Question

1. if ( f(x)=sqrt{x}+\frac{3}{sqrt{x}} ), then ( f^{prime}(4)= )

Explanation:

Step1: Rewrite the function

Rewrite \(f(x)=\sqrt{x}+\frac{3}{\sqrt{x}}\) as \(f(x)=x^{\frac{1}{2}} + 3x^{-\frac{1}{2}}\).

Step2: Differentiate using the power rule

The power rule is \((x^n)^\prime=nx^{n - 1}\).
For \(y = x^{\frac{1}{2}}\), \(y^\prime=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}\).
For \(y = 3x^{-\frac{1}{2}}\), \(y^\prime=3\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=-\frac{3}{2}x^{-\frac{3}{2}}\).
So \(f^\prime(x)=\frac{1}{2}x^{-\frac{1}{2}}-\frac{3}{2}x^{-\frac{3}{2}}\).

Step3: Substitute \(x = 4\)

Substitute \(x = 4\) into \(f^\prime(x)\):
\(f^\prime(4)=\frac{1}{2}\times4^{-\frac{1}{2}}-\frac{3}{2}\times4^{-\frac{3}{2}}\).
Since \(4^{-\frac{1}{2}}=\frac{1}{\sqrt{4}}=\frac{1}{2}\) and \(4^{-\frac{3}{2}}=\frac{1}{4^{\frac{3}{2}}}=\frac{1}{(\sqrt{4})^3}=\frac{1}{8}\).
\(f^\prime(4)=\frac{1}{2}\times\frac{1}{2}-\frac{3}{2}\times\frac{1}{8}\).
\(f^\prime(4)=\frac{1}{4}-\frac{3}{16}\).
\(f^\prime(4)=\frac{4 - 3}{16}\).

Answer:

\(\frac{1}{16}\)