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

  • #1
1,291
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Homework Statement


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

Homework Equations





The Attempt at a Solution




[tex]= \frac{\frac{-dy}{dx}} {(\frac{dy}{dx})^2}[/tex]
[tex]= \frac{\frac{-dy}{dx}} {\frac{dy}{dx} (\frac{dy}{dx})}[/tex]
[tex]= \frac{1}{\frac{dy}{dx}}[/tex]

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

[tex] 2 = \frac{1}{\frac{dy}{dx}}[/tex]
[tex] 2(\frac{dy}{dx}) = 1 [/tex]
 

Answers and Replies

  • #2
34,553
6,268

Homework Statement


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

Homework Equations





The Attempt at a Solution




[tex]= \frac{\frac{-dy}{dx}} {(\frac{dy}{dx})^2}[/tex]
[tex]= \frac{\frac{-dy}{dx}} {\frac{dy}{dx} (\frac{dy}{dx})}[/tex]
[tex]= \frac{1}{\frac{dy}{dx}}[/tex]
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:

[tex] 2 = \frac{1}{\frac{dy}{dx}}[/tex]
[tex] 2(\frac{dy}{dx}) = 1 [/tex]
Sure. And if the goal is to solve for dy/dx, divide both sides of the last equation by 2.
 
  • #3
1,291
0
Sure.
What happened to the minus sign?
Simple mis-type.

Note that you can simplify 1/(dy/dx).
Like this?
[tex]=\frac{dx}{dy}[/tex]

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

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

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:
  • #4
34,553
6,268
Simple mis-type.


Like this?
[tex]=\frac{dx}{dy}[/tex]

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

[tex]=\frac{dx}{dy} = lim_{Δx→0} \frac{Δx}{Δy}[/tex]
No, the limit would be as Δy→0. Here the assumption would be that x is some function of y.
[tex]= lim_{Δx→0} \frac{Δx}{f(x+Δx) - f(x)}[/tex]
[tex]= f'(y)[/tex]

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.
 

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