If v= 0.5(x^2+x'^2) why does the derivative v'=xx'+x'x''?

Question

The first is taken from a lyaponav equation. I can't see where the jump comes from? The apostrophe is shown as a dot in the original source, presumably a derivative. what do i need to use chain rule?? I'm lost.

I agree, very eh! yet i have seen the same written in many documents.

Answer 1

The derivative of x^2 is 2x*(dx/dt) = 2xx'. The derivative of (x')^2 is 2*(x')*(dx'/dt) which is 2x'x". Multiply by the 0.5 coefficient.

Answer 2

Others have already given you the respond (confirm you verify them out your self until now accepting them). in case you sense relaxed with trinomial, then turn it right into a trinomial. inspite of each little thing, that could be a "quadratic" (from "quadratum", Latin for sq.) purely like all different one. x^2 - 36 is the comparable as x^2 + 0x - 36 you opt for 2 numbers that multiply to -36 and upload as much as 0 +6 and -6 ought to artwork x^2 + 0x - 36 = (x + 6)(x - 6) ------------------------------ A shortcut is to renowned it as a "distinction of squares", that could be a recipe you opt for to memorize. a^2 - b^2 = (a-b)(a+b) I write the version first, because of the fact there's a greater common recipe. In a "distinction of powers", the version of roots is usually a piece: a^2 - b^2 = (a-b)(a + b) a^3 - b^3 = (a-b)(a^2 + ab + b^2) a^4 - b^4 = (a-b)(a^3 + a^2b + ab^2 + b^3) etc a^7 - b^7 = (a-b)(a^6 + a^5b + a^4b^2 + ...)

Answer 3

eh