find $f(x)$. \n$f(x)=e^{\\sqrt{x - 5}}$ \nchoose the correct setup below to start differentiating the…

find $f(x)$. \n$f(x)=e^{\\sqrt{x - 5}}$ \nchoose the correct setup below to start differentiating the function. \n$\\bigcirc$ a. $f(x)=\\frac{d}{dx}(e^{\\sqrt{x}})-\\frac{d}{dx}(e^{\\sqrt{5}})$ \n$\\bigcirc$ b. $f(x)=\\frac{\\frac{d}{dx}(e^{\\sqrt{x}})}{\\frac{d}{dx}(e^{\\sqrt{5}})}$ \n$\\bigcirc$ c. $f(x)=\\frac{d}{dx}(e^{\\sqrt{x}})-e^{\\sqrt{5}}$ \n$\\bigcirc$ d. $f(x)=e^{\\sqrt{x - 5}}\\frac{d}{dx}(\\sqrt{x - 5})$

find $f(x)$. \n$f(x)=e^{\\sqrt{x - 5}}$ \nchoose the correct setup below to start differentiating the function. \n$\\bigcirc$ a. $f(x)=\\frac{d}{dx}(e^{\\sqrt{x}})-\\frac{d}{dx}(e^{\\sqrt{5}})$ \n$\\bigcirc$ b. $f(x)=\\frac{\\frac{d}{dx}(e^{\\sqrt{x}})}{\\frac{d}{dx}(e^{\\sqrt{5}})}$ \n$\\bigcirc$ c. $f(x)=\\frac{d}{dx}(e^{\\sqrt{x}})-e^{\\sqrt{5}}$ \n$\\bigcirc$ d. $f(x)=e^{\\sqrt{x - 5}}\\frac{d}{dx}(\\sqrt{x - 5})$

Answer

Explanation:

Step1: Apply the chain rule

The chain rule states that if (y = e^{u}) and (u = \sqrt{x - 5}), then (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). For (y = e^{u}), (\frac{dy}{du}=e^{u}). For (u=\sqrt{x - 5}=(x - 5)^{\frac{1}{2}}), using the power rule (\frac{d}{dx}(x^{n})=nx^{n - 1}), we have (\frac{du}{dx}=\frac{1}{2}(x - 5)^{-\frac{1}{2}}). Substituting (u = \sqrt{x - 5}) back, we get (f^{\prime}(x)=e^{\sqrt{x - 5}}\cdot\frac{d}{dx}(\sqrt{x - 5})).

Answer:

D. (f^{\prime}(x)=e^{\sqrt{x - 5}}\frac{d}{dx}(\sqrt{x - 5}))