/sci/ - Science & Math

>>10696862
reply you niggers

>>10696862
They're differentials and yes, they're related to derivatives by dy = f(x) dx iff dy/dx = f(x)

>>10696862
(not OP but) I haven't really learned about partial differentials yet, but I think I have a rough idea of what they are (thanks 3blue1brown).
Anyway, why is there a separate notation for partial derivatives? Why not just use the normal "d" with respect to various variables? Isn't it pretty much representing the same thing?

>>10697091
Because it's not the same thing. With partial derivatives you are deriving some variables while others are left constant and don't matter. And if you write the total derivative you derive all of the variables.

>>10696862
>Was does the "d" even mean?
It means "differential", retard.

>>10696862
>How come d(fg)=f.dg+g.df ??
1/2

did you just fucking represent an integral with an S?

>>10696862
>>10697822
2/2
The other theorems referenced in the proof:
>Theorem 4.7 (limit of a product): if f(x) approaches a and g(x) approaches b as x approaches X then f(x)g(x) approaches ab as x approaches X
>Theorem 5.1: If f is differentiable at x, then f is continuous at x.
I can answer your other questions similarly if you're happy with this and still here OP.

>be mathfag putting off ODEs because we take the same class the engineers take and I know its gonna be just boring af plug and chug memorization
>come to a point where every class requires I have taken ODEs so I finally bite the bullet
>one class TA takes dy/dx=something and rewrites it as dx/dy=1/something
>'woah, why did you do that?'
>'oh, I just took the inverse of dy/dx'
>'we don't even know what y is, why are you assuming that it has an inverse AND that it is differentiable'
>'oh, I'm not, I mean the inverse as in 1/(dy/dx)'
>'that doesn't make any sense, derivatives aren't fractions'
>'but you can treat it as a ratio of differentials'
>'but why? Derivatives are not differentials'
>'They are. Derivatives are just a ratio of very small numbers'
>'How are you TAing a math class?!?!'
Later I found out he was an electrical engineer.

>>10696968
>>10697608
no shit, niggers... I meant what does the d in d(fg) mean not in general. kill yourselves.
>>10697825
yes, i am based
>>10697822
>>10697830
nice try although you are not derivating against other vairable when you simply say d(yx), you would need a diferential denominator for that an bot x and y should be in function of that varible you are derivating against.
So it doesnt really make sense execp if that variable is a constant or some thing alike.
>>10697947
>>10697947
hahahaa yes diferentials are for huge brainlets.
>>10697091
this is not a partial derivative, my friend. It shows you haven't seen it yet.

>>10696862
d is for distance you uneducated nigger

>>10696862
Ask your teacher you dumb first year

>>10697975
lmao not takin analishit this semester, pal. already did and wont ask a chemishit because he probably knows less then i do

>>10697968
Oh sorry I understand the question now. I thought you were asking for proof of the product rule. d(fg)=f.dg+g.df follows from the definition of the differential as the derivative multiplied by the increment (I can also explain that definition if you want). See pic related for derivation of d(fg)=f.dg+g.df
t. computational biochem phd student

>>10698142
>derivative multiplied by the increment
By the increment of x I mean.

Based.
Thanks it actually solved my doubt. I never thought i would have to use a variable both things were function of to later on discard. pretty based

In non retarded notation it would be
h=g*h
h(x)=f(x)*g(x)
(x is any variable you can write h, f and g in terms of)
therefor:
h(x)=S[f(x)g(x)]'dx

please continue OP