find ( f^{prime}(x) ).\n\n( f(x)=e^{sqrt{x - 18}} )\n\n( f^{prime}(x)= )

Explanation:

Step1: Let ( u = \sqrt{x - 18}=(x - 18)^{\frac{1}{2}})

The function (f(x)=e^{u}). By the chain rule (\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}). First, find (\frac{df}{du}). Since (y = e^{u}), then (\frac{df}{du}=e^{u}).

Step2: Find (\frac{du}{dx})

Using the power rule (\frac{d}{dx}(x^{n})=nx^{n - 1}), for (u=(x - 18)^{\frac{1}{2}}), (\frac{du}{dx}=\frac{1}{2}(x - 18)^{\frac{1}{2}-1}\cdot\frac{d}{dx}(x - 18)). Since (\frac{d}{dx}(x - 18)=1), then (\frac{du}{dx}=\frac{1}{2\sqrt{x - 18}}).

Step3: Apply the chain rule

(\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}). Substitute (\frac{df}{du}=e^{u}) and (\frac{du}{dx}=\frac{1}{2\sqrt{x - 18}}) into the chain - rule formula. Since (u = \sqrt{x - 18}), we have (f^{\prime}(x)=e^{\sqrt{x - 18}}\cdot\frac{1}{2\sqrt{x - 18}}).

Answer:

(f^{\prime}(x)=\frac{e^{\sqrt{x - 18}}}{2\sqrt{x - 18}})