# Dividing dy/dx by (dy/dx)^2

Nano-Passion

## Homework Statement

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

## 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:

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

Mentor

## Homework Statement

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

## 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.

Last edited:
Mentor
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.