# In-Class Activity 2.5

## Question 1

Write as a sum of power functions:

### a. $\dfrac{6x^3+2x^2-x}{x^2}$

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<summary> Example: </summary>

Write as a sum of power functions:

$\dfrac{5x^4+3x^3-2x^2}{x^3}$

\begin{align}
\dfrac{5x^4+3x^3-2x^2}{x^3}&=\dfrac{5x^4}{x^3}+\dfrac{3x^3}{x^3}-\dfrac{2x^2}{x^3} \\
&=5x^{4-3}+3x^{3-3} -2x^{2-3} \\
&=5x^1+3x^0-2x^{-1} \\
&=5x+3-2x^{-1}
\end{align}

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<summary> Solution: </summary>

\begin{align}
\dfrac{6x^3+2x^2-x}{x^2}&=\dfrac{6x^3}{x^2}+\dfrac{2x^2}{x^2}-\dfrac{x}{x^2} \\
&=6x^{3-2}+2x^{2-2}-x^{1-2} \\
&=6x^1+2x^0-x^{-1} \\
&=6x+2-x^{-1}
\end{align}

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### b. $\dfrac{-2x^3+5x^2-2x}{\sqrt{x}}$

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<summary> Example: </summary>

Write as a sum of power functions:

$\dfrac{-4x^4+2x^3-5x}{\sqrt{x}}$

\begin{align}
\dfrac{-4x^4+2x^3-5x}{\sqrt{x}} &= \dfrac{-4x^4}{\sqrt{x}} + \dfrac{2x^3}{\sqrt{x}} - \dfrac{5x}{\sqrt{x}} \\
&= \dfrac{-4x^4}{x^{\frac{1}{2}}} + \dfrac{2x^3}{x^{\frac{1}{2}}} - \dfrac{5x}{x^{\frac{1}{2}}} \\
&= -4x^{4-\frac{1}{2}} + 2x^{3-\frac{1}{2}} - 5x^{1-\frac{1}{2}} \\
&= -4x^{\frac{7}{2}} + 2x^{\frac{5}{2}} - 5x^{\frac{1}{2}} \\
\end{align}

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<summary> Solution: </summary>

\begin{align}
\dfrac{-2x^3+5x^2-2x}{\sqrt{x}}&=\dfrac{-2x^3+5x^2-2x}{x^{1/2}} \\
&=\dfrac{-2x^3}{x^{1/2}}+\dfrac{5x^2}{x^{1/2}}-\dfrac{2x}{x^{1/2}} \\
&=-2x^{3-1/2}+5x^{2-1/2}-2x^{1-1/2} \\
&=-2x^{6/2-1/2}+5x^{4/2-1/2}-2x^{2/2-1/2} \\
&=-2x^{5/2}+5x^{3/2}-2x^{1/2}
\end{align}

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## Question 2

Write as a sum of power functions.

### a. $\dfrac{1}{6}x(3x-4)(6x+2)$

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<summary> Example: </summary>

$\dfrac{1}{5}x(4x+3)(5x-2)$

\begin{align}
\dfrac{1}{5}x(4x+3)(5x-2)&=\dfrac{x(4x+3)(5x-2)}{5} \\
&=\dfrac{x(20x^2-8x+15x-6)}{5} \\
&=\dfrac{x(20x^2+7x-6)}{5} \\
&=\dfrac{20x^3+7x^2-6x}{5} \\
&=\dfrac{20}{5}x^3+\dfrac{7}{5}x^2-\dfrac{6}{5}x \\
&=4x^3+\dfrac{7}{5}x^2-\dfrac{6}{5}x
\end{align}

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<summary> Solution: </summary>

\begin{align}
\dfrac{1}{6}x(3x-4)(6x+2)&=\dfrac{x(3x-4)(6x+2)}{6} \\
&=\dfrac{x(18x^2+6x-24x-8)}{6} \\
&=\dfrac{x(18x^2-18x-8)}{6} \\
&=\dfrac{18x^3-18x^2-8x}{6} \\
&=\dfrac{18}{6} x^3-\dfrac{18}{6}x^2-\dfrac{8}{6}x \\
&=3x^3-3x^2-\dfrac{4}{3}x
\end{align}

