Are independent variable infinitesimals positive? A calculation I'm doing involves the inequality $$\frac{dt}{dx}\leq \frac{-kt}{x}$$ and I want to shift the dx and t to the other sides, so I can integrate.
Now t is known to be positive (it's $xf(x)$, where f is a function and both x and f are positive), so I assume I can shift it.
But I'm not sure about $dx$. I could consider it the independent variable on the graph of f, and say it's defined to be positive, while I can't say the same for dt, since as a function it may increase or decrease.
Is this assumption correct?
 A: Question
If I understand correctly, the main question is essentially:

If $f(y)\dfrac{\mathrm dy}{\mathrm dx}\le g(x)$, can we "multiply both sides by $\mathrm dx$" and then integrate, as we might with an equation?

Main Answer
It's a property of definite integrals that if $f(x)\le g(x)$ on $[a,b]$, then $\int_a^bf(x)\,\mathrm dx\le \int_a^bg(x)\,\mathrm dx$. Therefore, if we have $f(y)\dfrac{\mathrm dy}{\mathrm dx}\le g(x)$ for relevant values of $x$, we may write $\int_a^Xf(y)\dfrac{\mathrm dy}{\mathrm dx}\,\mathrm dx\le \int_a^Xg(x)\,\mathrm dx$ and use integration by substitution to rewrite the integral on the left as ${\displaystyle \int_{{\displaystyle y\mid_{x=a}}}^{{\displaystyle y\mid_{x=X}}}}f(y)\,\mathrm dy$. If $F$ and $G$ are antiderivatives for $f$ and $g$, respectively, then FTC says that the inequality becomes $F\left({\displaystyle y\mid_{x=X}}\right)-F\left({\displaystyle y\mid_{x=a}}\right)\le G\left(X\right)-G(a)$. Setting $C$ to be the contstant $F\left({\displaystyle y\mid_{x=a}}\right)-G(a)$, we could write this as $F\left({\displaystyle y\mid_{x=X}}\right)\le G\left(X\right)+C$. In more common notation, we have $F(y)\le G(x)+C$.
For comparison, if we were to "multiply both sides by $\mathrm dx$" and integrate as we would typically do for a separable differential equation, we'd have $\int f(y)\,\mathrm dy\le\int g(x)\,\mathrm dx$ and immediately arrive at $F(y)\le G(x)+C$.
Since we arrive at the same inequality as in the argument that did not rely on uncertain properties about multiplying through by $\mathrm dx$, this method is valid. Whether you want to think about it as "treating $\mathrm dx$ as positive" is probably a matter of personal taste.

Edit/Note
However, we can also do the same careful steps with $\int_X^b$ instead, to find that $F\left({\displaystyle y\mid_{x=b}}\right)-F\left({\displaystyle y\mid_{x=X}}\right)\le G\left(b\right)-G(X)$, and hence there is a constant $C_2=F\left({\displaystyle y\mid_{x=b}}\right)-G\left(b\right)$ for which $G(x)+C_2\le F(y)$.
This is the result we would have obtained by multiplying through by $\mathrm dx$ and flipping the inequality sign and integrating. In this sense, $\mathrm dx$ is perhaps best considered as "either positive and negative", since we have $G(x)+C_2\le F(y)\le G(x)+C_1$.
