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**absentinsomniac****Administrator**- Registered: 2012-06-09
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>> Covers shit day before test that's twice as hard as the basic concepts we've been covering for weeks.

What the fuck is this chain rule bullshit idek. I'll be kinda pissed if it' son this test.

The fuck can't I use a calculator to graph god damn 1/x^2 or like e^-x^2 wtf

Yeah what the fuck we didn't even go over the fucking chain rule or any of this derivative bullshit. ???? Better not be on test wtf.

Fucc

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**loon_attic****Banned**- Registered: 2012-06-08
- Posts: 10,272

don't die

the chain rule wasn't that hard wtf. but it's true professors tend to be pretty shit at teaching math idk. As are textbooks and online resources. :|

I swear, math could be taught so much better

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**absentinsomniac****Administrator**- Registered: 2012-06-09
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literally fuck learning how the math works when you can just do bullshit like this and not learn anything about the mechanics of it

[video]https://www.youtube.com/watch?v=apXayVtvCCc[/video]

Completely useless except for just getting by and passing, but fucked if I care at this point.

Fucc

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That looks pretty cumbersome. The derivatives for sin and cos are easy to remember and they're all you need to know. For secant and cosecant, you certainly learned the reciprocal rule by now (i.e. the derivative of 1/f is - f'/f^2), so you don't need to remember anything else (sec and csc are the reciprocals of sin and cos), and you can derive the derivative for tan and cotan by rewriting them as sin(x)/cos(x) and cos(x)/sin(x).

BTW, I can't remeber the last time I used the secant functions other than for some basic algebraic manipulation of some strange expressions in AC circuits. I was questioning my memory regarding their meaning, I had to look them up.

Studies show that learning exotic programming languages like Haskell, LISP/Scheme accelerates neckbeard & facial hair growth.

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**absentinsomniac****Administrator**- Registered: 2012-06-09
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I'm not sure if we learned the reciprocal rule or not by now lmao. I can't find any mention of it in the notes or anywhere. I don't remember ever hearing anything about it.

ugh. The way in the video seems a lot faster tbh, although not something I'd ever remember for more than the length of a test lol. Bleh. Like you have to remember random trig identities and shit sometimes to solve shit. I'm sure you're right and it'd be easier if I understood it though.

Edit: yeah the recipricol rule looks way easier, and I'm like 80% sure we did not cover that. What the fug.

There's so much fucking shit to memorize like fuck me. God damn derivatives and inverses and logarithms and blah blah.

Fucc

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**loon_attic****Banned**- Registered: 2012-06-08
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it was so easy in high school with a good teacher but I'm sure I'd do awful in university at this point, doing the same content

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**absentinsomniac****Administrator**- Registered: 2012-06-09
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Apparently a lot of kids come to Uni after taking calc in high school and fail in college. That's what my professor said anyway. It's not *too* hard, it's just a lot of work. I don't know.

I've been horrible at trig stuff for a while because there's so much to remember, and my memory is no good.

Fucc

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**loon_attic****Banned**- Registered: 2012-06-08
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I got, I think a B- in multivariable calc in uni [edit: not HS]. And I was doing fairly well, going to every class, sitting in the front like a good little boy, paying attention, taking notes, and doing the homework, and studying a bit. AND the professor was nice and lenient that time.

Fucking shit m8, fuckin smart kids or whatever

Then when I stopped giving a fuck I got an F in stats :)

*Last edited by loon_attic (2016-07-19 14:31:47)*

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**absentinsomniac****Administrator**- Registered: 2012-06-09
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Yeah it's mainly about effort. I mean, there are *some* very few people who can put in a fuckload of effort and still fail, but that's pretty rare. It's usually most of the kids in the class are smart enough, but they either 1.) didn't try hard enough or 2.) didn't have the algebra / precalc background and trying to learn all that while learning calc was too much.

I think, anyway. Very few people who can even make it to calc are simply incapable of passing it. If any. Precalc is pretty hard in it's own right if you aren't well up on all of the algebra.

Fucc

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**absentinsomniac****Administrator**- Registered: 2012-06-09
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Why the fuck does the derivative of some function set to zero find the critical points? Like, conceptually, a derivative is both the tangent line to a curve, and the instantaneous rate of change.

