# 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?

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#### 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.

callculus
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.

GreginGre
October 09, 2019 15:17 PM