Total differential

by pluton   Last Updated October 09, 2019 15:20 PM

I am considering a function $f(x,y)$ with all the appropriate assumptions so that what comes next is well defined. I think we have the equivalence: $$\mathrm{d}f=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y=0\Leftrightarrow \frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$$ What is the exact mathematical argument behind this? How should the quantities $\mathrm{d}x$ and $\mathrm{d}y$ be treated?

Tags : derivatives

Answers 2

Paritial answer: Your equivalence is not true in general.

Let´s say $f(x,y)=x^{0.5}\cdot y^{0.5} \ \ \forall x,y>0$, Then we have the partial derivatives

$$\frac{\partial f}{\partial x}=0.5\cdot x^{-0.5}\cdot y^{0.5}, \frac{\partial f}{\partial y}=0.5\cdot x^{0.5}\cdot y^{-0.5} $$

Let us inspect the point at $x=y=1$

$$\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y=0\Rightarrow 0.5dx+0.5dy=0\Rightarrow \boxed{dx=-dy}$$

So here $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0.5\neq 0$.

And for $dx$ and $dy$ there are several combinations possible.

October 09, 2019 14:25 PM

You have to understand the real nature of $df, dx$ and $dy$, and to be precise in your statement.

First of all, you have to understand that $df$ is the differential at a point $(x_0,y_0)\in\mathbb{R}^2$, so at least you should write

$df= \frac{\partial f}{\partial x}(x_0,y_0)\mathrm{d}x+\frac{\partial f}{\partial y}(x_0,y_0)\mathrm{d}y$.

Now $df,dx$ and $dy$ are linear forms on $\mathbb{R}^2$.

The notation $dx$ denotes the linear from $(h,k)\in\mathbb{R}^2\mapsto h\in\mathbb{R}$, while $dy$ is the linear form $(h,k)\in\mathbb{R}^2\mapsto k\in\mathbb{R}$.

Then $df$ is a linear combination of the linear forms $dx$ and $dy$. It is easy to check that $dx$ and $dy$ are linearly independant. In particular, if $df=0$, it is equivalent taht the coefficients of the combinations are both $0$.

If you prefer a direct argument (which amounts to what I said previously), we may rewrite the equality as follows:

$$df(h,k)=\frac{\partial f}{\partial x}(x_0,y_0)h+\frac{\partial f}{\partial y}(x_0,y_0)k=0$$ for all $(h,k)\in\mathbb{R}^2$.

Setting $(h,k)=(1,0)$, then $(h,k)=(0,1)$ yields the desired equivalence.

October 09, 2019 15:17 PM

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