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Hello! This is a general Math question. Based on my past experience, am I ready for Calculus?

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@ParthKohli, without a doubt.
you are! haha-

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Hmm, thanks for flattering. I want more answers with a detail.
And I doubt if you guys know my past experience :P
Parth, I will tutor you on Calculus for free if you want a good introduction and help.
calculus is only an expansion of algebra, guess that includes your answer already.
@ParthKohli To learn calculus you should be well versed in Logarithms, Trigonometry, and Sets. Have you learned all these?
I felt Calc I was like 75% Algebra and 20% Trig...pretty sure it will be easy for someone like you
Thank you. Yes, @ash2326
Calculus is MOSTLY Algebra, just FYI. ;-)
Speaking from direct experience.
Then you are ready:D
Parth, do you have knowledge in Trig, Algebra I-II, and working with functions?
I went to a website and learned all these concepts very thoroughly. What is the starting in Calculus? Limits?
math is hard keke ayiyi
It depends on the school. At MIT, we started with Integrals first and defined limits based on that.
Calculus will come next year in my syllabus lol :P
You're ready, Parth. If you wouldn't mind taking a few minutes out of your time today, and send me what you know and give me some examples. I can take that information and place you at a level. :) And like I said, "I can tutor you."
And when I say that it is mostly, I mean that you'll REALLY need to be good with it. Everything from factoring, to trig identities, to partial fractions, to series and sequences, to things like this: |dw:1342968425801:dw|
Derivative is small change in \(y\) - known as \(dy\) - over the small change in \(x\) - known as \(dx\).
You see this stuff over and over in Calculus
@mathdumbo That's a nice way:D
Integral is a number of which something is a derivative of(as far as I know).
not a number but function.. :P
@ParthKohli You'll get a good start here
Average slope: \[m=\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\] Turns into... \[m=\frac{d y}{d x}=\text{instantaneous slope}\]
So, \[\int \limits {2xdx } = x^2\]?
Yes, @agentx5. I saw a video on Khanacademy on the intuition of differentiation.
And I believe that you are allowed to 'skip' some of the topics in Mathematics; that's how beautiful it is :)
Parth, how would you solve this: \[{2 - 5x \sqrt23}{-3x \sqrt2}\]
^ You need to have a background with working on radicals.
I believe that we should stick to the question, and yes I do know that :)
There is a lot of really small division in calculus.
I don't think we can simplify that.
And by small, I mean integrals.
It was just an example.
I have heard that we can calculate "nth" root of any number by differentiation. Am I correct ?
You can move the constant out front: \(\large\int \limits {2x \ \ dx } = 2\int x \ \ dx = 2[\frac{x^{1+1}}{1+1}] + C = \cancel{2} * \frac{x^2}{\cancel{2}} + C = x^2 + C\)
Parth, don't forget the C
Yep, because the derivative of any constant is 0, and \(x + 0 = x\).
The "C" means some constant. Why put it on indefinite integrals? Because \(\frac{d}{dx}\) constant = 0
You got it lol :-D
I am just starting it ;)
Thank you, all! You are a great help!
What concept should I start with?
Any resource?
parth,I learn it on MIT , and Khanacedemy
That is a phenomenal reference.
also , this awesome book
Hah! Paul's Notes! I find it a little difficult to start with the basics on that.
P.S. I learned my logs and trig on there. :)
Try E-book "Calculus for dummies"
That book help me a lot
Parth try this one! ^_^ \[\huge \lim_{x \rightarrow 0} \frac{1}{x-1}\] and \[\huge \lim_{x \rightarrow 1} \frac{1}{x-1}\] and \[\huge \lim_{x \rightarrow \infty} \frac{1}{x-1}\] Do it visually :-D (draw a graph with arrows & stuff)
Try it @ParthKohli :-D It's not a hard problem, it just makes you think and once you get the concepts down it just goes back to Algebra tricks for the most part. hint: What happens if the x = 1 to the denominator?
I find these as the answers: 1) \(-1\) 2) \(\infty \) 3) \(\infty\) You use L'Hopital's Rule.
You may do it visually by taking the numbers closer to the number which it is approaching.
Aww come on draw it! ^_^ Make a sketch
Hmm. There's a thing that I am missing: CONIC SECTIONS.
Would I be able to do Calculus without conic sections?
Incorrect @ParthKohli |dw:1342970148488:dw| Correct answers are: \(\huge \lim_{x \rightarrow 0} \frac{1}{x-1} = -1\) \(\huge \lim_{x \rightarrow 1} \frac{1}{x-1} = \text{Does NOT exist.}\) \(\huge \lim_{x \rightarrow \infty} \frac{1}{x-1} = 0\) You cannot use the l'Hospital's rule unless after simplification you still have a \(\large \frac{\pm\ \infty}{\pm\ \infty}\) or \(\large \frac{0}{0}\) form when you go to substitute.
This is why sketching/graphing things visually is so helpful in Calculus :-D
Hmm—you should have included that I am a beginner too :)
I had to get lunch, sry about the delay
Hey no worries, we all have to learn to fly sometime right? ^_^
Isn't 'Does NOT Exist' just infinity?
Actually, no, take a closer look. The limit is as if you're taking your right and left hand and tracing the function from both points towards whatever the limit is approaching. In the case of the first one, it's clearly approaching -1 from both the right and left. But in the second one, it goes on way up to infinity, and the other way down to negative infinity. It creates a kind of a paradox, it can't be both and yet it is. So the limit simply doesn't exist. Here's an example where the limit is infinity: \[\huge \lim_{x \rightarrow \infty} x^2 = \infty\] |dw:1342970876711:dw|
\(\huge \lim_{x \rightarrow 1} | \frac{1}{x-1} | = \infty\) |dw:1342971089127:dw|
^_^ Make more sense now?
I, for once, believe in visual techniques :)
Amen brotha! ;-P
Heh! That was so clear :')
And don't forget you can't just use l'Hostpital's rule whenever you feel like it, it's got to meet the prerequisites of the theorem that I mentioned above :-)

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