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843
844

I choose you Calculus! (i.imgur.com)

submitted ago by jalvarado

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[–]braggart1 42 points43 points ago

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appearded?

[–]experts_never_lie 17 points18 points ago

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They don't seem to understand proper mathematical notation either.

[–]septimoamante 6 points7 points ago

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exactly. I wondered if it was only me that was bothered by the notation.

[–]souleh 5 points6 points ago

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And now a bearded appearded

[–]DrHenryPym 0 points1 point ago

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This comment deserves more upvotes.

[–][deleted] ago

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[deleted]

[–]braggart1 1 point2 points ago

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well when calculus has its own language in which to communicate your comment will become valid.

[–]tty2 106 points107 points ago

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Not to mention, d/dx isn't a quantity being multiplied. It's a linear operator being applied to the function, so it goes in front.

[–]ChaosCon 51 points52 points ago

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It's not wrong, it's just postfix notation!

[–]septimoamante 7 points8 points ago

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but.. no.

[–]dustinechos 0 points1 point ago

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What did the english major say to his girlfriend? Oh wait, I'm thinking of juxtaposition.

[–]spiraldroid 2 points3 points ago

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I was thinking tits.

[–]aqualis 7 points8 points ago

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I twitched when I saw that. I believe I've been nerdsniped.

[–][deleted] ago

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[deleted]

[–]tty2 7 points8 points ago

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Except for when you're taking the derivative with respect to a different variable.

[–]schobel94 0 points1 point ago

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I'm glad I wasn't the only one who noticed this.

[–]eyelessfade -2 points-1 points ago

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f(x)'

[–]alexgeek 11 points12 points ago

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*f'(x)

[–]douglasr007 2 points3 points ago

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f'

(it's just an alternative method, the above one posted is still correct as well)

[–]citizenthom 27 points28 points ago

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Use the natural log!!

[–]deadpansnarker 4 points5 points ago

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It's super effective!

[–]DAVENP0RT 10 points11 points ago

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Calculus courses should go and ahead and say up front, "Hi, my name is Natural Logarithm and I'm going to be fucking your brain this semester."

Seriously, if anyone here is just getting into natural logs in school, don't miss a single day.

[–]eserikto 7 points8 points ago

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I'm not criticizing, but curious: What was hard to understand about natural log? Did your precalc classes cover exponents and logarithms?

[–]DAVENP0RT 5 points6 points ago

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I missed the day where e was introduced, that's why I said don't miss a single day. It was a bit difficult trying to understand (ex )' = ex from a textbook.

[–]gwot 1 point2 points ago

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I think its simple to say that by definition e makes this so.

[–]r_u_srs_srsly 0 points1 point ago

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negative, (ex )' = ex *x'

[–]DAVENP0RT 3 points4 points ago

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My point exactly.

But really, x' = 1 * x0 = 1, so saying (ex )' = ex * x' = ex is technically accurate, right?

[–]r_u_srs_srsly 2 points3 points ago*

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no one said x MUST be a constant ...say x represents sin(x)

(ex )' = ex * x' = (esin(x) )cos(x)

then it is important to make sure you dont forget it.

edit: added example

[–]DAVENP0RT 3 points4 points ago

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Ah, entirely correct. Bringing back memories of all the stupid mistakes I made on tests...

[–]javajunkie314 1 point2 points ago

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Well, if ()' is taken to mean d/dx, then x' = dx/dx := 1. That said, friends don't let friends use implicit notations in change-of-variable problems.

[–]eserikto -5 points-4 points ago

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I'm not criticizing, but curious: What was hard to understand about natural log? Did your precalc classes cover exponents and logarithms?

[–]nthgthdgdcrtdtrk 1 point2 points ago

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I have one of those! I just dropped one in the toilet!

[–]christopsy666 55 points56 points ago*

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Using d/dy Operator. It is very effective.

edit: for the people getting anal: i assumed x is no implicit function of y and d/dy is the total derivative, which doesn't matter because it would still be true for the partial derivative (math grad here)

also: δ/δy does not stand for a partial derivative, δy is a variation of a function, a complete different approach used in finding extremal curves in functionals

[–]essex23 10 points11 points ago*

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Today I learned that apparently many people don't understand calculus. (edit: in regards to the other responses)

if f = ex then df/dy =0 since there is no variable y. It would be non-standard notation to has x(y) without explicitly stating it. Also don't tell me that partials are correct here. You can easily have functions whose total derivative is equal to its partial.

