Calculus Examples

$f (t) = \frac{9}{25 - t^{2}}$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate using the Constant Multiple Rule.

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Step 1.1.1.1

Since $9$ is constant with respect to $t$ , the derivative of $\frac{9}{25 - t^{2}}$ with respect to $t$ is $9 \frac{d}{d t} [\frac{1}{25 - t^{2}}]$ .

$9 \frac{d}{d t} [\frac{1}{25 - t^{2}}]$

Step 1.1.1.2

Rewrite $\frac{1}{25 - t^{2}}$ as ${(25 - t^{2})}^{- 1}$ .

$9 \frac{d}{d t} [{(25 - t^{2})}^{- 1}]$

Step 1.1.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{- 1}$ and $g (t) = 25 - t^{2}$ .

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Step 1.1.2.1

To apply the Chain Rule, set $u$ as $25 - t^{2}$ .

$9 (\frac{d}{d u} [u^{- 1}] \frac{d}{d t} [25 - t^{2}])$

Step 1.1.2.2

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

$9 (- u^{- 2} \frac{d}{d t} [25 - t^{2}])$

Step 1.1.2.3

Replace all occurrences of $u$ with $25 - t^{2}$ .

$9 (- {(25 - t^{2})}^{- 2} \frac{d}{d t} [25 - t^{2}])$

Step 1.1.3

Differentiate.

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Step 1.1.3.1

Multiply $- 1$ by $9$ .

$- 9 ({(25 - t^{2})}^{- 2} \frac{d}{d t} [25 - t^{2}])$

Step 1.1.3.2

By the Sum Rule, the derivative of $25 - t^{2}$ with respect to $t$ is $\frac{d}{d t} [25] + \frac{d}{d t} [- t^{2}]$ .

$- 9 {(25 - t^{2})}^{- 2} (\frac{d}{d t} [25] + \frac{d}{d t} [- t^{2}])$

Step 1.1.3.3

Since $25$ is constant with respect to $t$ , the derivative of $25$ with respect to $t$ is $0$ .

$- 9 {(25 - t^{2})}^{- 2} (0 + \frac{d}{d t} [- t^{2}])$

Step 1.1.3.4

Add $0$ and $\frac{d}{d t} [- t^{2}]$ .

$- 9 {(25 - t^{2})}^{- 2} \frac{d}{d t} [- t^{2}]$

Step 1.1.3.5

Since $- 1$ is constant with respect to $t$ , the derivative of $- t^{2}$ with respect to $t$ is $- \frac{d}{d t} [t^{2}]$ .

$- 9 {(25 - t^{2})}^{- 2} (- \frac{d}{d t} [t^{2}])$

Step 1.1.3.6

Multiply $- 1$ by $- 9$ .

$9 {(25 - t^{2})}^{- 2} \frac{d}{d t} [t^{2}]$

Step 1.1.3.7

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$9 {(25 - t^{2})}^{- 2} (2 t)$

Step 1.1.3.8

Multiply $2$ by $9$ .

$18 {(25 - t^{2})}^{- 2} t$

Step 1.1.4

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

$18 \frac{1}{{(25 - t^{2})}^{2}} t$

Step 1.1.5

Simplify.

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Step 1.1.5.1

Combine terms.

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Step 1.1.5.1.1

Combine $18$ and $\frac{1}{{(25 - t^{2})}^{2}}$ .

$\frac{18}{{(25 - t^{2})}^{2}} t$

Step 1.1.5.1.2

Combine $\frac{18}{{(25 - t^{2})}^{2}}$ and $t$ .

$\frac{18 t}{{(25 - t^{2})}^{2}}$

Step 1.1.5.2

Reorder terms.

$f' (t) = \frac{18 t}{{(- t^{2} + 25)}^{2}}$

Step 1.2

The first derivative of $f (t)$ with respect to $t$ is $\frac{18 t}{{(- t^{2} + 25)}^{2}}$ .

$\frac{18 t}{{(- t^{2} + 25)}^{2}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{18 t}{{(- t^{2} + 25)}^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$\frac{18 t}{{(- t^{2} + 25)}^{2}} = 0$

Step 2.2

Set the numerator equal to zero.

$18 t = 0$

Step 2.3

Divide each term in $18 t = 0$ by $18$ and simplify.

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Step 2.3.1

Divide each term in $18 t = 0$ by $18$ .

$\frac{18 t}{18} = \frac{0}{18}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $18$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{18 t}{18} = \frac{0}{18}$

Step 2.3.2.1.2

Divide $t$ by $1$ .

$t = \frac{0}{18}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $0$ by $18$ .

