Derivative of (v)^2 with respect to position

[TheWonderer1]

I forgot where I came across this and why I got so determined to figure it out but I wanted to ask about this d/dx(v^2) business.

My question is to solidify my understanding of the chain rule with physics equations (sorry for crap terminology). Therefore, I know I use it and do the math as
d(v^2)/dv * dv/dx = 2v x dv/dx = 2 dx/dt * dv/dx= 2 dv/dt = 2a. I just sort of do that automatically but I’m unsure of the “why”.

Basically, if possible, could you explain to me the chain rule being used for these sort of equations? I understand the use for something like the derivative of (3x+1)^7 = 21(3x+1)^6.

 

[kuruman]

It's easier to see what's going on if you go backwards from what you have. The definitions of velocity and acceleration are
$$v=\frac{dx}{dt}~;~~~a=\frac{dv}{dt}$$
Apply the chain rule
$$a=\frac{dv}{dt}=\frac{dv}{dx} \frac{dx}{dt}=v\frac{dv}{dx}=\frac{1}{2}\frac{d}{dx}(v^2)$$Does this make sense?

 

[TheWonderer1]

It is clear. So I think I used the chain rule correctly in my case. I know I got to the right answer just want to be sure my reasoning was on point and I didn’t misuse the chain rule.

Could I have done the chain rule differently? Just trying to get further concrete understanding.

 

[kuruman]

Could I have done the chain rule differently?

You did do it differently. You started from d(v²)/dx and you showed that it is equal to 2a. I started from a and showed that it is equal to (1/2)d(v²/dx. Both ways are equally correct and lead to the same equation.

 

[TheWonderer1]

Oops, to ask my question more explicitly we have two ways are there other ways?

 

[kuruman]

Oops, to ask my question more explicitly we have two ways are there other ways?

I can't think of any. You have to start from the definition of the acceleration and there is a limited choice of what to do next if you want to introduce the velocity in the expression and use the chain rule.