[–]akunin 7 points8 points ago

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Isn't that exactly why it's super effective?

[–]essex23 -1 points0 points ago

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Yeah. That's the joke. A standard first year calculus joke. However read some of the other responses trying to explain this joke. Most of them are just horribly wrong and display a lack of understanding of basic calculus.

[–]ccampo 2 points3 points ago

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What you say is true IF the variable y is defined to not exist OR y is independent of x. Instead in this case, if f = ex, then df/dy = ∂f/∂x × dx/dy = ex × dx/dy where x can be a function of y, x = x(y), because of the ambiguity.

Chain rule brah.

[–]Manhattan0532 0 points1 point ago

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If my recently aquired math skills don't betray me this makes it even worse, since you're now dealing with an implicit differential: d/dy * ex = ex * dx/dy

[–]ryegye24 17 points18 points ago

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Your new found skills kind of do fail you. ex is a constant when differentiated by d/dy. Imagine that instead of ex it's eC, and instead of d/dy it's d/dx. Then you have d/dx of eC. It is literally the exact same situation, only changing the names of the variables. Now it's much more obvious how ex is a constant in this situation, meaning d/dy of ex = 0.

[–]Timmmmbob 9 points10 points ago

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I'm pretty sure he is right, and you are only maybe right. If x is constant, dx/dy=0, but what if x=f(y)? His solution still works. Yours doesn't.

Edit: yeah he is definitely right. Look up the chain rule.

[–]ryegye24 8 points9 points ago

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That's assuming that x = f(y), which we have no reason to believe is true. Nowhere is x defined to be in any way dependent on the value of y, and there is no reason to assume that the value of x is dependent of the value of y. They are just variable names. If it was written eC, where C is some constant then it becomes obvious why deriving it by y is zero. Also, notice I never said dx/dy, I only said d/dy.

[–]Timmmmbob 1 point2 points ago

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Y is a function of x in the general case, even if f(x)=C. We haven't ben told the relationship between x and y, so we can't just assume they are independent like you are.

[–]jorapello 4 points5 points ago

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Yep... taking the derivative with respect to a variable that isn't present in the operand gives you 0.

d/dy(y2) = 2y

d/dy(x2) = 0

d/dx(ex) = ex

d/dy(ex) = 0

[–]necroforest 4 points5 points ago

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You're thinking of partial derivatives.

[–][deleted] 2 points3 points ago

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You are correct. Formally, jorapello is incorrect. But if ex = f(q) or any variable (besides y), then the operations amount to the same thing.

[–]boinGfliP14 1 point2 points ago

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Exactly! d/dy (ex) would only be zero if it was a partial deriv! instead we use the chain rule! I've been outa class for awhile so forgive me... d/dy (ex) = ex dx/dy dy/dx

[–]christopsy666 1 point2 points ago

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d/dy (ex) is also zero if d/dy is the total derivative assuming x is no implicit function of y

[–][deleted] 1 point2 points ago

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You are correct. Formally, jorapello is incorrect. But if ex = f(q) or any variable (besides y), then the operations amount to the same thing.

[–]jorapello 0 points1 point ago

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Thanks for the corrections. :)

[–][deleted] ago

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[deleted]

[–]jorapello 1 point2 points ago

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Thanks for the corrections.

[–]Tulki 0 points1 point ago

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Yet again assuming y is a function of x which was never stated.

[–][deleted] ago

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[deleted]

[–]NYKevin 3 points4 points ago

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Y is not mentioned in the original at all. Thus when it was introduced in the comments, it was a free variable unless it was explicitly defined otherwise.

[–]christopsy666 0 points1 point ago

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You got to be kidding me. You assume that it is a implicit function of y even it is not stated. Could be implicit function of t or z also and those could be implicit functions of y as well. Thats not how it works. Independent variables have to be classified at least once, preferably in the beginning.

[–]feureau 0 points1 point ago

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You lost me by the second sentence.... :(

[–]TBMonkey 5 points6 points ago

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It's actually magic; he's a witch.

[–]aarongoodermuth 2 points3 points ago

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he turned me into a newt

[–]CorporateGiant 1 point2 points ago

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A newt?

