Operations with $\frac{dy}{dx}$.

Question

While working on a problem I have found a solution. I am curious about a clean and correct way to write it down. I want to find the derivative of $y(x)=:y$.

$$(5y^4+1)\frac{dy}{dx} + 1 = 0\\ \frac{dy}{dx} = -\frac{1}{5y^4+1}$$

Is it mathematically correct to divide by $(5y^4+1)$? I am aware that it is greater than zero, however is it really multiplied with $\frac{dy}{dx}$? Is it an operator? What is the correct way to write it down?

The denom is positive definite, you should be fine, but have you made an error writing $y(x)= y$? — Demetri Pananos, Dec 19 '16 at 21:18
$\frac{dy}{dx}$ is a function. Yes, what you're doing is fine. — MathematicsStudent1122, Dec 19 '16 at 21:18
@DemetriP I just wanted to indicate that from now on I will say $y$ instead of $y(x)$. — B.Swan, Dec 19 '16 at 21:21
Why do you think it might not be correct to divide by what you did? — Git Gud, Dec 19 '16 at 21:25
Perhaps $\frac{dy}{dx}$ is an operator, but it has been said that it is a function. — B.Swan, Dec 19 '16 at 21:26
@B.Schnebbler What's your definition of operator? It certainly is a function. — Git Gud, Dec 19 '16 at 21:27
$\frac{dy}{dx}$ is a function, $\frac{d}{dx}$ is the derivative operator. — Dave, Dec 19 '16 at 21:39
I'm confused about the problem. Are you solving an implicit differentiation problem or a differential equation? — Cure, Dec 19 '16 at 21:41

score 1 · Accepted Answer · answered Dec 20 '16 at 02:03

Yes, what you're doing is perfectly "legal". Note that $\frac{dy}{dx}$ is the differentiation operation $\frac{d}{dx}$ being applied to $y$, and is thus the result of applying an operator; which is a function. Meanwhile $\frac{d}{dx}$ on its own is the differential operator, which maps functions to functions.

Operations with $\frac{dy}{dx}$.

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