differentiate the following function.\n\n$f(x)=e^{-3x + 3}$\n\nchoose the correct setup below to start…

differentiate the following function.\n\n$f(x)=e^{-3x + 3}$\n\nchoose the correct setup below to start differentiating the function.\n\n$\\bigcirc$ a. $\\frac{d}{dx}(e^{-3x + 3})=\\frac{d}{dx}(e^{-3x})+\\frac{d}{dx}(e^{3})$\n\n$\\bigcirc$ b. $\\frac{d}{dx}(e^{-3x + 3})=e^{-3x + 3}\\frac{d}{dx}(-3x + 3)$\n\n$\\bigcirc$ c. $\\frac{d}{dx}(e^{-3x + 3})=e^{3}+\\frac{d}{dx}(e^{-3x})$\n\n$\\bigcirc$ d. $\\frac{d}{dx}(e^{-3x + 3})=\\frac{d}{dx}(e^{-3x})\\cdot\\frac{d}{dx}(e^{3})$\n\n$\\frac{d}{dx}(e^{-3x + 3})=\\square$\n

differentiate the following function.\n\n$f(x)=e^{-3x + 3}$\n\nchoose the correct setup below to start differentiating the function.\n\n$\\bigcirc$ a. $\\frac{d}{dx}(e^{-3x + 3})=\\frac{d}{dx}(e^{-3x})+\\frac{d}{dx}(e^{3})$\n\n$\\bigcirc$ b. $\\frac{d}{dx}(e^{-3x + 3})=e^{-3x + 3}\\frac{d}{dx}(-3x + 3)$\n\n$\\bigcirc$ c. $\\frac{d}{dx}(e^{-3x + 3})=e^{3}+\\frac{d}{dx}(e^{-3x})$\n\n$\\bigcirc$ d. $\\frac{d}{dx}(e^{-3x + 3})=\\frac{d}{dx}(e^{-3x})\\cdot\\frac{d}{dx}(e^{3})$\n\n$\\frac{d}{dx}(e^{-3x + 3})=\\square$\n

Answer

Explanation:

Step1: Apply the chain rule

The chain rule states that if (y = e^{u}), then (\frac{dy}{dx}=e^{u}\cdot\frac{du}{dx}). Here (u=-3x + 3), so (\frac{d}{dx}(e^{-3x + 3})=e^{-3x + 3}\cdot\frac{d}{dx}(-3x + 3))

Step2: Differentiate the inner function

Differentiate (u=-3x + 3) with respect to (x). Using the power rule (\frac{d}{dx}(ax^{n})=nax^{n - 1}) (for (n = 1) in (-3x) and the derivative of a constant (3) is (0)), we get (\frac{d}{dx}(-3x+3)=-3)

Step3: Multiply the results

Substitute (\frac{d}{dx}(-3x + 3)=-3) into (e^{-3x + 3}\cdot\frac{d}{dx}(-3x + 3)), so (e^{-3x + 3}\cdot(-3)=-3e^{-3x + 3})

Answer:

(-3e^{-3x + 3})