Question

if \(f=f(x,y)\)
the total differential is
\[\text df= \frac{\partial f}{\partial x}\text dx+\frac{\partial f}{\partial y}\text dy\] right?

Answer 1

or is the total differential
\[\frac{\text df}{\text dx}= \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\text dy}{\text dy}\]

Answer 2

\[ (xyz) = c \\
\implies yz dx + xz dy + xy dz = 0\]

Answer 3

whta?

Answer 4

yes it is as what you said.

Answer 5

the first

Answer 6

is the total differential

\(\text df\) or \(\frac{\text df}{\text dx}\),

Answer 7

ah ok, the total differential is just

\[\text df\]

that's all i was checking

Answer 8

thanks

Answer 9

does the other thing have a name ?

Answer 10

i think so let me check

Answer 11

the total differential in the x direction must have a better name

Answer 12

the second one you did looks sorta like teh chain rule for partial derivatives.. but i'm not sure

Answer 13

ops typo
\[\frac{\text df}{\text dx}= \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\text dy}{\text dx}\]

Answer 14

i suppose \( \frac{\text df}{\text dx}\) is the total derivative ,

Answer 15

yeah but that's not the actual equation correct?

Answer 16

the question was

\[f(x,y)=y^3(1-x^2)-\cos^2x-3\], find the total differential

Answer 17

but i think i can do it

Answer 18

yeah that's for that but the total derivative for an f(x,y) would be

\[\frac{dz}{dt}=\frac{\delta f}{\delta x}\frac{dx}{dt}+\frac{\delta f}{\delta y}\frac{dy}{dt}\]

Answer 19

\partial 's ?

Answer 20

\[f(x,y)=y^3(1-x^2)-\cos^2x-3\]

\[\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\left(y^3(1-x^2)-\cos^2x-3\right)=-2xy^3+2\sin x\cos x\]\[\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left(y^3(1-x^2)-\cos^2x-3\right)=3y^2(1-x^2)\]
\[\text df= \frac{\partial f}{\partial x}\text dx+\frac{\partial f}{\partial y}\text dy\]\[\text df=\left(-2xy^3+2\sin x\cos x\right)\text dx+\left(3y^2(1-x^2)\right)
\text dy\]

Answer 21

if the queston had asked fot the total derivavative

\[\frac{\text df}{\text dx}=\left(-2xy^3+2\sin x\cos x\right)+\left(3y^2(1-x^2)\right)
\frac{\text dy}{\text dx}\]

Answer 22

getting differentiation equation out of the given equation is not that difficult ... getting the surface out of differential is challenging.

Answer 23

how'd ya do that exactly?

Answer 24

there are bunch of ways ... if one doesn't work then other will ...(if satisfies the condition of integrability) ... easiest one is inspection.
\[ yzdx+xzdy+xydz=0 \]
this will give \[ xyz = c\]

Answer 25

some kinda hyperbolic something?