[–]aarongoodermuth 2 points3 points ago

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i got better

[–]ryegye24 2 points3 points ago

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Ok, are you aware of the fact that if C is a constant then d/dx(C)=0, right (the derivative in terms of x of C equals zero)? (If not, then I won't be able to explain this to you at all). This means that when you derive a constant, "C", by x the derivative is 0 (because there is no slope). Now, "C" and "x" are just variable names, there isn't anything special about them, and "C" can be any constant. So we can say " C = ex ", where C, e, and x are all constants. Now, we can say d/dy (ex) = 0, or in english, the derivative in terms of y of e to the x is equal to zero. This is because there is no 'y' in the equation ex, so all of those variables are constants. No matter what the value of y is, ex will always have the same value. Therefore we can treat it as a constant.

[–]feureau 1 point2 points ago

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Oh, now I get it! Thanks!

[–]Mattieohya -1 points0 points ago

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You are thinking it is d/dy ex(y)

x isn't a function of y so you treat it as a constant.

[–][deleted] ago

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[deleted]

[–]antiproton 2 points3 points ago*

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You're being unnecessarily pedantic. Simply stating 'x' as a variable implies it is independent of all other variables precisely to avoid confusion that you're forcing upon this discussion. There's no reason to say "x (where x may or may not depend on y)" when you can simply say "f(y)".

Furthermore,

First, if y = ex then d/dy of ex = dy/dy = 1, not 0.

is just plain incorrect. You are conflating a function and a variable. Setting y(x) = ex =/=> d/dy[y(x)] = 1. You can substitute y for f(x) and then take d/dy[f(x)] which does equal 0.

In short, you have to assume many things in mathematics. In this case, assuming x is independent of y is totally reasonable becuase it's convention. Your entire argument is not technically incorrect, but it provides no insights and does not invalidate ryegye24's comment at all.

[–]Zantier 1 point2 points ago

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I'm fairly certain that dy/dy would equal 1, whether y is a function of x or not. When you consider that d/dy represents the rate of change of the expression with respect to that variable, it wouldn't make sense to say that y isn't changing as y changes.

Thinking about functions and free variables is confusing the hell out of me right now... I need a lie down.

[–]antiproton 1 point2 points ago

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It's important to remember that you take derivatives of functions, not equations. The statement y=ex in the context we're discussing, is an assignment of a function.

One of the definitions of a derivative is 'the measure of how a function changes as its input changes'. In this case, you cannot say that y is an input variable but simply setting y = ex and then taking the derivative with respect to y.

[–]Zantier 0 points1 point ago

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I see what you're getting at. Is d/dy a valid operator when y is a function? I need to go learn some maths, brb.

[–]antiproton 0 points1 point ago

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Sure. You can say f(y) = y and then take d/dy[f(y)] = d/dy[y] = 1. The entire problem with the above conversation is people are using y and y(x) interchangably, when they actually mean different things. That's why we usually say f(x) instead of y(x)... to avoid this kind of confusion.

[–]ryegye24 0 points1 point ago*

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You are assuming that y has been set to y = ex. We have no reason to believe this. Everything after that is based on this unverified assumption. Stealing from jorapello's comment:

Yep... taking the derivative with respect to a variable that isn't present in the operand gives you 0.

d/dy(y2 ) = 2y

d/dy(x2 ) = 0

d/dx(ex ) = ex

d/dy(ex ) = 0

And this part

(d/dy)ex = (d/dx)ex dx/dy = ex dx/dy

Is just plain wrong.

[–][deleted] ago

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[deleted]

[–]BCSteve 1 point2 points ago

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Your argument rests on y being a function of x.

While in the course of a formal proof, it might be necessary to explicitly declare that x and y are independent variables, I think in the context of a leisurely internet discussion it is fair to assume that they are independent. If the intention at the outset had been to discuss the cases of y being a function of x, it would have initially been declared as y(x), which would then cover all cases where y is dependent and independent.

So while you might be right on a technicality, ryegye24's original comment is valid. It's nitpicking.

[–]christopsy666 1 point2 points ago

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who says you are dealing with an implicit function? there is no indication.

[–][deleted] 1 point2 points ago

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Ignore the naysayers, you're totally correct: textbook application of the chain rule. There's no reason to assume x is independent of y, and if it is than dx/dy = 0 and you're still right!

[–]Quantris 0 points1 point ago

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Please don't listen to ryegye24; as some others have pointed out your math is exactly right. What christopsy666 was suggesting (I think) was that by making the conventional choice that 'y' is equal to whichever function of 'x' we're talking about (conventional because this is what you do when you graph a function), then d/dy ex = d/dy y = 1 (or: x = log y => dx/dy = 1/y = 1/ex; which you can sub into your expression)

Alternatively, without any other information, one may argue that x and y should be considered as independent variables (e.g. dx/dy = 0; or a change in y doesn't correspond with a change in x). This seems to be an unstated assumption in ryegye24's response.

