Help a calculus babby out.
I don't see where Feynman gets this extra "v" from. What am I doing wrong?
I've never done calculus outside of a pure math context before, and im finding it pretty tough.
d/dt v^2 = 2dv/dt * v
>>7971539
Sorry to be so stupid, but like I said I have only done differentiation in pure maths before.
In maths, I would do pic related. Why can't I do the same for Physics?
>>7971542
Because you aren't applying the chain rule properly
D/dx of f^2 = derivative of outer function times derivative of inner function = 2 f . Df/dx
D/dx x^2 = 2x = 2x .1 because d/dx of x = 1
>>7971542
because d/dt(x) = 1; ergo d/dt(x^2) = 2x(d/dt(x)) = 2x
