Question

hw 8 - derivative rules section 2.3: problem 9 (1 point) if $f(x)=\frac{3x^{2}+8x + 28}{sqrt{x}}$, find $f(x)$. $f(x)=$

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\frac{3x^{2}+8x + 28}{\sqrt{x}}$ as $f(x)=3x^{\frac{3}{2}}+8x^{\frac{1}{2}}+28x^{-\frac{1}{2}}$ using the rule $\frac{a + b + c}{d}=\frac{a}{d}+\frac{b}{d}+\frac{c}{d}$ and $\frac{x^{n}}{\sqrt{x}}=x^{n-\frac{1}{2}}$.

Step2: Apply the power - rule for derivatives

The power - rule states that if $y = x^{n}$, then $y^\prime=nx^{n - 1}$.
For $y_1 = 3x^{\frac{3}{2}}$, $y_1^\prime=3\times\frac{3}{2}x^{\frac{3}{2}-1}=\frac{9}{2}x^{\frac{1}{2}}$.
For $y_2 = 8x^{\frac{1}{2}}$, $y_2^\prime=8\times\frac{1}{2}x^{\frac{1}{2}-1}=4x^{-\frac{1}{2}}$.
For $y_3 = 28x^{-\frac{1}{2}}$, $y_3^\prime=28\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=-14x^{-\frac{3}{2}}$.

Step3: Find the derivative of $f(x)$

$f^\prime(x)=y_1^\prime + y_2^\prime+y_3^\prime=\frac{9}{2}\sqrt{x}+\frac{4}{\sqrt{x}}-\frac{14}{x\sqrt{x}}$.

Answer:

$\frac{9}{2}\sqrt{x}+\frac{4}{\sqrt{x}}-\frac{14}{x\sqrt{x}}$