/sci/ - Science & Math

[Anonymous No.5359917]

Does this work?
<div class="math">\left(\frac{\partial}{\partial x}f(x,y)\right)^2=\left(\frac{\partial}{\partial x}\right)^2~f(x,y)^2</div>
<div class="math">=\frac{\partial ^2}{\partial x^2}~f(x,y)^2=f''(x,y)^2</div>

[Anonymous No.5359932]

that is one ugly bitch

[Anonymous No.5359939]

>>5359917


I think that should give you f'(x,y)^2

But I could be wrong

[Anonymous No.5359940]

>>5359917
Rather, is it a true proof of squaring the derivative equals a second derivative?

[Anonymous No.5359942]

>>5359932

[Anonymous No.5359950]

No it doesn't.

Some fields like to play fast and loose with differential operators, but if you don't know what you're doing it's easy to screw up like you have.

Go back to the definition of the derivative of a function, then read all the relevant theorems (chain rule, etc).

If you can't derive your formula explicitly using these theorems at every step, then your formula is wrong (which it is).

[Anonymous No.5359955]

>>5359940
try it on a function, here's one: f(x) = x^3

[Anonymous No.5359965]

>>5359940
No, and that should be apparent if you think about it in terms of a plotted graph.

Multiplying a first derivate function by itself will not give the second derivative, obviously

[Anonymous No.5361889]

But it is true that if we want to solve
<span class="math">\int_0^x\!f(t)\,dt = f(x) - 1[/spoiler]
we can let
<span class="math">\int f = \int_0^x\!f(t)\,dt[/spoiler]
and then
<span class="math">1 = f - \int f [/spoiler]
<span class="math"> 1 = (1 - \int) f [/spoiler]
<span class="math"> f = (1 - \int)^{-1} 1 [/spoiler]
<span class="math"> f = (1 + \int + \int^2 + \int^3 + \ldots)1[/spoiler]
<span class="math"> f = 1 + \left(\int 1\right) + \left(\int^2 1\right) + \ldots [/spoiler]
<span class="math"> f(x) = 1 + x + \int x + \int^2 x + \ldots [/spoiler]
<span class="math"> f(x) = 1 + x + \frac{x^2}{2} + \int\frac{x^2}{2} + \ldots [/spoiler]
<span class="math"> f(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots [/spoiler]
<span class="math"> f(x) = e^x [/spoiler]
believe it

[Anonymous No.5361894]

>>5359942
i'd fuck her, but that doesn't change the fact that she's ugly.

[Anonymous No.5362452]

>>5361889
what is this<span class="math">\int[/spoiler] symbol you're using as a variable?

[Anonymous No.5362496]

>>5361889
https://www.youtube.com/watch?v=loa-sMsS6ck#t=06s

[Anonymous No.5362527]

You're wrong and you should feel wrong. Just because you square something doesn't mean you take the derivative twice.

[Anonymous No.5362530]

>>5362452
Just factored out the integral
ain't no thang

[Anonymous No.5362541]

>>5362530
but you forgot to factor in the dt's as well.

[Anonymous No.5362554]

>>5362541
dt is just a formality. just write <span class="math">\int f[/spoiler] instead of <span class="math">\int f(t) dt[/spoiler], same information gets across. Save time and money

[Anonymous No.5362556]

>>5362554
But when you square the integral you also need to square the dt's which messes it all up.

[Anonymous No.5362563]

>>5362556
just think of it as a linear functional

[Anonymous No.5362566]

>>5362556
you just have to BELIEVE