Question

section 2.5: product and quotient
score: 30/80 answered: 3/8
question 4
if $f(x)=\frac{5x^{2}+8x + 7}{sqrt{x}}$, then:
$f(x)=$
$f(2)=$

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\frac{5x^{2}+8x + 7}{\sqrt{x}}$ as $f(x)=5x^{\frac{3}{2}}+8x^{\frac{1}{2}}+7x^{-\frac{1}{2}}$.

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$.
For $y = 5x^{\frac{3}{2}}$, $y^\prime=5\times\frac{3}{2}x^{\frac{3}{2}-1}=\frac{15}{2}x^{\frac{1}{2}}$.
For $y = 8x^{\frac{1}{2}}$, $y^\prime=8\times\frac{1}{2}x^{\frac{1}{2}-1}=4x^{-\frac{1}{2}}$.
For $y = 7x^{-\frac{1}{2}}$, $y^\prime=7\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=-\frac{7}{2}x^{-\frac{3}{2}}$.
So, $f^\prime(x)=\frac{15}{2}\sqrt{x}+\frac{4}{\sqrt{x}}-\frac{7}{2x\sqrt{x}}$.

Step3: Evaluate $f^\prime(2)$

Substitute $x = 2$ into $f^\prime(x)$.
$f^\prime(2)=\frac{15}{2}\sqrt{2}+\frac{4}{\sqrt{2}}-\frac{7}{2\times2\times\sqrt{2}}$.
First, make a common denominator of $2\sqrt{2}$:
$\frac{15}{2}\sqrt{2}=\frac{15\times2}{4}\sqrt{2}=\frac{30}{4}\sqrt{2}$, $\frac{4}{\sqrt{2}}=\frac{8}{2\sqrt{2}}$, and $-\frac{7}{4\sqrt{2}}$.
$f^\prime(2)=\frac{30\sqrt{2}}{4}+\frac{8}{2\sqrt{2}}-\frac{7}{4\sqrt{2}}=\frac{30\sqrt{2}}{4}+\frac{16}{4\sqrt{2}}-\frac{7}{4\sqrt{2}}=\frac{30\sqrt{2}}{4}+\frac{9}{4\sqrt{2}}=\frac{60 + 9}{4\sqrt{2}}=\frac{69}{4\sqrt{2}}=\frac{69\sqrt{2}}{8}$.

Answer:

$f^\prime(x)=\frac{15}{2}\sqrt{x}+\frac{4}{\sqrt{x}}-\frac{7}{2x\sqrt{x}}$; $f^\prime(2)=\frac{69\sqrt{2}}{8}$