Calculus Examples

$q (p) = \frac{1.2}{p} + 3.5$ , $q' (2)$

Step 1

Find the derivative of $q (p) = \frac{1.2}{p} + 3.5$ .

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Step 1.1

By the Sum Rule, the derivative of $\frac{1.2}{p} + 3.5$ with respect to $p$ is $\frac{d}{d p} [\frac{1.2}{p}] + \frac{d}{d p} [3.5]$ .

$\frac{d}{d p} [\frac{1.2}{p}] + \frac{d}{d p} [3.5]$

Step 1.2

Evaluate $\frac{d}{d p} [\frac{1.2}{p}]$ .

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Step 1.2.1

Since $1.2$ is constant with respect to $p$ , the derivative of $\frac{1.2}{p}$ with respect to $p$ is $1.2 \frac{d}{d p} [\frac{1}{p}]$ .

$1.2 \frac{d}{d p} [\frac{1}{p}] + \frac{d}{d p} [3.5]$

Step 1.2.2

Rewrite $\frac{1}{p}$ as $p^{- 1}$ .

$1.2 \frac{d}{d p} [p^{- 1}] + \frac{d}{d p} [3.5]$

Step 1.2.3

Differentiate using the Power Rule which states that $\frac{d}{d p} [p^{n}]$ is $n p^{n - 1}$ where $n = - 1$ .

$1.2 (- p^{- 2}) + \frac{d}{d p} [3.5]$

Step 1.2.4

Multiply $- 1$ by $1.2$ .

$- 1.2 p^{- 2} + \frac{d}{d p} [3.5]$

Step 1.3

Since $3.5$ is constant with respect to $p$ , the derivative of $3.5$ with respect to $p$ is $0$ .

$- 1.2 p^{- 2} + 0$

Step 1.4

Simplify.

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Step 1.4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- 1.2 \frac{1}{p^{2}} + 0$

Step 1.4.2

Combine terms.

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Step 1.4.2.1

Combine $- 1.2$ and $\frac{1}{p^{2}}$ .

$\frac{- 1.2}{p^{2}} + 0$

Step 1.4.2.2

Move the negative in front of the fraction.

$- \frac{1.2}{p^{2}} + 0$

Step 1.4.2.3

Add $- \frac{1.2}{p^{2}}$ and $0$ .

$- \frac{1.2}{p^{2}}$

Step 2

Replace the variable $p$ with $2$ in the expression.

$- \frac{1.2}{{(2)}^{2}}$

Step 3

Simplify.

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Step 3.1

Raise $2$ to the power of $2$ .

$- \frac{1.2}{4}$

Step 3.2

Divide $1.2$ by $4$ .

$- 1 \cdot 0.3$

Step 3.3

Multiply $- 1$ by $0.3$ .

$- 0.3$