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I have a question regarding partial differentiation of a function of $x,y$ with respect to another function of $x,y$.

Specifically, I was wondering whether my logic or technique would hold true for most stuations.

$$ \begin{align*} \frac{\partial(x-y)}{\partial y} & = -1 \\ & = -1\cdot\frac{\partial f(x-y)}{\partial f(x-y)} \end{align*} $$

Then, cross multiplying, we get: $$ \begin{align*} \frac{\partial f(x-y)}{\partial (x-y)} = -1 \cdot \frac{\partial f(x-y)}{\partial y} \end{align*} $$

My technique consisted of first constructing a partial derivative with respect to a variable, which equaled a quantity (it happened to be a scalar in the above example). Then, I multiplied by $1 = \frac{\partial f(x-y)}{\partial f(x-y)}$. Then, I rearranged the terms.

Is this a valid technique? I checked my calculus book under partial derivatives, under the chain rule, and also did a general search on derivative with respect to a function. I'm not sure if what I am doing relates closely to calculus of variations.

Thanks.

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2 Answers 2

up vote 5 down vote accepted

Your calculation is correct, but I'm hesitant to call it a "valid technique." Let me explain a bit:

In the context you're working in, we have a function $f(u)$, which is a function of a single variable, and also $u(x,y) = x - y$ is a function of two variables. In this setup, the chain rule reads as follows:

$$\frac{\partial f}{\partial x} = \frac{df}{du}\frac{\partial u}{\partial x}$$ $$\frac{\partial f}{\partial y} = \frac{df}{du}\frac{\partial u}{\partial y}$$

Since $\frac{\partial u}{\partial y} = -1$, we can conclude that $\frac{\partial f}{\partial y} = -1\cdot \frac{df}{du}$, and therefore that $$\frac{df}{du} = -1\cdot \frac{\partial f}{\partial y},$$ which, in your (somewhat non-standard) notation reads $\frac{\partial f(x-y)}{\partial (x-y)} = -1\cdot \frac{\partial f(x-y)}{\partial y}$. So, this equation is true, yes.


The reason I hesitate to call your method a "valid technique" is because one usually cannot manipulate the symbols $\partial x$ and $\partial y$ as independent entities.

For instance, if $f$ were instead a function of two variables, say $f(u,v)$, where both $u$ and $v$ were themselves functions of two variables (say $u = u(x,y)$ and $v = v(x,y)$), then the chain rule would read

$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}$$ $$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y}.$$ If you'll notice, we really can't interpret the $\partial u$ and $\partial v$ signs as canceling without getting false identities like $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial x}$.

Amusing Example: Just to really drive the point home, consider the ideal gas law (from chemistry) $PV = nRT$, where $n$ and $R$ are constants. We can consider $P$, $V$, and $T$ as functions $$P = P(V,T) = nR\frac{T}{V}$$ $$V = V(P,T) = nR \frac{T}{P}$$ $$T = T(P,V) = \frac{1}{nR}PV.$$ One can then check that, in fact: $$\frac{\partial P}{\partial V} \frac{\partial V}{\partial T}\frac{\partial T}{\partial P} = -1.$$ So much for canceling.

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Thank you. I have learned I cannot treat $\partial x$ as an independent identity in algebraic manipulation, the formula $\frac{\partial g(u)}{\partial x} = \frac{\mathrm{d} g(u)}{\mathrm{d} u} \frac{\partial u}{\partial x}$, when $u = f(x,y)$, and an interesting example using the ideal gas law to dispel any misconceptions about $\partial x$ as individual entities. –  jrand Nov 15 '12 at 1:58
    
:-) Exactly. And you're very welcome. –  Jesse Madnick Nov 15 '12 at 2:01

Generally speaking it is not ok to use cross-multiplication for partial derivatives, because unlike full derivatives they are not represented as ratios of differential forms, so $\frac{\partial}{\partial x}$ must be regarded as one symbol (a vector in fact).

Not sure which connection to calculus of variations you had in mind, however there is only one delicate case that I am aware of where full and partial derivative get entangled in a subtle way. That is, variation of the functional $$v[z(x,y)]=\iint_D F\left(x,y,z,p,q\right)dxdy$$ where $p=\frac{\partial z}{\partial x}$, $q=\frac{\partial z}{\partial x}$ results in the following expression $$\delta v=\iint_D\left(F_z-\frac{\partial}{\partial x}\{F_p\}-\frac{\partial}{\partial y}\{F_q\}\right) \delta z dxdy$$ where $\frac{\partial}{\partial x}\{F_p\}$ is the so-called "full partial derivative by x" (L.E. Elsgoltz, "Calculus of Variations", 1958). It means that $y$ is considered constant, however the dependence of $z$, $p$ and $q$ on $x$ is taken into account. Which results in the following expression $$\frac{\partial}{\partial x}\{F_p\}=F_{px}+F_{pz}\frac{\partial z}{\partial x}+F_{pp}\frac{\partial p}{\partial x}+F_{pq}\frac{\partial q}{\partial x}$$

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@julianfernandez: I don't think Valentin is suggesting that at all, but simply that the symbols $\partial z$ are not in themselves meaningful (and therefore can't be manipulated as if they were meaningful). –  Jesse Madnick Nov 15 '12 at 1:45
    
As to whether full derivatives can be treated as "ratios of differential forms," well... math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio –  Jesse Madnick Nov 15 '12 at 1:46
    
good point, but that's why i specified they are thought of a ratios of differential forms, not real numbers. in fact one of my tutors mentioned this briefly without going into details. perhaps i need to give it a deeper thought –  Valentin Nov 16 '12 at 17:16

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