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

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

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

Answer

Explanation:

Step1: Recall the chain rule

The chain rule for differentiating (y = e^{u(t)}) is (\frac{dy}{dt}=e^{u(t)}\cdot u^{\prime}(t)). For the function (y = 9e^{0.05t}-16e^{0.02t}), we differentiate each term separately. The derivative of (y = 9e^{0.05t}) using the chain rule: let (u = 0.05t), then (\frac{d}{dt}(9e^{0.05t})=9e^{0.05t}\cdot\frac{d}{dt}(0.05t)) The derivative of (y=- 16e^{0.02t}) using the chain rule: let (u = 0.02t), then (\frac{d}{dt}(-16e^{0.02t})=-16e^{0.02t}\cdot\frac{d}{dt}(0.02t))

Step2: Check each option

  • Option A: Incorrect application of the chain rule. The formula for the derivative of (a e^{u(t)}) is (a e^{u(t)}\cdot u^{\prime}(t)), not the product - rule - like form shown in this option.
  • Option B: Correct. Using the chain rule (\frac{d}{dt}(9e^{0.05t}-16e^{0.02t}) = 9e^{0.05t}\frac{d}{dt}(0.05t)-16e^{0.02t}\frac{d}{dt}(0.02t))
  • Option C: Wrong substitution of the inner - functions for the chain rule.
  • Option D: Incorrect application of the chain rule. The formula for the derivative of (a e^{u(t)}) is (a e^{u(t)}\cdot u^{\prime}(t)), not the form shown in this option.

Answer:

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