Can you do my 12th grade homework?

Oh wait this is 12th grade, so you haven't actually done any multi variable calculus right? Is this about implicit function form?

It's not a function, its an equation, you nigger.

If it's implicit functions try rewriting the expression on the form y=...., then do d/dx2 at the point x=3

yes. It's implicit derivation. Maybe there is something I don't know yet that solves this.
I didn't write the problem. Blame the idiots who made the website.
ill give it a try

61/10

1.225

i skipped it. fehk

here is the next one. should be easier

>tfw highschool dropout
cant help you, friend.

not going to college is one thing. Not finishing high school is just terrible. Get a ged. easy af.

-22/9

d/dx * both sides
d/dx(y^2)+d/dx(3)+d/dx(2x)=d/dx(14x^3)
2y*dy/dx+0+2=42x^2
dy/dx=(21*x^2 - 1)/y
substitute x=1,y=3:
dy/dx=(21*1^2-1)/3
=20/3

I haven't been in a classroom since 1998.

Yes, you are retarded.

3x^2+5+dy/dx=3y^2*dy/dx
3x^2+5=(3y^2-1) * dy/dx
dy/dx = (3x^2+5)/(3y^2-1)
=(3*1+5)/(27-1)
=8/26
=4/13
why are they asking for decimal answers? is your school for retards?

*Disclaimer Cheat Sheet*
Make it scientific notation on variables first. So Y is 10 x 3. 30. there are 2 x's so make sure it is times by 2, and then 10, so 20. But we have to use it step by step. SO just start with Y.

So. (30x30)+3+2(20)=14(20x20x20)
thats 9003+40=8000(14)
or 9043=320,000.

Now do it again except solve each of the efficients first. then make it notations on just variables.
Remember there are 2x variables so its 2x everytime.

3x3+3+2(2x)=14(2x^3)
or 9+3+(4)=14(8)
so then...9*10+3+(4x10)=14(8*10)
Remember just notations on the variables. Ok.

90+3+40=14*80
133=1120. Of course it looks wrong but...

133=1120 AND 9043=320,000 filter into reverse notations. So then.

13.3=112 and 904.3=32,000
so then by comparitive scaling (remember to mark as x1 and x2 and y1 and y2.
(13.3, 112) (904.3, 32000) we can chart on the graph the slope of each as:
13.3/112 and 904.3/32000.
And Simplify as ratio 1.33:11.2 and 9.043:320
Keeping it on notation as 1:10 (very important, we can use the same differential graphing index.

Which is roughly a 1:10 and 1:30 just in this example. If you have to use basing in case of function binary.
This means you can notorize y1 and y2 as 11.2:320 and use decimal placements to read as 1.12:32 and then x1 and x2 as 1.33:9.043

Add them up and you have 33.12 for y and 10.373 for x. Or (10.373,33.12) Simply compare to Binary Notations as 1.0373:3.312 for x:y and add original coordinations depending on the number of deritives you need, repeating the process each time. So 2.0373 for X:B Deritive and 6.312 for Y:B Deritive in Binary.

says that's not right. 0.308 or 4/13.

im gonna fucking kill my self

Math thread?

Take the implicit derivative (chain rule) and solve for dy/dx. Then do it again (you'll have another equation with x and y) to get (dy/dx)'.

< or > or = 1

Look either you can drop the attitude RIGHT NOW or I will take matters into my own hands. DON'T TICK ME OFF

check mate

-0.032

Been stuck on this for like 10 min.
I know I can use the fact a^2 - b^2 = (a - b) (a+b) to simplify the denominator further but that doesn't seem to do anything for me. Now 1/a^2 - b^2 looks sort of like a trig identity but I'm not sure I can use that here

oh it said second derivative lmao
cbf doing that. just do what I said a second time. You're a big boy and I've told you how to do it.

how tf did you get that?

3.141

Can't be asked to do this question but I'm assuming

1: Isolate equation for y=

2: Plug in the first principles formula to get the derivative, then plug in the points and that should be your new one

is it right? I don't want to bother typing it out if it's wrong.

...

it was correct. I am absolutely blown away. I couldn't figure it out for shit.

What's the question? Is it asking you to prove that one ax^2+bx+c=0 can be written x(x+d)^2+e=0

What's the catch?

Do what he:
did but do it twice, not once. it's second derivative

Nah, evaluate the integral of 1/sqrt(x^2 - 2x)

Veri qualiti hi watches. You give good price for watch I give away for free. Please enter card details for checks

not really, i tried that and didnt get the answer

Yes it's a badly formed question. Here's one perfectly valid solution given the question:

y^2 + 3 + 2x = 14x^3
2y + 2 = 42x^2
2 = 48x
x = 24

All you have to do to take this kind if derivative, is multiply the exponent by the coeffecient of each variable and subtract one from the exponent. Example: Find the slope (first derivative) of a line where y = 2x + 1, slope = 2, which means (2*1)x^(1-1) = 2x^0 = 2

Profs / quizzes / people trying to catch people out often ask badly formed questions as though there's one "secret" answer you're supposed to get if you're 1337.