[–]MBAHelper -4 points-3 points ago*

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Let f(x,y) = ex

Then (δ/δy) f(x,y) = (δ/δy) ex = 0

Note that f(x,y) is constant with respect to y.

EDIT: Sorry, I meant 0, not 1.

[–]Dembrogogue 2 points3 points ago

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You also meant ∂/∂y, not δ/δy.

[–]MBAHelper 1 point2 points ago

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Yes, thank you. I admit, I was too lazy to figure out how to type the "der" symbol, so I just copied the lower case delta from another comment. Maybe I should stop half-assing things.

[–]GarbageDemon 4 points5 points ago

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(δ/δy) ex = 0

[–]MBAHelper 1 point2 points ago

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Thank you for catching my error.

[–]Manhattan0532 1 point2 points ago

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I can't yet read this notation, so I'm not sure wether your correcting me or restating my equation.

[–]jalvarado[S] 1 point2 points ago

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It's the partial derivative with respect to y. He's kind of teaching you something new.

[–]Manhattan0532 2 points3 points ago*

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Maybe I'll get it some day.

Edit: On second thought, I think I know what he's doing. But he's assuming we're dealing with a plane, while I was working with a curve.

[–]MBAHelper 1 point2 points ago*

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Draw the curve ex and visualize extruding it off of the paper so that it looks like a quarter pipe ramp. Going up and down the ramp would indicate changes in the x direction. The ramp has a slope in the x direction. If you place a skate board in the x direction, the skateboard will roll down the slope.

Going side to side on the ramp would indicate changes in the y direction. The ramp has no slope in the y direction. If you place a skate board in the y direction, the skateboard will not roll anywhere.

We also use the same idea while skiing or snowboarding. If you want to stop sliding down the mountain, stand in the y direction, not the x direction.

[–]MBAHelper 0 points1 point ago

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Sorry, I was just writing out the partial derivative.

f(x,y) means a function of two variables. Obviously, our function ex has only 1 variable (x), but if we're going to claim a derivative with respect to y, we have to change our function notation from f(x) to f(x,y).

δ/δy is very similar to d/dy: d/dy is when you only have 1 variable, δ/δy is when you may have more than one variable. d/dy is called the "ordinary derivative", δ/δy is called the "partial derivative".

[–]chowriit 2 points3 points ago

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It wouldn't do anything. δ/δy, on the other hand...

[–]Dembrogogue 10 points11 points ago

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You mean ∂/∂y?

δ is the Greek letter "delta" meaning "a small quantity". It's not the partial differential symbol.

[–]ccampo 1 point2 points ago

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He could mean the variational derivative.

[–]chowriit 1 point2 points ago

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Yes, yes I did. Serve me right for being lazy when flicking through the character map.

[–]notGivingAwayinfo 19 points20 points ago

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3 and ex are walking down the street in math land shooting the shit, nothing special. Then 3 sees a differential operator walking towards them and screams bloody murder, and runs the fuck away. The cocky ex calls 3 a pussy and goes up to the differential operator. "sup bitch, I am ex" The operator replies, " And I am a partial differential operator with respect to y mother fucker"

Lesson learned: Pay close attention when in math land?

[–]christopsy666 2 points3 points ago

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The lesson here should be: Be careful on what a function does depend on and to what respect you are differentiating.

[–][deleted] 1 point2 points ago

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I laughed at this way more than I should.

Have an upvote, kind sir.

[–]zengenesis 4 points5 points ago

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It took me a while when I first took partial derivatives to realize that this wasn't actually a Greek symbol but a made up symbol.

[–]indexspartan 4 points5 points ago

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You use integrate, It's not very effective

[–]dik-dik 12 points13 points ago

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Don't forget about the + C

[–]indexspartan 1 point2 points ago

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i always seem to forget

[–]Elquinis 6 points7 points ago

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I would send out Natural Log!

[–]espotoaster 11 points12 points ago

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http://i.imgur.com/RTWc7.jpg

[–]Elquinis 1 point2 points ago

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Now! Quick, use divide by zero!

[–][deleted] 9 points10 points ago

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http://i.imgur.com/u3Khp.png

[–]Elquinis 1 point2 points ago

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You can't fool me, that's just the Korean version.