Instantaneous rate of change can be thought of like the speed limit at any given instant while driving. It's like, you're going 60 mph, that means at that exact instant you're traveling fast enough to travel 60 miles in one hour. If you kept that instantaneous rate of change over 1 hour, you'd have gone 60 miles. Your average rate of change would have been 60 mph, which is just distance per unit of time, aka distanced traveled / time of travel. Same concept with the derivative, it's called the instantaneous rate of change because it's literally just the slope of the tangent line, the tangent line being the closest approximation at that point calculated by Newton's difference quotient. It's often called the instantaneous rate of change because if the function (containing a curve) is modeling something with the variable of time, the derivative would be the rate at some given point where you calculated it per whatever the other variable is. For example if you had a curve modeling acceleration, the speed could be found at any given point along the distance by finding the slope of the tangent line and using a limit to make that small enough to be a good approximation, aka the derivative.

So with that in mind, I can't conceptualize why the fuck setting some derivative of a function equal to zero somehow automatically gives you the critical points of the original function? Like, if we have a graph of acceleration x^2 or whatever, the derivative of that function... Wait.

OH FUCK. So the derivative of a function isn't the instantaneous rate of change or the slope of the tangent line at some point, it's actually a function that you can plug any point on the original graph into and get the slope of the tangent line at that point. I mean, it *is* the ratio of the slope expressed via variables, but it's not the actual slope at any given point. So the derivative of some function is just a function "containing" all the tangent lines, waiting for a point to be plugged in. That doesn't even make fucking sense, because if we have like, y = x^2, the derivative is 2x, 2(any point in original function) doesn't give you the fucking slope of the tangent line, does it? It fucking must... Let's see:

f(x) = x^2

f'(x) = 2x

Derivative of x^2 at x=3 is 6.

2(3) = 6.

What the fuck, it does lol. w0w.

https://en.wikipedia.org/wiki/Derivativ … a_function

Ok, so when you set the derivative equal to zero, what it's doing is finding all the places where the slope of the tangent line is zero, which is where the curve levels off. So you can find critical points because it's finding places where the curve levels off and thus has to be turning. Woe. Yeah that makes sense. Fucking math, man. Fucking math. And that's why you have to check the endpoints, because they may never "level off" to produce a tangent slope of zero.

Fucc

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**V.R.****receive {_, _} -> void.**- Registered: 2013-04-02
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You're still doing the Calc I basics? So am I, but I'll have to get to much more.

"Humanity Is Overrated" - Shrek

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**absentinsomniac****Administrator**- Registered: 2012-06-09
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Yes. I'll also probably do calc 2 and linear algebra or whatever, plus I already passed discrete math and Stats 1 and 2. Plus business applications of calc in community college plus micro and macro econ. Oh and I have to take a course that involves computational logic soon.

Plus no I'm not "just now" learning the basics as outlined in these questions I'm just now clarifying and understanding some of the things I learned years ago. I'm in analytical geometry and calc right now because my school fucking lied to me and said I would be fine with the business applications course, but hey whatever.

Fucc

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**TheWake****Illuminatus Sacerdos**- From: Yankee-Occupied South
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**absentinsomniac****Administrator**- Registered: 2012-06-09
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Implicit differentiation forces you to recognize that we operate under an abstraction in Algebra, y is actually abstracting f(x). Y is a dependant variable, x is an independent variable. Aka y is a function of x. It's super easy to forget this because most problems are explicitly written as y= x. Implicit differentiation forces you to remember y is actually a function and this is reflected in the derivative.

Edit: holy shit I completely forget everything about inverse trig functions oh shit

Fucc

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**V.R.****receive {_, _} -> void.**- Registered: 2013-04-02
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TheWake wrote:

I'll probably post in this thread more when I start hitting math econ hard.

Utility and production theory give an intuitive application of calculus, I agree.

"Humanity Is Overrated" - Shrek

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**absentinsomniac****Administrator**- Registered: 2012-06-09
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http://tutorial.math.lamar.edu/

Probably my favorite resource right now. Has everything from like Algebra through Calc III and differential equations.

Fucc

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**absentinsomniac****Administrator**- Registered: 2012-06-09
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There is a distinction to be made between "antiderivative" (also known as "indefinite integral" aka the "primative") and a related but different concept the "definite integral". If you read this thread, you'll see that I discussed the difference between the derivative of a function and the instantaneous rate of change at some point. The derivative of a function encompasses *all of the possible instantaneous rates of change on a curve / function*, and the instantaneous rate of change **at a point** is one specific point plugged into that derivative function. Now, the "antiderivative" or "indefinite integral" is similarly a function. It is the original function from which the derivative came from. That's why you have to tack a "+ c" on the end of antiderivatives, the derivative of a constant is zero, so for any given derivative function it's possible that any constant was tacked on and derived away. What we are really doing is defining a whole family of possible functions when we give a antiderivative.

