differentiate the following function.\n$f(t)=7e^{0.04t}-18e^{0.02t}$\n$\frac{d}{dt}(7e^{0.04t}-18e^{0.02t})=s…

Explanation:

Step1: Differentiate (7e^{0.04t})

Use the chain rule (\frac{d}{dt}(ae^{bt}) = abe^{bt}). For (a = 7) and (b=0.04), we have (\frac{d}{dt}(7e^{0.04t})=7\times0.04e^{0.04t}=0.28e^{0.04t})

Step2: Differentiate (- 18e^{0.02t})

For (a=-18) and (b = 0.02), using the chain rule (\frac{d}{dt}(ae^{bt})=abe^{bt}), we get (\frac{d}{dt}(-18e^{0.02t})=-18\times0.02e^{0.02t}=-0.36e^{0.02t})

Step3: Combine the results

(\frac{d}{dt}(7e^{0.04t}-18e^{0.02t})=\frac{d}{dt}(7e^{0.04t})-\frac{d}{dt}(18e^{0.02t})=0.28e^{0.04t}-0.36e^{0.02t})

Answer:

(0.28e^{0.04t}-0.36e^{0.02t})