Derivative When Substituting Variables

[Drakkith]

I'm working through a proof in my differential equations book, but I think I'm hung up on a basic calculus derivative.

If we have a function ##f(x,y)## and we substitute ##v=\frac{y}{x}## , rearrange to get ##y=vx##, and then take the derivative, supposedly by the product rule we get $$\frac{dy}{dx}=v+x\frac{dv}{dx}$$
I'm not quite sure how this works since ##v## is a function of both y and x and y itself is a function of x. What's going on here?

 

[cnh1995]

I think it should be dy/dx=v+x*dv/dx as per the product rule.

 

[Drakkith]

I think it should be dy/dx=v+x*dv/dx as per the product rule.

I'm sorry, I forgot to add the X in the 2nd term. My mistake.
I'm still not sure what's going on though.

 

[Mark44]

I'm working through a proof in my differential equations book, but I think I'm hung up on a basic calculus derivative.

If we have a function ##f(x,y)## and we substitute ##v=\frac{y}{x}## , rearrange to get ##y=vx##, and then take the derivative, supposedly by the product rule we get $$\frac{dy}{dx}=v+x\frac{dv}{dx}$$
I'm not quite sure how this works since ##v## is a function of both y and x and y itself is a function of x. What's going on here?

##v = \frac y x##, so in a sense v is a function of both x and y, but the assumption is that y is a function of x. This means that v is also a function of x alone.

Starting with the equation y = vx, differentiate both sides with respect to x. This gives you y' = v + v'x, just using the product rule.

 

[Drakkith]

Okay. I figured it was something easy I was missing. Thanks guys.