Question

which of the following equations would represent $f(x)$ if $f(x)=\frac{x^{2}+2x}{\sqrt{x}}$?
$\frac{3\sqrt{x}+1}{2\sqrt{x}}$
$4\sqrt{x^{3}}+4\sqrt{x}$
$\frac{3x + 1}{\sqrt{x}}$
$\frac{3x+2}{2\sqrt{x}}$
clear my selection

Explanation:

Step1: Simplify the function

First, rewrite \(f(x)=\frac{x^{2}+2x}{\sqrt{x}}\) as \(f(x)=x^{\frac{3}{2}} + 2x^{\frac{1}{2}}\) using the rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\) (\(a=x\), for \(x^{2}\div x^{\frac{1}{2}}=x^{2-\frac{1}{2}}=x^{\frac{3}{2}}\) and \(2x\div x^{\frac{1}{2}}=2x^{1-\frac{1}{2}}=2x^{\frac{1}{2}}\)).

Step2: Differentiate term - by - term

Use the power rule \((x^{n})^\prime=nx^{n - 1}\).
For \(y = x^{\frac{3}{2}}\), \(y^\prime=\frac{3}{2}x^{\frac{3}{2}-1}=\frac{3}{2}x^{\frac{1}{2}}\).
For \(y = 2x^{\frac{1}{2}}\), \(y^\prime=2\times\frac{1}{2}x^{\frac{1}{2}-1}=x^{-\frac{1}{2}}\).

Step3: Combine the derivatives

\(f^\prime(x)=\frac{3}{2}x^{\frac{1}{2}}+x^{-\frac{1}{2}}\).
Then, get a common denominator: \(f^\prime(x)=\frac{3x + 2}{2\sqrt{x}}\) (since \(x^{\frac{1}{2}}=\sqrt{x}\) and \(x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}\), \(\frac{3}{2}\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{3x + 2}{2\sqrt{x}}\)).

Answer:

\(\frac{3x + 2}{2\sqrt{x}}\) (the fourth option)