Will taking abstract algebra help me become a better programmer? I've looked into the material they teach in the course and read up on some of the basics (groups rings, and fields, and examples of them) but it doesn't seem like I will ever use anything from it. Yet I've heard numerous people say it will actually make me better at coding.

Nice.

x^3+5x+y=y^3
x^3+5x=y^3-y | derive wrt x on the left and use the chain rule on the right
3x^2+5 = (dy/dx)*3y^2-(dy/dx) (*)
Let's call this last one (*), from this one you can get the first derivative at the point (1,2) (you'll need it later)
3x^2+5=(dy/dx)*(3y^2-1)
dy/dx=(3x^2+5)/(3y^2-1) ={x=1,y=2} =8/11
Now you can go back to (*) and continue by deriving once again wrt to x and using the chain rule
6x=d^2y/dx^2*3y^2+(dy/dx)^2 * 6y - d^2y/dx^2 | note that d^2y/dx^2 is the second derivative and (dy/dx)^2 is the first derivative squared
6x=d^2y/dx^2 *(3y^2-1) + (dy/dx)^2 * 6y
From here on out it's just moving the second derivative to the left and everything else to the right
d^2y/dx^2 = (6x-6y*((dy/dx)^2 ))/(3y^2-1) = {x=1,y=2, dy/dx =8/11} = (6-12*(64/121))/11=-0.032

All programming is basically abstract algebra in a sense. If you got the mathmatical aptitude it can be worth it.

So this is what STEMlords do for fun.

I think I'll stick to saving the world, please.

>Or should I say: your welcome.

Neat. I'm getting a minor in math and that course is one of the upper level electives I can can take. Thanks.

you'll probably need to do some change of variables, I'd just go to Wolfram Alpha to check what they get and reverse engineer it from there

you're
Humanties can't even fucking spell right.

extremely low quality b8

When you are creating something new, it is clearly good to have people in your team, who have a deeper understanding of the underlying principles. During my short stint in telcomm industry I met this guy from Nokia-Siemens Networks. I was told that he is the top patent inventor of the team from the Siemens side. I also recognized him as one of the guys who competed at IMO for West Germany. A modest dude who explained the recipe of his success as "Oh, most of my inventions are trivial applications of modular arithmetic - programmers and engineers don't understand it." I will testify that his last sentence is true. They learn about the binary mod and various rules around it, but they don't learn the bidirectional power of congruences, and don't learn to think in terms of the residue class rings - it's all remainders of divisions of integers to them. So, by being at the right place at the right time, you can dine out simply because you can work swiftly with periodically repeating discrete structures and patterns.

Something clearly not all the CS majors need to know, but "in the land of blind people one-eyed man is the king".

thanks man

got through a couple more and this one is really hard

Algebra and Calculus have nothing to do with programming unless you work in aerospace or something. The rest of us just import a math library and have that do all the work.

Don't waste your money.

The only reason programmers are forced to take so much math is so colleges can make more money making people retake classes and shit.

I've worked in multi-million-dollar web and database stuff for years. Most math I need is adding, subtracting, and multiplication.

This one is basically the same, just move all x to one side and all y to another then use the chain rule and the product rule. Do you need a reminder how those work?

Most appreciated bro

kek

Actually it's more if you work on something that ain't frontend or databases. As soon as you do real stuff you need the math. Yes I am looking down on your life choices. Deal with it.

I remember doing this long ago, I don't remember how.

That's how much you actually "learn".

Part of abstract algebra is all about working within the confines of discrete structures. Thighs like RSA encryption rely on principles from abstract algebra and number theory.
Techniques from real analysis (called advanced calculus some places) are used for numerical approximations in computers.
Just because YOU don't use them doesn't mean these subjects won't be useful to others.

y = bout tree fiddy

I don't think I used implicit functions at any point outside of high school and I'm one year away from an MSc in Applied Physics.

The answer is 9.787 if I didn't mess up inserting the values.

Ooops I only derivated once, it's -0.815.

When you start using chain rule, you have to notarize it based on every increment you use chain rule, so when you go to second deritive you have to put notation the first deritive by base 10 otherwise you end up with results that counter the affirmative chain cummulatively. That makes it so generatives make implicit false rounding on the 2nd deritive to original variable notations and further deritives.

So per chain you have to add notations for every dx/dy as dx(10) and dy(10) by the 2nd deritive or the original notations are off decimal. And Have a alternating reciprocal that throws the whole variable notation off. And you end up with half-repeating numbers.

lol wut

Based random poster

you have to Times 10 every dx/dy that comes around by the 2nd deritive chain otherwise the next deritive gets thrown off whether you use notation or not. So on 3 deritives you have to use notation twice. on the 2nd deritive instead of just once because it is cummulative.

I don't get why you think you need to multiply anything by 10.

After the first chain, you have to make an extra dx/dy by 10^1 per derivitive or you only end up counting the original variables for half the process (starting at the 2nd deritive when it gets pushed to the right).

You have to use 10^1 notation to begin with or it doesn't work on graphing and generates false rounds!!

>derivated
Differentiated