Question

if (f(x)=sqrt{x}+\frac{3}{sqrt{x}}), then (f(4)=)
(a) (\frac{1}{16})
(b) (\frac{5}{16})
(c) 1
(d) (\frac{7}{2})
(e) (\frac{9}{4})

Explanation:

Step1: Rewrite function

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

Step2: Differentiate using power - rule

The power - rule is $\frac{d}{dx}(x^n)=nx^{n - 1}$. So $f'(x)=\frac{1}{2}x^{\frac{1}{2}-1}+3\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}-\frac{3}{2}x^{-\frac{3}{2}}$.

Step3: Substitute $x = 4$

Substitute $x = 4$ into $f'(x)$. We have $f'(4)=\frac{1}{2}(4)^{-\frac{1}{2}}-\frac{3}{2}(4)^{-\frac{3}{2}}$. Since $(4)^{-\frac{1}{2}}=\frac{1}{\sqrt{4}}=\frac{1}{2}$ and $(4)^{-\frac{3}{2}}=\frac{1}{(\sqrt{4})^3}=\frac{1}{8}$, then $f'(4)=\frac{1}{2}\times\frac{1}{2}-\frac{3}{2}\times\frac{1}{8}=\frac{1}{4}-\frac{3}{16}=\frac{4 - 3}{16}=\frac{1}{16}$.

Answer:

A. $\frac{1}{16}$