Question

A tip on how to become good in Differential Calculus:

MEMORIZE as LITTLE as possible, and learn to MANIPULATE as MUCH as possible.

Answer 1

give MEDALS as much as possible :P just joking

Answer 2

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.

Answer 3

that will be beneficial for me .. thanks lgba

Answer 4

welcome

Answer 5

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

Answer 6

POINT NOTED.

Answer 7

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

Answer 8

it goes like that right?

Answer 9

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

Answer 10

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

Answer 11

GooD work.

Answer 12

oh. ill keep those points in mind. thanks!

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

@lgbasallote Dude, you're a hero of OpenStudy. I look forward to seeing more tips from ya :) and believe me, I really needed them!

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

@ParthKohli, pop quiz!

Evaluate\[\int\frac{2x}{x^2+6}\,dx.\]:)

Answer 24

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

Answer 25

No... not trig substitution

Answer 26

jst substitute x^2 = u

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

ah yes...@hartnn

Answer 33

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

Answer 34

then it'll just be integration of 1/u

Answer 35

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

Answer 36

Oh well, nope.

Answer 37

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

Answer 38

oops

Answer 39

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

Answer 40

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

Answer 41

Protect me from Calculus, O Heaven!

Answer 42

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

Answer 43

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

Answer 44

I need more practice. Ugh!

Answer 45

parth u learning limits also ??

Answer 46

Already learnt, yes.

Answer 47

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

Answer 48

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

Answer 49

*noted

Answer 50

....wow

Answer 51

... wow * 2

Answer 52

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

Answer 53

That's @TuringTest's intuition too!

Answer 54

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

Answer 55

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

Answer 56

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.

Answer 57

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

Answer 58

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.