Now, the terminology is a mess. The "antiderivative" is sometimes referred to as the "integral", and so is the "definite integral", but you will run into a lot of problems if you don't think of these two concepts as related, but different. The "definite integral" is defined as the signed area under a curve or under the bounds of a function on a graph. Signed meaning it's either positive or negative depending depending on weather there's more area in the positive region or the negative. The two subtract out. See this picture:

The definite integral comes out to be an actual number, it is not a function. The number comes from the LIMIT of a integral sum over some *specific domain* [a, b] bounded on the range by the line. It's on a closed interval. As you can see in the picture of above, it's the *area* under the line and above the line on a specific domain.

Now, the fundamental theorem of calculus "links" the definite integral and the antiderivatives. All it really does is show that using the antiderivative of the function we're integrating in a specific way allows us to find the value of the definite integral e.g. the area under the curve. It does not say that the antiderivative and the definite integral are the same thing. It just shows there's a way to use the antiderivative and the definite integral to solve a definite integral problem without using Rieman's sums or adding a lot of stuff up. I won't get into the mechanics of how this happens here (don't have time), but it do. The second part of the theorem also shows that the derivative of the primitive in an indefinite definite integral that is unbounded (e.g. has an x in the top as a bound) is equal to the original function, which is helpful for some reason in higher math.

Fucc

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**V.R.****receive {_, _} -> void.**- Registered: 2013-04-02
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Surprisingly, I didn't see much of any integration in my entry exam examples even though there's just about everything else, including probability theory.

"Humanity Is Overrated" - Shrek

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**absentinsomniac****Administrator**- Registered: 2012-06-09
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I do not understand why, for problems like e^(2x) or 5^(3x), the exponent is considered the "inner function", whereas with problems like (x^2 + 4)^3, the inner function is considered x^2+4 and the outer function is the exponent. Is it because for (x^2 + 4)^3 the exponent is applied to the inner function whereas e^(2x) is interpreted as (e^(2x))? But that's the same thing as ((x^2 + 4)^3)... Then I thought it was because you evaluate e^(2x) by evaluating (2x) first and then e^u, except that's pretty much the opposite. That would make it the "outer" function as it's being evaluated first.

I know when you do the derivative, you evaluate the entire function d/dx (e^(2x)) as equal to (e^(2x) * d/dx(2x), but I'm pretty sure it's just the fact that the method breaks the fuck down and they kind of just were like "Ok well for a constant to a power we're just going to call the exponent times a variable the "inner" function even though there's no reason to actually call it that, because it's convenient". That "convenience" has been confusing me.

I mean, once you get down into the rational, it's because when we do d/dx (2^x) we're actually automatically doing d/dx (e^(ln(2))^x), because it's the only way afaik to get it to work. If you think of 2^x as (e^(ln(2)))^x (because e^(ln(2)) = 2 so they're equivalent so you can do this legally without changing the value), then the actual operation is clearly a composition where x is the outer function and e^(ln(2)) is the inner function. If you go through the derivative of (e^ln(2))^x (by knowing that d/dx(e^x) = e^x) which ends up as e^(ln(2)x) * ln(2) then you end up with what you'd expect based on how we're taught to do it, which is 2^x * ln2...

So it kind of makes sense, but when you look at it just as 2^(x) it really looks like you should evaluate it just like you'd evaluate (x^2 + 4)^3. That means the "rules" of calculus are fucking broken in this instance. Even when we have an obvious composition like (e^ln(2))^x, you still don't just get the derivative by normal rules like x(e^(ln(2))^1 * d/dx(e^(ln(2)), because then you'd end up with fucking 0. You have no choice but to, at some point, recognize the standard rules break down unless you fork from the rules and use facts like d/dx(e^x) = e^x and x^(2x) = e^(ln(2)) or whatever.

( e^(lnx) = x and ln(e^x) = x (they're inverse))

Fucc

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**loon_attic****Banned**- Registered: 2012-06-08
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maybe the inner function is always the one that has x or whatever variable you're deriving with respect to>??!0iewnf

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**absentinsomniac****Administrator**- Registered: 2012-06-09
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I can't think of a counterexample for that, but I've never seen "inner function" defined that way. That would probably have made my life easier.

