4(8+h)3-3(8+h)2+12=
(8+h)2(32+4h-3)+12=
(64+16h+h2)(29+4h)+12=
1856+256h
+464h+64h2
+29h2+4h3+12=
1856+720h+93h2+4h3+12=
4h3+93h2+720h+1868.
Consider x(t)=t4-t3+12t+k where k is an unknown constant.
dx/dt=4t3-3t2+12 which has a similar structure to the given expression.
x(t+h)=(t+h)4-(t+h)3+12(t+h)+k=t4+4t3h+6t2h2+4th3+h4-t3-3t2h-3th2-h3+12t+12h;
x(t+h)-x(t)=4t3h+6t2h2+4th3+h4-3t2h-3th2-h3+12h. Note that k disappears.
Now, consider h to be very small, so that terms involving h2, h3 and h4 are negligibly small, then:
x(t+h)-x(t)=4t3h-3t2h+12h.
x(t+h)-x(t) is the change in value of x when t changes in value by a very small amount h.
The rate of change is the gradient of the curve x(t)=(x(t+h)-x(t))/h=4t3-3t2+12.
This is the limit of the slope as h→0 at the point (t,x(t)). All we have to do now is plug in t=8 to find the instantaneous rate of change:
4(83)-3(82)+12=2048-192+12=1868.
Note that this is exactly what we get when we plug h=0 into 4h3+93h2+720h+1868.
I suspect that the question you posed was part of a different, but related, question. You were given a function of t and asked to find the derivative from first principles using h as a small increment of t. This is how the calculus works when working out a derivative (rate of change at a point).