[–]gigitrix 0 points1 point ago

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I̭̜̹͎̠̥̱ͩ͊͑̏͛͝ͅt͚͂ͦ̎̊'̢̛̩͓̍̉͋̆̒̎͟s̩̱͔̼͗ͥ͂ͮ̈̒̇̑ ̧̒̌̿ͩ̑ͬ́͜͏̯̝̬̥̺̯̺ṉ̖͙̽͘ỏ̵̶̟̠̤̝̎͠t̊͟҉̤̗̝ ̶ͭͪ͂̎͏̧̹̼͙̣̩̪v͔̺̥͖̮̼͙͆̎͗ͨͤ͊ͬͩ̾͜è̘͇͍͓̫̭̹ͅr̶̙̟̬̳̻̙̥͕͍̉ͤy̠̬̯ͤͧ͆̽ͯ̕ ̸̽̔ͬ͛ͩ͋҉̵̝̼̤̺̠e̝̟̓ͭ͡f̢͓̪͎͓̣̬͕̎̀̆̃ͅf̹̪̫̗͚̠̂͛ͨ̃ͭ͗̽̃ͅe̮̟̗̬̤̟̼̯̤̋̎ͫ̓̕͟c̬͉̮̗̰̪̰ͩ̐̓͊̿ͭ͆̿͠t̆ͪ͒ͫ͂́̑͛҉̶̵͇̠͎͓̫̳̭̘ͅi͗̉͏̥̟̖͎̜v̶̺̀ͥ̏̅̋ͣ͊͜e͍̬̺̣͔̍̈́̚

[–]jimjimgreen 0 points1 point ago

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A wild Cthulhu appears. He's gonna drag your algebraic ass to Azathoth.

[–]Pants536 1 point2 points ago

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You use Self Destruct!

[–][deleted] 3 points4 points ago

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my last week of Cal 2 right now :) Final this thursday

[–]TheTopStrap 2 points3 points ago

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Don't forget... +C!

[–]gwot 1 point2 points ago

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not for differentiation ಠ_ಠ

[–]nome0009 1 point2 points ago

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Ah no way! Me too! Good luck :)

[–]I_Wont_Draw_That 3 points4 points ago

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My biggest problem with this: you don't use moves. Your pokemon uses moves.

[–]AbusingVitaminK 3 points4 points ago

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I don't understand math, so upvote for 'appearded.'

[–][deleted] 2 points3 points ago

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The derivative of ex is, actually, ex. According to my high-school Calculus textbook, it was "one of the most integral properties of derivatives".

[–]neerg 6 points7 points ago

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Hah, a math textbook made a pun!

[–]AbusingVitaminK 1 point2 points ago

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Oh! Alright, now that makes a lot more sense, haha. I never took calculus in high school, and I don't need it for my major... so I never really learned this, lol.

[–]mko1991 2 points3 points ago

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how about just multiply by zero?

[–]dydxexisex 2 points3 points ago

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OMG That's me!!!!!

[–]neerg 2 points3 points ago

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"Dyd xex I sex", you ask?

[–]zengenesis 2 points3 points ago

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haha, wow a calculus debate. Reddit just became amazing in my eyes.

[–]lycergus 2 points3 points ago

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I lol'd. What level does Derivitine learn that move, anyway?

[–]SpeakMouthWords 7 points8 points ago

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OPERATORS DO NOT WORK THAT WAY, GOODNIGHT.

[–]GRX13 1 point2 points ago

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Take the limit as x -> -∞ !

[–]element420 1 point2 points ago

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Idempotency, fuck yeah

[–]bluecalm 1 point2 points ago

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This is great!

[–]fxpstclvrst 1 point2 points ago

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Now I feel sad about how poorly I did in precal in high school all over again.

[–]FloppyMcPrplHat 1 point2 points ago

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Mathematical!

[–]shamecamel 1 point2 points ago

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I can guarantee that if this was tweaked into an actual video game as a sort of time-attack against math, it'd be huge. We all know how awesome Math Munchers was back in the day, give it some sparkle and release it on DS and it'd be amazing. We can't let Professor Layton have all the glory.

[–]polishedbullet 1 point2 points ago

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We must do what any other intellectual would do... turns on Dino Puzzle

[–]BeefPieSoup 1 point2 points ago

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Next time, choose Natural Logarithm

[–]GregOttawa 1 point2 points ago

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It's only natural.

