Sure.Nano-Passion said:Homework Statement
Am I right to treat dy/dx as a fraction in this scenerio?
What happened to the minus sign?Nano-Passion said: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]
Sure. And if the goal is to solve for dy/dx, divide both sides of the last equation by 2.Nano-Passion said: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]
Simple mis-type.Mark44 said:Sure.
What happened to the minus sign?
Like this?Note that you can simplify 1/(dy/dx).
Sure. And if the goal is to solve for dy/dx, divide both sides of the last equation by 2.
No, the limit would be as Δy→0. Here the assumption would be that x is some function of y.Nano-Passion said: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]
Nano-Passion said:[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.
The purpose of dividing dy/dx by (dy/dx)^2 is to simplify the expression and to eliminate any possible division by zero.
No, you cannot divide by zero. In this case, the expression is undefined.
No, dividing dy/dx by (dy/dx)^2 is not the same as taking the derivative. The derivative of dy/dx is equal to (d^2y)/(dx^2).
No, dividing dy/dx by (dy/dx)^2 will not change the value of dy/dx. It will only simplify the expression.
Yes, you can divide dy/dx by (dy/dx)^2 even if dy/dx is a complex number. The result will also be a complex number.