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### b. $\dfrac{1}{2}\sqrt{x}(4-2x)(3x^2-1)$

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<summary> Example: </summary>

$\dfrac{1}{3} \sqrt{x}(5-3x)(2x^2-7)$

\begin{align}
\dfrac{1}{3} \sqrt{x}(5-3x)(2x^2-7) &= \dfrac{\sqrt{x}(5-3x)(2x^2-7)}{3} \\
&= \dfrac{\sqrt{x} \left(10x^2 - 35 - 6x^3 + 21x \right)}{3} \\
&= \dfrac{\sqrt{x} \left(-6x^3 + 10x^2 + 21x - 35 \right)}{3} \\
&= \dfrac{x^{\frac{1}{2}} \left(-6x^3 + 10x^2 + 21x - 35 \right)}{3} \\
&= \dfrac{-6x^{3 + \frac{1}{2}} + 10x^{2 + \frac{1}{2}} + 21x^{1 + \frac{1}{2}} - 35x^{\frac{1}{2}}}{3} \\
&= \dfrac{-6x^{\frac{7}{2}}}{3} + \dfrac{10x^{\frac{5}{2}}}{3} + \dfrac{21x^{\frac{3}{2}}}{3} - \dfrac{35x^{\frac{1}{2}}}{3} \\
&= -2x^{\frac{7}{2}} + \dfrac{10}{3}x^{\frac{5}{2}} + 7x^{\frac{3}{2}} - \dfrac{35}{3}x^{\frac{1}{2}}
\end{align}

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<summary> Solution: </summary>

\begin{align}
\dfrac{1}{2}\sqrt{x}(4-2x)(3x^2-1)&=\dfrac{x^{1/2}(4-2x)(3x^2-1)}{2} \\
&=\dfrac{x^{1/2}(12x^2-4-6x^3+2x)}{2} \\
&=\dfrac{12x^{2+1/2}-4x^{1/2}-6x^{3+1/2}+2x^{1+1/2}}{2} \\
&=\dfrac{12x^{5/2}-4x^{1/2}-6x^{7/2}+2x^{3/2}}{2} \\
&=\dfrac{12}{2}x^{5/2}-\dfrac{4}{2}x^{1/2}-\dfrac{6}{2}x^{7/2}+\dfrac{2}{2}x^{3/2} \\
&=6x^{5/2}-2x^{1/2}-3x^{7/2}+x^{3/2}
\end{align}

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## Question 3

Find the indicated derivative:

### a. \begin{align} y&=x^{-3} \\ y'&= \end{align}

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<summary> Example: </summary>

If $y=x^{-4}$, then the derivative is found by the power rule $(x^n)'=nx^{n-1}$

\begin{align}
y'&=-4x^{-4-1} \\
&=-4x^{-5} \\
&=-\dfrac{4}{x^5}
\end{align}

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<summary> Solution: </summary>

If $y=x^{-3}$, then the derivative is found by the power rule $(x^n)'=nx^{n-1}$

\begin{align}
y'&=-3x^{-3-1} \\
&=-3x^{-4} \\
&=-\dfrac{3}{x^4}
\end{align}

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### b. \begin{align} f(x)&=x^{5/2} \\ f'(x)&= \end{align}

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<summary> Example: </summary>

If $y=x^{7/3}$, then the derivative is found by the power rule $(x^n)'=nx^{n-1}$

\begin{align}
y'&=\dfrac{7}{3} x^{7/3-1} \\
&=\dfrac{7}{3}x^{7/3-3/3} \\
&=\dfrac{7}{3}x^{4/3}
\end{align}

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<summary> Solution: </summary>

If $y=x^{5/2}$, then the derivative is found by the power rule $(x^n)'=nx^{n-1}$

\begin{align}
y'&=\dfrac{5}{2} x^{5/2-1} \\
&=\dfrac{5}{2}x^{5/2-2/2} \\
&=\dfrac{5}{2}x^{3/2}
\end{align}

