/sci/ - Science & Math

>x = 1 + ... + 1 [x times]
>2x = 1 + ... + 1 [x times]

>>1120609
>D(x^2) = D(x + ... + x) [x times]
>D(x^2) = D(x) + ... + D(x) [x times]
This is where you are taking a wrong turn: the derivative is not properly defined for the right hand side of the equation.

>2x = x
>2 = 1
You divided by 0.

>x = 1 + ... + 1 [x times]
>whatthefuckamireading.jpg

>>1120628
Are you saying the derivative doesn't work? Or that it isn't linear?

>>1120607

"w" + "u" - "u" + "w" + "z" - "z" + "w" + "z" - "z" + "." + "d" - "d" + "a" + "b" - "b" + "n" + "y" - "y" + "o" + "e" - "e" + "n" + "u" - "u" + "t" + "c" - "c" + "a" + "f" - "f" + "l" + "o" - "o" + "k" + "a" - "a" + "." + "l" - "l" + "s" + "r" - "r" + "e" + "z" - "z"

>>1120636
x = 1 * x = 1 (x times)

durr

>>1120639
Are you saying that this makes any sense?

God.
Learn your laws.

>>1120650
wat

>>1120652
Well, where's the flaw then? And x doesn't have to be zero.

>>1120660
where is the sense then?
-1 = 1 + 1 + 1 - 4
potato = D(hurr)
1 = 2

>>1120629
This man is right

>>1120628
This man is wrong, as derivatives and sums commutate in the example given.

>>1120639
The problem lies in the "x times" part: the shape of the expression is determined by the variable you're differentiating with respect to:
<div class="math">\frac{mathrm{d}}{\mathrm{d}x}\left [ x + x + \cdots + x\right ] = \frac{mathrm{d}}{\mathrm{d}x}\left [x^2\right ] = 2x</div>

The first line is only true for positive integer values, which means that the function you are trying to use is discontinuous over the reals, which means that it's not differentiable anywhere except on the set of integers.

But if you try to take the limit of x as x goes to a, you will get that it doesn't exist, which means that the function is not continuous anywhere except 0, which means that it's only differentiable at 0, which means that 2x = x only when x = 0, which makes your last step impossible.

>>1120687
mind = blown

>>1120672
Division by zero...? I don't see that... Also, are you sure <span class="math">\frac{\mathrm{d}}{\mathrm{d}x}\left [\sum_{n=1}^x x \right ] = \sum_{n=1}^x \frac{\mathrm{d}}{\mathrm{d}x}\left [ x\right ][/spoiler]?

>>1120687
/thread

>>1120687
nice

>>1120702
>Division by zero...? I don't see that...
2x=x
subtract x
x=0

What is there not to see?

Protip: You never divide by x, you can only factor it out.

>hurr hurr
>derp
>durrahurr
>I am a faggot
/thread

>D(x^2) = D(x) + ... + D(x) [x times]
>2x = 1 + ... + 1 [x times]

D(x) = 0

right?

cauz the derivate of a constant is zero.

<div class="math">x=3</div>
<div class="math">x^2=9</div>
<div class="math">\frac{d}{dx}x^2=\frac{d}{dx}9</div>
<div class="math">2x=0</div>
<div class="math">6=0</div>

DOHOHO

>>1120885

WHAT THE FUCK???

op wrongly assumes

<div class="math">\frac{d}{dx}(x^2) = x\frac{d}{dx}(x)

bitches don't know bout my product rule</div>

>>1120945
fuck
I can't eqn

you know what I meant

>>1120945

D(x^2) = 2x

SHUT THE FUCK UP

herp, product rule:
d/dx[u*v]=u*d/dx[v] + v*d/dx[u]

So d/dx[x^2]=x*d/dx[x]+x*d/dx[x]=x*1+x*1=x+x=2x.

>>1120958
lol

what about "OP wrongly assumes" don't you fucking understand?

>>1120978

We know, faggot.

Alternatively; <span class="math">D(x^n) = n*x^(n-1)<div class="math">[/spoiler]</div>

>>1120978
I bet you'd use the quotient rule for x/2.

>>1120885

wat

>>1120885

wtf? is that true?

>>1120989

>>1120945

Apparently not everybody.

>>1121003

I wouldn't use it, but that doesn't stop it from being applicable.

>>1120605

"w" + "j" - "j" + "w" + "r" - "r" + "w" + "e" - "e" + "." + "c" - "c" + "a" + "o" - "o" + "n" + "v" - "v" + "o" + "i" - "i" + "n" + "a" - "a" + "t" + "b" - "b" + "a" + "s" - "s" + "l" + "r" - "r" + "k" + "u" - "u" + "." + "u" - "u" + "s" + "b" - "b" + "e" + "z" - "z"

>>1121027
It's just not efficient :s

>>1121017

Yeah but how could that faggot prove that 6 = 0 ?

>>1121042

Which is why I obviously wouldn't use it. I was originally just pointing out product rule to >>1120945 before people started assuming I used product rule for everything.

>>1121045

By doing it wrong.

>>1121058
>implying you dont use the product rule for everything

>>1121045
Why did you include that image with your post?

This kinda sums it up:
.
.
.
.
http://tinyurl.com/2wth3c5

>>1120885

><div class="math">\frac{d}{dx}x^2</div>

You're not derivating a variable here, <div class="math">x^2</div> is constant thus <div class="math">\frac{d}{dx}x^2=0</div>

Have a nice day.

>>1121078
TEN POINTS TO GRYFFENDOR

Why do you americunts call it "calculus" ?

>>1120605

"w" + "g" - "g" + "w" + "e" - "e" + "w" + "r" - "r" + "." + "z" - "z" + "a" + "s" - "s" + "n" + "e" - "e" + "o" + "p" - "p" + "n" + "e" - "e" + "t" + "t" - "t" + "a" + "f" - "f" + "l" + "g" - "g" + "k" + "c" - "c" + "." + "n" - "n" + "s" + "i" - "i" + "e" + "v" - "v"

>>1121102

It's called calculus everywhere when it comes to university level

get the fuck out you highschooler

>>1120608

"w" + "x" - "x" + "w" + "j" - "j" + "w" + "s" - "s" + "." + "g" - "g" + "a" + "g" - "g" + "n" + "g" - "g" + "o" + "p" - "p" + "n" + "c" - "c" + "t" + "s" - "s" + "a" + "f" - "f" + "l" + "p" - "p" + "k" + "q" - "q" + "." + "t" - "t" + "s" + "r" - "r" + "e" + "b" - "b"

>>1121123
Bitches don't know 'bout my Analysis.

when you say [x times] that implies x is an integer value. However you can only take derivatives at accumulation points. as the integers are discrete in the reals, this proof fails.