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My understanding of this is that it is a rate of change and in my mind I think of it like a vector.. you go some units in the $x$ direction and some units in the $y$ direction giving you the total rate of change for a unit movement of $x$ and a unit movement of $y$. Is this correct ?

So if we start at some origin then the total differential is basically the distance we moved multiplied by the differential in that direction ?

When I solve this problem myself I get

$$(2x + y)dx + (2y + x)dy$$ where wolfram says :

$$x\frac{dy}{dx} +2 y \frac{dy}{dx}+2 x+y$$

Are the two answers the same and could you please explain and could you please explain if my understanding of total differential is correct ?

Thank you,

Gerry Myerson
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user3528592
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Please go through this playlist for a clearer understanding of differentiation. You can learn about implicit derivatives like the one you've tried to do.

I'm not sure how you solved it but here's how I would it:

$$\frac{d}{dx} (x^2 + xy + y^2)\\ =\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2)\\ = 2x + x\frac{dy}{dx} + y\frac{dx}{dx} + 2y\frac{dy}{dx}\\ \equiv x\frac{dy}{dx} +2 y \frac{dy}{dx}+2 x+y \tag{W$\alpha$ is right!}$$

Here we've used two important concepts:

  • $(uv)' = u'v + uv' \quad\dots$ (used in $\frac{d}{dx} xy$)
  • $(f(g(x)))' = f'(g(x))\cdot g'(x) \quad\dots$ (used in $\frac{d}{dx} y^2$)

Now there's a rough way of differentiating it without respect to any variable:

$$\Delta (x^2 + xy + y^2) \\ = 2x\Delta x + y\Delta x + x\Delta y + 2y\Delta y \\ = (2x + y)\Delta x + (2y + x)\Delta y \tag{You're right!}$$

So, basically you did do it right. Wolfram alpha (and I) did it with respect to $x$ and considered $y$ to be a function of $x$.
You've basically differentiated with respect to a third variable. The variable itself is irrelevant in the way we've done it.

Mhh, maybe your understanding of differentiation is better than mine since you've got an answer with a self-made intuitive explanation :D
Good Job!

Nick
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    The OP's notation is common in certain subfields of physics, including thermodynamics. In particular, in undergraduate thermodynamics, one writes things like $d(E-TS)=dE-d(TS)=TdS-pdV+mdN-TdS-SdT=-SdT-pdV+mdN.$ From this they conclude that $E-TS$ is a "natural" function of $T,V,N$, whereas $E$ is a "natural" function of $S,V,N$. In graduate thermodynamics a better way of thinking about it (using a Legendre transform) is presented. – Ian Aug 27 '14 at 12:50
  • @Ian: That's nice but notation is irrelevant really. I could have used $\star$ instead of $\Delta$, the process is what matters. Ofcourse, readability and coherency do matter but hopefully my usage can be overlooked. I use $\Delta$ for sentimental reasons. – Nick Aug 27 '14 at 12:56
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    I did not mean the $d$ notation specifically. I meant the use of a total differential, rather than only differentiating with respect to one variable at a time, or explicitly differentiating with respect to a "hidden variable" that governs all of the others. I agree that the actual choice of the symbol is irrelevant. – Ian Aug 27 '14 at 13:09
  • Hi, Thanks. I have no doubt that it was my fault. My notation probably wasn't very clear. Looking over my notes it does say that : $z = f(x,y) = x^2+xy+y^2$ – user3528592 Aug 27 '14 at 13:12