If $\mathrm df=M_1\mathrm dx+N_1\mathrm dy=M_2\mathrm dx+N_2\mathrm dy$ ,then why $M_1=M_2$ and $N_1=N_2$?

Question

If $\mathrm df=M_1\mathrm dx+N_1\mathrm dy=M_2\mathrm dx+N_2\mathrm dy$ ,then why can we compare the terms involving $\mathrm dx$ and $\mathrm dy$ and write $M_1=M_2$ and $N_1=N_2$ ?

My intuition is in the expression $\mathrm df=\dfrac{\partial f}{\partial x}\mathrm dx+\dfrac{\partial f}{\partial y}\mathrm dy$

the term $\dfrac{\partial f}{\partial x}$ represents the rate of change of $f$ along the $x$ direction and thus $\dfrac{\partial f}{\partial x}\mathrm dx$ is the change of $f$ along that direction and similarly for the second term. So by comparing the terms involving $\mathrm dx$ and $\mathrm dy$ we are actually comparing the change of $f$ along the individual directions.

But I'm not sure whether $\dfrac{\partial f}{\partial x}$ can be considered as the rate of change because it doesn't behave like a fraction: if $f$ is a function of $x$ and $y$ and $x$ and $y$ are functions of $t$ then $\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial x}{\partial t}+\dfrac{\partial f}{\partial y}\cdot\dfrac{\partial y}{\partial t}=2\dfrac{\mathrm df}{\mathrm dt}$ which is absurd.

So please help to clear my doubts. Thanks in advance.

Answer 1

The last expression is not completely correct as $x$ and $y$ only depend on $t$. It is better:

$$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\partial f}{\partial x}\cdot\dfrac{\mathrm d x}{\mathrm d t}+\dfrac{\partial f}{\partial y}\cdot\dfrac{\mathrm d y}{\mathrm d t}$$

Maybe, for better understanding a well chosen example is useful. Consider the path $x(t)=t^2$ and $y(t)=C$. It is a straight line with the $y$ coordinate constant. Now, calculate the variation along this line,

$$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\partial f}{\partial x}\cdot 2t+\dfrac{\partial f}{\partial y}\cdot0=\dfrac{\partial f}{\partial x}\cdot 2t$$

Remaining $y$ constant, the partial derivative multiplied the rate of change for $x$ is the rate of change for $f$, as establishes the chain rule for a single variable functions. In the general case, simply occurs that the variations for $x$ and $y$ are superimposed, as vector components.

The example is useful to explain why $M_1=M_2$ It has to be so if the differential holds for any $\mathrm dx$ and $\mathrm dy$, from in my example we can set $\mathrm dx=2t$ and $\mathrm dy=0$ and the equality is mandatory.

Answer 2

Both $$\mathrm df=\dfrac{\partial f}{\partial x}\mathrm dx+\dfrac{\partial f}{\partial y}\mathrm dy$$ and $$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial x}{\partial t}+\dfrac{\partial f}{\partial y}\cdot\dfrac{\partial y}{\partial t}$$ are correct.

The implication $$=2\dfrac{\mathrm df}{\mathrm dt}$$ is not correct because you are assuming $$\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial x}{\partial t}=\dfrac{\partial f}{\partial t}$$ which is not true in multi-variable calculus.