An office machine is purchased for $6000. Under certain assumptions, its salvage value, V, in dollars, is depreciated according to a method called double declining balance, by basically 71% each year, and is given by V(t) = 6000(0.71)^t, where t is the time in years after purchase.
Find V'(t)
should be easy once you recall that
d/dx a^x = a^x ln(a)
To find V'(t), we need to take the derivative of the function V(t) = 6000(0.71)^t with respect to t.
The derivative of an exponentiation of a constant, like (0.71)^t, can be found using the chain rule. The general form of the chain rule states that if we have a function g(x) raised to the power of another function f(x), then the derivative can be found using the following formula:
(d/dx) [g^f(x)] = g^f(x) * (d/dx)[f(x)] * ln(g)
Let's apply this to our function V(t) = 6000(0.71)^t.
First, we'll find the derivative of the exponent part, which is t: (d/dt)(t) = 1.
Next, we'll find the derivative of the base part, which is 0.71. Since it's a constant, its derivative is 0.
Now, we can apply the chain rule formula: V'(t) = 6000(0.71)^t * 1 * ln(0.71).
Simplifying this expression further: V'(t) = 6000(0.71)^t * ln(0.71).
Therefore, the derivative of V(t) with respect to t, V'(t), is given by V'(t) = 6000(0.71)^t * ln(0.71).
Note: Since the salvage value is decreasing over time, it is expected that the derivative will be negative, indicating a decreasing rate of change.