a pationt recives an injection of 1.5 milligrams of a medication, and the amount remaining in the bloodstream t hours later is A(t)= 1.5e^-0.08t .find the instantaneous rate of change of this amount:

a) Immediately after the injection (time t=0)

To find the instantaneous rate of change immediately after the injection (t=0), we need to find the derivative of the function A(t).

The derivative of A(t) with respect to t can be found by applying the chain rule of differentiation.

First, let's rewrite the function A(t) using the exponential function's natural base (e) and simplifying it:

A(t) = 1.5 * e^(-0.08t)

Now, to find the derivative, differentiate each term of the function:

dA(t)/dt = (d/dt)(1.5) * (d/dt)(e^(-0.08t))

The derivative of a constant term (1.5) is zero, so the first term disappears.

dA(t)/dt = 0 * (d/dt)(e^(-0.08t))

Now, let's differentiate the exponential term, e^(-0.08t). The derivative of e^(-kt) with respect to t is -k * e^(-kt), where k is a constant.

Using this derivative rule, the derivative of e^(-0.08t) is:

dA(t)/dt = 0 * (-0.08) * e^(-0.08t)

Simplifying further:

dA(t)/dt = 0

The instantaneous rate of change immediately after the injection (t=0) is zero. This means that there is no change in the amount of medication in the bloodstream right after the injection.

To find the instantaneous rate of change of the amount remaining in the bloodstream immediately after the injection (time t=0), we need to take the derivative of the given function A(t) with respect to time (t).

A(t) = 1.5e^(-0.08t)

To find the derivative, we can use the power rule and chain rule:

dA/dt = -0.08 * 1.5e^(-0.08t)

The constant -0.08 comes from the exponent -0.08t, and the derivative of e^(-0.08t) is itself multiplied by -0.08.

Now, let's simplify this expression:

dA/dt = -0.12e^(-0.08t)

Therefore, the instantaneous rate of change of the amount remaining in the bloodstream immediately after the injection is -0.12e^(-0.08t).