# Chain rule say what?

1. Sep 5, 2012

### vish22

ok stupid question probably-
take v(velocity) to be a function of x and x to be a function of t(time).
then dv/dt=vdv/dx thats cool
but in the hint in problem 2.12 classical mechanics by john r taylor he equates vdv/dx and 1/2(dv^2)/dx
that is- vdv/dx=1/2(dv^2)/dx

2. Sep 5, 2012

### universal_101

Just assume, v2 = y , then differentiate both sides w.r.t x, you have,
d(v2)/dx = dy/dx ...........(1)
2v dv/dx = dy/dx....................(2)

and then eliminate dy/dx to solve 1 and 2, and you have your result.....and I think this should be moved to homework or something.

3. Sep 5, 2012

### vish22

thanks universal i had just got that part,also in the next step of hint he separates the differentials like so-
md(v^2)=F(x)dx
now how is it possible to integrate the lhs wrt v^2??

4. Sep 6, 2012

### HallsofIvy

For any function, f, the integral of df is just f (plus the "constant of integration) - that's the "Fundamental Theorem of Calculus".

$$\int m d(v^2)= m\int d(v^2)= mv^2+ C$$
assuming that m is a does not depend on v.