Question

hw 8 - derivative rules section 2.3: problem 1 (1 point) write the following expressions as $x^{p}$ for some constant $p$. $sqrt{x}=square$ $sqrt3{x}=square$ $sqrt{x^{3}}=square$ $sqrt3{x^{2}}=square$ $\frac{1}{sqrt{x}}=square$ $\frac{1}{sqrt3{x}}=square$

Explanation:

Step1: Recall radical - exponent rule

The rule is $\sqrt[n]{x^m}=x^{\frac{m}{n}}$, and $\frac{1}{x^n}=x^{-n}$.

Step2: Rewrite $\sqrt{x}$

$\sqrt{x}=x^{\frac{1}{2}}$ since $n = 2$ and $m = 1$ in $\sqrt[n]{x^m}$.

Step3: Rewrite $\sqrt[3]{x}$

$\sqrt[3]{x}=x^{\frac{1}{3}}$ as $n = 3$ and $m = 1$ in $\sqrt[n]{x^m}$.

Step4: Rewrite $\sqrt{x^3}$

$\sqrt{x^3}=x^{\frac{3}{2}}$ with $n = 2$ and $m = 3$ in $\sqrt[n]{x^m}$.

Step5: Rewrite $\sqrt[5]{x^2}$

$\sqrt[5]{x^2}=x^{\frac{2}{5}}$ as $n = 5$ and $m = 2$ in $\sqrt[n]{x^m}$.

Step6: Rewrite $\frac{1}{\sqrt{x}}$

$\frac{1}{\sqrt{x}}=\frac{1}{x^{\frac{1}{2}}}=x^{-\frac{1}{2}}$ using $\frac{1}{x^n}=x^{-n}$.

Step7: Rewrite $\frac{1}{\sqrt[3]{x}}$

$\frac{1}{\sqrt[3]{x}}=\frac{1}{x^{\frac{1}{3}}}=x^{-\frac{1}{3}}$ using $\frac{1}{x^n}=x^{-n}$.

Answer:

$\sqrt{x}=x^{\frac{1}{2}}$
$\sqrt[3]{x}=x^{\frac{1}{3}}$
$\sqrt{x^3}=x^{\frac{3}{2}}$
$\sqrt[5]{x^2}=x^{\frac{2}{5}}$
$\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$
$\frac{1}{\sqrt[3]{x}}=x^{-\frac{1}{3}}$