if f(x) = x^n
then d f(x)/dx = n x^(n-1)
if f(x) = x^2
then d f(x)/dx = 2 x
so if f(t) = t^2
df(t)/dt = 2 t
the constant multiple remains (three times the function has three times the slope) .
if f(t) = 2.1 t^2
d f(t)/dt = 4.2 t
Now about that constant B
The slope of a constant is zero.
d/dt (A t^2 + B)
= 2 A t + 0
the end except to prove that d/dx (x^n) = n x^(n-1)
f(x+dx) = (x+dx)^n
(x+dx)^n = x^n + C(n,1) x^(n-1) dx + C(n,2) x^(n-2) (dx)^2 ....... +dx^n
C(n,k) = binomial coefficient
= n!/[(n-k!) k!]
so for Cn,1)
C(n,1) = n!/[(n-1)!] = n (lo and behold)
f(x+dx) = x^n + n x^(n-1)dx + terms in dx^2 through dx^n
f(x+dx) -f(x) = n x^(n-1) dx + terms in dx^2 to dx^n
df(x)/dx = [f(x+dx)-f(x)]/dx
as dx ---> 0
= n x^(n-1)
f(x) = x^n
Ignore the last two lines, I was thinking out loud.
You may not need that proof at the end that d x^n/dx = n x^(n-1) but it is not a bad idea to know where the derivative comes from.
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