Question

solve the problem.
a business estimates that the salvage value v of a piece of machinery after t years is given by
v(t) = $36,000e^{-42t}.
find the formula for the rate of change of the salvage value.
\\( v(t) = 36,000e^{-42t} \\)
\\( v(t) = -1,512,000e^{-t} \\)
\\( v(t) = -1,512,000e^{-42t} \\)
\\( v(t) = 1,512,000e^{-42t} \\)

Explanation:

Step1: Recall Derivative of Exponential Function

The derivative of \( e^{u} \) with respect to \( t \) is \( e^{u} \cdot u' \) (chain rule), where \( u \) is a function of \( t \). Here, \( V(t) = 36000e^{-42t} \), so \( u = -42t \).

Step2: Differentiate \( V(t) \)

First, the derivative of the constant multiple \( 36000 \) times a function is \( 36000 \) times the derivative of the function. The derivative of \( e^{-42t} \) is \( e^{-42t} \cdot (-42) \) (since \( u' = -42 \) for \( u = -42t \)). So, \( V'(t)=36000\times(-42)e^{-42t} \).

Step3: Calculate the Constant Multiple

Calculate \( 36000\times(-42) \): \( 36000\times42 = 1512000 \), so \( 36000\times(-42)= - 1512000 \). Thus, \( V'(t)=-1512000e^{-42t} \).

Answer:

\( V'(t) = -1,512,000e^{-42t} \) (the third option)