Answer:
See Below.
Step-by-step explanation:
We are given that:
[tex]\displaystyle I = I_0 e^{-kt}[/tex]
Where I₀ and k are constants.
And we want to prove that:
[tex]\displaystyle \frac{dI}{dt}+kI=0[/tex]
From the original equation, take the derivative of both sides with respect to t. Hence:
[tex]\displaystyle \frac{d}{dt}\left[I\right] = \frac{d}{dt}\left[I_0e^{-kt}\right][/tex]
Differentiate. Since I₀ is a constant:
[tex]\displaystyle \frac{dI}{dt} = I_0\left(\frac{d}{dt}\left[ e^{-kt}\right]\right)[/tex]
Using the chain rule:
[tex]\displaystyle \frac{dI}{dt} = I_0\left(-ke^{-kt}\right) = -kI_0e^{-kt}[/tex]
We have:
[tex]\displaystyle \frac{dI}{dt}+kI=0[/tex]
Substitute:
[tex]\displaystyle \left(-kI_0e^{-kt}\right) + k\left(I_0e^{-kt}\right) = 0[/tex]
Distribute and simplify:
[tex]\displaystyle -kI_0e^{-kt} + kI_0e^{-kt} = 0 \stackrel{\checkmark}{=}0[/tex]
Hence proven.
