Is it okay to treat dy/dx as a fraction when solving for derivatives?

[Nano-Passion]

Homework Statement

Am I right to treat dy/dx as a fraction in this scenerio?

Homework Equations

The Attempt at a Solution

= \frac{\frac{-dy}{dx}} {(\frac{dy}{dx})^2}
= \frac{\frac{-dy}{dx}} {\frac{dy}{dx} (\frac{dy}{dx})}
= \frac{1}{\frac{dy}{dx}}

Also, would we be able to multiple out dy/dx if we had for example:

2 = \frac{1}{\frac{dy}{dx}}
2(\frac{dy}{dx}) = 1

 

[Mark44]

Homework Statement

Am I right to treat dy/dx as a fraction in this scenerio?

Sure.

Homework Equations

The Attempt at a Solution

= \frac{\frac{-dy}{dx}} {(\frac{dy}{dx})^2}
= \frac{\frac{-dy}{dx}} {\frac{dy}{dx} (\frac{dy}{dx})}
= \frac{1}{\frac{dy}{dx}}

What happened to the minus sign?

Note that you can simplify 1/(dy/dx).

Also, would we be able to multiple out dy/dx if we had for example:

2 = \frac{1}{\frac{dy}{dx}}
2(\frac{dy}{dx}) = 1

Sure. And if the goal is to solve for dy/dx, divide both sides of the last equation by 2.

 

[Nano-Passion]

Sure.
What happened to the minus sign?

Simple mis-type.

Note that you can simplify 1/(dy/dx).

Like this?
=\frac{dx}{dy}

Which would read change of x with respect to y.. Hmm, I'm not used to it being like that.

=\frac{dx}{dy} = lim_{Δx→0} \frac{Δx}{Δy}
= lim_{Δx→0} \frac{Δx}{f(x+Δx) - f(x)}
= f'(y)

Is what I stated correct?

Sure. And if the goal is to solve for dy/dx, divide both sides of the last equation by 2.

Oh okay, thank you! I just wanted to make sure I wasn't violating any rules here pertaining to derivatives.

 

[Mark44]

Simple mis-type.


Like this?
=\frac{dx}{dy}

Which would read change of x with respect to y.. Hmm, I'm not used to it being like that.

=\frac{dx}{dy} = lim_{Δx→0} \frac{Δx}{Δy}

No, the limit would be as Δy→0. Here the assumption would be that x is some function of y.

= lim_{Δx→0} \frac{Δx}{f(x+Δx) - f(x)}
= f'(y)

Is what I stated correct?




Oh okay, thank you! I just wanted to make sure I wasn't violating any rules here pertaining to derivatives.