Question

find f(x).
f(x)=e^{\sqrt{x - 4}}
f(x)=\square

Explanation:

Step1: Let \( u = \sqrt{x - 4}=(x - 4)^{\frac{1}{2}}\)

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 \(u=(x - 4)^{\frac{1}{2}}\), \(\frac{du}{dx}=\frac{1}{2}(x - 4)^{\frac{1}{2}-1}\cdot\frac{d}{dx}(x - 4)\) (by the chain rule for \(y=(x - 4)^{n}\)). Since \(\frac{d}{dx}(x - 4)=1\), then \(\frac{du}{dx}=\frac{1}{2\sqrt{x - 4}}\).

Step3: Apply the chain rule \(\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}\)

Substitute \(\frac{df}{du}=e^{u}\) and \(\frac{du}{dx}=\frac{1}{2\sqrt{x - 4}}\) into the chain - rule formula. Since \(u = \sqrt{x - 4}\), we have \(f^{\prime}(x)=e^{\sqrt{x - 4}}\cdot\frac{1}{2\sqrt{x - 4}}\).

Answer:

\(f^{\prime}(x)=\frac{e^{\sqrt{x - 4}}}{2\sqrt{x - 4}}\)