[–]FluffyPillow 1 point2 points ago

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fuck this shit, fuck summer school.. im so blown

[–]thefloppydog 1 point2 points ago

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Use integrate! .... wait, no, that doesn't effect it either. Might just have to substitute in natural log.

[–]kungffffuuuuu 1 point2 points ago

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Did anyone ever look at the back of your "pokemon trainer" and constantly see mario wearing a backpack?....

[–]Imperial10 1 point2 points ago

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Too bad I don't understand math.

[–]Zimyver 1 point2 points ago

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Use your special attack: differentiate wrt y.

[–]intensenerd 1 point2 points ago

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Read the title as "I choose you Calculon!" Was very confused at the comic.

[–]r_u_srs_srsly 1 point2 points ago

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didn't bother reading comments but the answer is really (ex ) (x')
dont forget the x'

[–]flinxsl 1 point2 points ago

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quick, try integrate over all real numbers!

[–]kobel4k3r5 1 point2 points ago

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shouldn't it be that it has no effect... too much math, not enough pokemon for you.

[–]gunitsniper2700 1 point2 points ago

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im doing this shit right now. can someone explain to me when to use it?

[–]stereosanctity 2 points3 points ago

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i'll bet the person that created this image gets all KINDS of pussy.

[–]JonAmazon 2 points3 points ago

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Operator after the operand... twitch

[–]prsndudethng 5 points6 points ago

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it would be better if it was "it doesnt effect ex"

[–]gwot 2 points3 points ago

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but it does!

[–]zelani 1 point2 points ago

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I don't know anything about calculus, and this still made me laugh.

[–]magister0 4 points5 points ago

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Then you laugh too easily

[–][deleted] ago

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[deleted]

[–]ForTheLazyOnes 1 point2 points ago

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Keep this stuff on youtube ಠ_ಠ

[–]advice_is_for_pusies 2 points3 points ago

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i like it. I'm also a degree holder in mathematics and pursuing a masters in math. it's nice

[–]jalvarado[S] 0 points1 point ago

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Awesome. I'm a math lover myself. Although just starting college, I've delved into the marvel of Differential Equation and started grazing Multivariable stuff. It's pretty awesome. Computer Science major though.

[–]advice_is_for_pusies 1 point2 points ago

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glad you like it. study hard even when it seems easy. Also, take linear algebra. That has been my favorite class. It got me going into the algebraic side of mathematics versus the differential, number side. I love ideas of math, not just working with numbers.

Although conversely I'll be at UNO soon getting a degree in financial mathematics. Trying to see how far a 1430 GRE score will take me

[–]jnethery 0 points1 point ago

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I'm taking linear algebra in Fall semester! Going for a math degree, myself (been doing mech engineering up til now).

[–]Ali-Sama 0 points1 point ago

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I like it and i don't hold any math degrees;) i think it is funny to anyone who took calculus and remember it;) I got up to differential;)

[–]WiseHalmon 0 points1 point ago

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Guys, ... this is silly. Divide by infinity, multiply by 0, super effective. Or not, I don't really know.

troll face

[–]jnethery 0 points1 point ago

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You can't divide by infinity, as it isn't a number. Instead, you must take the limit of the function divided by an arbitrary variable as said variable approaches infinity.

[–]dino340 0 points1 point ago

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THESE COMMENTS MAKE MY BRAIN HURT!

[–]LongJohnMcVenturson 0 points1 point ago

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So glad that I will never have to know what any of this shit is.

[–]Petardfoozer -1 points0 points ago

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It's funny cus the derivative of ex is ex

[–][deleted] ago

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[deleted]

[–]Kintobor 14 points15 points ago

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  • C? On a differential? Really?

[–]Zantier 6 points7 points ago

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True, but that's only when you integrate.

[–]scb8mp 5 points6 points ago

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It's not an integral, there is no C

[–]snoochnooch 1 point2 points ago

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It's true. Probably where I've lost the most points in my calculus classes.

[–]antraxx -4 points-3 points ago

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The notation is horrible.

ex by itself has no point of reference. its just a term that makes no sense

x is not defined

if you want to be very anal about it you should do this

Let f: R -> R be a function defined as f(x) = e ^ x

Now you can differentiate using either Leibniz' notation where y = f(x), s.t. the derivative is

dy/dx = e ^ x

or using standard mathematical notation where the derivative is

f'(x) = e ^ x

[–]darmon -2 points-1 points ago

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This was so unfunny, I literally gasped when I saw it had 840 upvotes.

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