differentiate the following function.\nf(t)=9e^{0.05t}-16e^{0.02t}\nchoose the correct setup below to start…

differentiate the following function.\nf(t)=9e^{0.05t}-16e^{0.02t}\nchoose the correct setup below to start differentiating the function.\na. \\( \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } - 16 e ^ { 0.02 t } \\right) = 9 e ^ { 0.05 t } \\frac { d } { d t } \\left( 16 e ^ { 0.02 t } \\right) - 16 e ^ { 0.02 t } \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } \\right) \\)\nb. \\( \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } - 16 e ^ { 0.02 t } \\right) = 9 e ^ { 0.05 t } \\frac { d } { d t } ( 0.05 t ) - 16 e ^ { 0.02 t } \\frac { d } { d t } ( 0.02 t ) \\)\nc. \\( \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } - 16 e ^ { 0.02 t } \\right) = 9 e ^ { 0.05 t } \\frac { d } { d t } ( 0.02 t ) - 16 e ^ { 0.02 t } \\frac { d } { d t } ( 0.05 t ) \\)\nd. \\( \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } - 16 e ^ { 0.02 t } \\right) = 16 e ^ { 0.02 t } \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } \\right) - 9 e ^ { 0.05 t } \\frac { d } { d t } \\left( 16 e ^ { 0.02 t } \\right) \\)\n\\( \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } - 16 e ^ { 0.02 t } \\right) = \\square \\)

differentiate the following function.\nf(t)=9e^{0.05t}-16e^{0.02t}\nchoose the correct setup below to start differentiating the function.\na. \\( \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } - 16 e ^ { 0.02 t } \\right) = 9 e ^ { 0.05 t } \\frac { d } { d t } \\left( 16 e ^ { 0.02 t } \\right) - 16 e ^ { 0.02 t } \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } \\right) \\)\nb. \\( \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } - 16 e ^ { 0.02 t } \\right) = 9 e ^ { 0.05 t } \\frac { d } { d t } ( 0.05 t ) - 16 e ^ { 0.02 t } \\frac { d } { d t } ( 0.02 t ) \\)\nc. \\( \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } - 16 e ^ { 0.02 t } \\right) = 9 e ^ { 0.05 t } \\frac { d } { d t } ( 0.02 t ) - 16 e ^ { 0.02 t } \\frac { d } { d t } ( 0.05 t ) \\)\nd. \\( \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } - 16 e ^ { 0.02 t } \\right) = 16 e ^ { 0.02 t } \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } \\right) - 9 e ^ { 0.05 t } \\frac { d } { d t } \\left( 16 e ^ { 0.02 t } \\right) \\)\n\\( \\frac { d } { d t } \\left( 9 e ^ { 0.05 t } - 16 e ^ { 0.02 t } \\right) = \\square \\)

Answer

Explanation:

Step1: Apply the derivative rule of sum/difference

The derivative of (u - v) is (u' - v'), where (u = 9e^{0.05t}) and (v=16e^{0.02t}). So (\frac{d}{dt}(9e^{0.05t}-16e^{0.02t})=\frac{d}{dt}(9e^{0.05t})-\frac{d}{dt}(16e^{0.02t}))

Step2: Apply the chain - rule for exponential functions

The chain - rule states that if (y = e^{ax}), then (y'=ae^{ax}). For (y = 9e^{0.05t}), let (a = 0.05), then (\frac{d}{dt}(9e^{0.05t})=9\times0.05e^{0.05t}) For (y = 16e^{0.02t}), let (a = 0.02), then (\frac{d}{dt}(16e^{0.02t})=16\times0.02e^{0.02t})

Step3: Calculate the derivatives

(\frac{d}{dt}(9e^{0.05t})=9\times0.05e^{0.05t}=0.45e^{0.05t}) (\frac{d}{dt}(16e^{0.02t})=16\times0.02e^{0.02t}=0.32e^{0.02t})

Step4: Find the final derivative

(\frac{d}{dt}(9e^{0.05t}-16e^{0.02t})=0.45e^{0.05t}-0.32e^{0.02t})

Answer:

(0.45e^{0.05t}-0.32e^{0.02t})