Confusion over calculus notation (differentials/derivatives)

Question

I have read from multiple sources that dy/dx is not to be interpreted as a ratio as the idea of 'dy' and 'dx' themselves will lead to logical difficulties.

However, I have seen in many areas (e.g. thermodynamics) where notations involving dx etc. are used constantly. For example, in http://en.wikipedia.org/wiki/Total_differential, it seems to tread df/dt as a ratio by multiplying dt on both sides to cancel out the denominator.

So in what cases are we allowed to manipulate the notation as if they are quotients?

The concept of differential may help : if $y=f(x)$ then $\ dy=f'(x);dx\ $ this means that $dy$ will be 'proportional' to $dx$ with a constant $f'(x)$ at the point $x$ (without having to suppose that $dx$ is infinitesimal). — Raymond Manzoni, May 12 '13 at 11:34
Actually, it is a ratio; and the $d$ in $dx$ is a 'difference operator' i.e. $dx$ means $x_2-x_1$. — , Feb 18 '17 at 23:54

score 2 · Answer 1 · answered May 12 '13 at 11:41

Counterexample:

Consider the polar coordinates:

$$x=r\cos\theta$$ $$y=r\sin\theta$$ Then $$\dfrac{\partial x}{\partial r}=\cos\theta ,\dfrac{\partial x}{\partial \theta}=-r\sin\theta $$ However, what is $\dfrac{\partial r}{\partial x}$? is it $\dfrac{1}{\frac{\partial x}{\partial r }}$?

No:

$$x^2+y^2=r^2\Rightarrow 2x+0=2r\dfrac {\partial r}{\partial x}\Rightarrow $$ $$\dfrac {\partial r}{\partial x}=\frac{x}{r}=cos\theta $$ Can you find the other partial derivatives?

Generally, you are not allowed to treat differentials as fractions. Even in the example in Wikipedia, the rule that holds is the Chain Rule, and not a rule of fractions(http://en.wikipedia.org/wiki/Chain_rule).

In case of 1 variable, the differentials are used like fractions, but in more varialbes it is completely wrong to do that.

Confusion over calculus notation (differentials/derivatives)

1 Answers1