How do you differentiate #23000(0.93)^t#?
1 Answer
Jun 1, 2018
If you mean differentiate w.r.t
#z(t) = alpha beta^t, qquad {(alpha = 23000),(beta = 0.93):}#
Then maybe use log rules:
#ln z = ln alpha + t ln beta#
Differentiate (?):

#1/ z dz = 0 + dt * ln beta# 
#implies (dz)/(dt) = z * ln beta = alpha beta^t * ln beta#
This means that:
# (dz)/(dt) = ln(0.93) * 23000 * 0.93^t #
A calculator kills that one stone dead.