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### c. $$\dfrac{d}{dx} \left(\dfrac{5x^3}{4}-\dfrac{2}{5x^3}\right)$$

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<summary> Example: </summary>

$\dfrac{d}{dx} \left( \dfrac{3x^5}{2}-\dfrac{6}{7x^5}\right)$

Rewrite the inside as sum of power functions:

\begin{align}
\dfrac{3x^5}{2}-\dfrac{6}{7x^5}&=\dfrac{3}{2}x^5 - \dfrac{6}{7}x^{-5}
\end{align}

Then we can take the derivative using the power rule:

\begin{align}
\dfrac{d}{dx} \left(\dfrac{3}{2}x^5 - \dfrac{6}{7}x^{-5}\right)&=\dfrac{3}{2} \left( 5x^{5-1}\right) - \dfrac{6}{7} \left(-5x^{-5-1}\right) \\
&=\dfrac{15}{2}x^4 + \dfrac{30}{7}x^{-6} \\
&=\dfrac{15}{2}x^4+\dfrac{30}{7x^6}
\end{align}

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<summary> Solution: </summary>

$\dfrac{d}{dx} \left(\dfrac{5x^3}{4}-\dfrac{2}{5x^3}\right)$

Rewrite the inside as sum of power functions:

\begin{align}
\dfrac{5x^3}{4}-\dfrac{2}{5x^3}&=\dfrac{5}{4}x^3-\dfrac{2}{5}x^{-3}
\end{align}

Then we can take the derivative using the power rule:

\begin{align}
\dfrac{d}{dx} \left(\dfrac{5}{4}x^3-\dfrac{2}{5}x^{-3}\right)&=\dfrac{5}{4}(3x^{3-1}-\dfrac{2}{5}(-3x^{-3-1}) \\
&=\dfrac{15}{4}x^2+\dfrac{6}{5}x^{-4} \\
&=\dfrac{15}{4}x^2+\dfrac{6}{5x^4}
\end{align}

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### d. \begin{align} f(x)&=\dfrac{2x^5-4x^3+2x}{x^3} \\ f'(x)&= \end{align}

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<summary> Example: </summary>

If $$f(x)=\dfrac{3x^6-2x^4+3x}{x^4}$$ then we must rewrite as sum of power functions:

\begin{align}
\dfrac{3x^6-2x^4+3x}{x^4}&=3x^{6-4} -2x^{4-4}+3x^{1-4} \\
&=3x^2-2x^0+3x^{-3}
\end{align}

Now we can take the derivative by the power rule $(x^n)'=nx^{n-1}$:
\begin{align}
f'(x)&=\left(3x^2-2x^0+3x^{-3} \right)' \\
&=3(2x^{2-1}) -2(0x^{0-1})+3(-3x^{-3-1}) \\
&=6x^1-0-9x^{-4} \\
&=6x-9x^{-4} \\
&=6x- \dfrac{9}{x^4}
\end{align}

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<summary> Solution: </summary>

If $$f(x)=\dfrac{2x^5-4x^3+2x}{x^3}$$ then we must rewrite as sum of power functions:

\begin{align}
\dfrac{2x^5-4x^3+2x}{x^3}&=\dfrac{2x^5}{x^3}-\dfrac{4x^3}{x^3}+\dfrac{2x}{x^3} \\
&=2x^{5-3}-4x^{3-3} +2x^{1-3} \\
&=2x^2-4x^0 +2x^{-2}
\end{align}

Now we can take the derivative by the power rule $(x^n)'=nx^{n-1}$:
\begin{align}
f'(x)&=\left(2x^2-4x^0 +2x^{-2} \right)' \\
&=2(2x^{2-1}) -4(0x^{0-1})+2(-2x^{-2-1}) \\
&=4x^1-0-4x^{-3} \\
&=4x-4x^{-3} \\
&=4x-\dfrac{4}{x^3}
\end{align}

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