Of course, there's shit like d/dx sine(x)^(3x). The only way to do this afaik is to raise it to e^(ln(sine(x)^(3x)) and do some fiddling with the log power rule or whatever. If you just have (sine(x))^(3x) the obvious thing to do is use the power rule for derivatives and try 3x (sine(x))^(3x)-1 * cosine(x) but of course that's not right because the power rule is only defined for the power of a constant or whatever. So the "power rule" is pretty limited in scope. I guess "inner function" and "outer function" are kind of just useful things to teach people when you're beginning or something, because they don't seem very rigorous at all. I'm pretty bad at leibniz notation ugh.

I mean it's all composite functions, so it *should* be easy to identify which function is inside of the other function. e^(3x) is a composite of 3^u and u which = 3x, which in and of itself is confusing because how tf is e^u it's own function if it's literally just a constant to the power of a function? It doesn't seem like it should even be defined in the context of compositions. Like I said it's only defined that way because behind the abstraction you are expected to automatically do e^log shit. But whatever. It's just a lot of really bad jargon I guess, it all makes sense behind the scenes, but the way they teach it is fucking loaded with shitty abstractions that have a bunch of rules you have to remember so you don't go off course. It's way too easy to forget some obscure limitation of some rule or apply a rule out of it's proven context because some teacher a while ago said the "inner function" is the one that is evaluated last or something. For e^(2x) 2x is probably going to be evaluated first, so that's just a shitty way of defining the process imo.

Fucc

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**Princess-Retard****Aspiring-Camwhore**- Registered: 2016-05-09
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Could you do a little calcul for me, please?

What is the exact level of virginity in this thread? Thank you!

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**absentinsomniac****Administrator**- Registered: 2012-06-09
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2/5 or if we include you && assuming you've been stabbed then 2/6. So converting that to a "level" with percentages it's about 33 percent virgin itt.

But that's arithmetic not calculus.

Fucc

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I'm away for like two months and people start messing around with Calculus. Crap.

So in case anyone else has this curiousity:

absentinsomniac wrote:

I do not understand why, for problems like e^(2x) or 5^(3x), the exponent is considered the "inner function", whereas with problems like (x^2 + 4)^3, the inner function is considered x^2+4 and the outer function is the exponent. Is it because for (x^2 + 4)^3 the exponent is applied to the inner function whereas e^(2x) is interpreted as (e^(2x))?

It may very well be that I learned this and have long forgotten it, but I cannot for the life of me remember any instance of this whole inner function and outer function thing. It may be due to the different curricula. We probably called it by other names.

There's a decent-ish explanation here: http://oregonstate.edu/instruct/mth251/ … nRule.html . This isn't really a problem in Calculus per se.

I do use the chain rule given at the end of the article (and certainly use composition and decomposition), which suggests that I probably learned this stuff, but fuck me if I remember "inner" and "outer" or the pointless drilling related to it.

The "inner" function, of course, is always the one that has x in it (oh, but see below), because it's the g in f(g(x)). F is a function of the "expression" g(x). So, for instance, e^(2x) is the function f(u) = e^u applied to g(x) = 2x (which suggests that 2x is the "inner" function). (x^2 + 3)^2 is the function f(u) = u^2 applied to the function g(x) = x^2 + 3, which suggests that x^2 + 3 is the "inner" function (i.e. it's the one "in" f0.

As for this one:

Of course, there's shit like d/dx sine(x)^(3x).

You were right about the track you chose, see http://www.analyzemath.com/calculus/Dif … ative.html for related example. It's a rather pointless exercise in symbolic manipulation. The good news is that functions like sin(x)^(3x) (or anything of the form f(x)^g(x)) don't really occur in real-life processes. I don't think I've ever had to do something like that in more than a decade now. The technique itself a useful trick sometimes, but nothing else.

Don't sweat *too* hard. Especially if you want to program computers for a living, you won't have to take the derivative of anything more complex than a polynomial, simple sin/cos and/or power functions for a long, long time, and when you *do* have to take a more complex one, it will be perfectly acceptable to ask a real mathematician.

___

THE NOTE BELOW: After you'll be done with high school, where everything is dumbed down in the form of an endless lesson in taxonomy, terminology, fancy names and gazillion of rules, you'll get to university-level Calculus, where everything suddenly gets simpler for some reason, despite the actual subject being a lot more complex. One of the fancy things that you'll learn is how to take the derivative of functions that depend of *several* variables (e.g. 2 variables, which describe not a curve, but a surface). So you shouldn't take the "has x in it" as anything more than a rule of thumb.

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**absentinsomniac****Administrator**- Registered: 2012-06-09
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I'm a senior at uni lmao

Fucc

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