. that + C.

- ParthKohli

Protect me from Calculus, O Heaven!

- hartnn

your answer was correct ln |x^2+6| + c

- ParthKohli

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

- ParthKohli

I need more practice. Ugh!

- hartnn

parth u learning limits also ??

- ParthKohli

Already learnt, yes.

- hartnn

ok, i was gonna give a tutorial on that....and definite integration...which u want first ?

- ParthKohli

I'd like a tutorial on the applications of definite integrals!

- hartnn

*noted

- lgbasallote

....wow

- ParthKohli

... wow * 2

- Kainui

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

- ParthKohli

That's @TuringTest's intuition too!

- lgbasallote

but remember @Kainui i did not put in chain rule in what i wrote

- Kainui

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.
\[d/dx[2y^0(x)^3]=6y^0x^2(d/dx(x))+0=6x^2\]

- Kainui

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.

- lgbasallote

this is differential calculus....how the heck did integration by parts come in o.O

- Kainui

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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