Explaining Chain Rule: vdv/dx=1/2(dv^2)/dx

[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 that's 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
Could someone please explain this?

 

[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.

 

[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??

 

[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.