$t = 0$

Step 3

The values which make the derivative equal to $0$ are $0$ .

$0$

Step 4

Find where the derivative is undefined.

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Step 4.1

Set the denominator in $\frac{18 t}{{(- t^{2} + 25)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(- t^{2} + 25)}^{2} = 0$

Step 4.2

Solve for $t$ .

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Step 4.2.1

Factor the left side of the equation.

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Step 4.2.1.1

Factor $- 1$ out of $- t^{2} + 25$ .

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Step 4.2.1.1.1

Factor $- 1$ out of $- t^{2}$ .

${(- (t^{2}) + 25)}^{2} = 0$

Step 4.2.1.1.2

Rewrite $25$ as $- 1 (- 25)$ .

${(- (t^{2}) - 1 \cdot - 25)}^{2} = 0$

Step 4.2.1.1.3

Factor $- 1$ out of $- (t^{2}) - 1 (- 25)$ .

${(- (t^{2} - 25))}^{2} = 0$

Step 4.2.1.2

Rewrite $25$ as $5^{2}$ .

${(- (t^{2} - 5^{2}))}^{2} = 0$

Step 4.2.1.3

Factor.

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Step 4.2.1.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = t$ and $b = 5$ .

${(- ((t + 5) (t - 5)))}^{2} = 0$

Step 4.2.1.3.2

Remove unnecessary parentheses.

${(- (t + 5) (t - 5))}^{2} = 0$

Step 4.2.1.4

Apply the product rule to $- (t + 5) (t - 5)$ .

${(- (t + 5))}^{2} {(t - 5)}^{2} = 0$

Step 4.2.1.5

Apply the product rule to $- (t + 5)$ .

${(- 1)}^{2} {(t + 5)}^{2} {(t - 5)}^{2} = 0$

Step 4.2.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

${(t + 5)}^{2} = 0$

${(t - 5)}^{2} = 0$

Step 4.2.3

Set ${(t + 5)}^{2}$ equal to $0$ and solve for $t$ .

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Step 4.2.3.1

Set ${(t + 5)}^{2}$ equal to $0$ .

${(t + 5)}^{2} = 0$

Step 4.2.3.2

Solve ${(t + 5)}^{2} = 0$ for $t$ .

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Step 4.2.3.2.1

Set the $t + 5$ equal to $0$ .

$t + 5 = 0$

Step 4.2.3.2.2

Subtract $5$ from both sides of the equation.

$t = - 5$

Step 4.2.4

Set ${(t - 5)}^{2}$ equal to $0$ and solve for $t$ .

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Step 4.2.4.1

Set ${(t - 5)}^{2}$ equal to $0$ .

${(t - 5)}^{2} = 0$

Step 4.2.4.2

Solve ${(t - 5)}^{2} = 0$ for $t$ .

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Step 4.2.4.2.1

Set the $t - 5$ equal to $0$ .

$t - 5 = 0$

Step 4.2.4.2.2

Add $5$ to both sides of the equation.

$t = 5$

Step 4.2.5

The final solution is all the values that make ${(- 1)}^{2} {(t + 5)}^{2} {(t - 5)}^{2} = 0$ true.

$t = - 5, 5$

Step 4.3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$t = - 5, t = 5$

Step 5

Split $(- \infty, \infty)$ into separate intervals around the $t$ values that make the derivative $0$ or undefined.

$(- \infty, - 5) \cup (- 5, 0) \cup (0, 5) \cup (5, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 5)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $t$ with $- 6$ in the expression.

$f' (- 6) = \frac{18 (- 6)}{{(- {(- 6)}^{2} + 25)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $18$ by $- 6$ .

$f' (- 6) = \frac{- 108}{{(- {(- 6)}^{2} + 25)}^{2}}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Raise $- 6$ to the power of $2$ .

$f' (- 6) = \frac{- 108}{{(- 1 \cdot 36 + 25)}^{2}}$

Step 6.2.2.2

Multiply $- 1$ by $36$ .

$f' (- 6) = \frac{- 108}{{(- 36 + 25)}^{2}}$

Step 6.2.2.3

Add $- 36$ and $25$ .

$f' (- 6) = \frac{- 108}{{(- 11)}^{2}}$

Step 6.2.2.4

Raise $- 11$ to the power of $2$ .

$f' (- 6) = \frac{- 108}{121}$

Step 6.2.3

Move the negative in front of the fraction.

$f' (- 6) = - \frac{108}{121}$

Step 6.2.4

The final answer is $- \frac{108}{121}$ .

$- \frac{108}{121}$

Step 6.3

At $t = - 6$ the derivative is $- \frac{108}{121}$ . Since this is negative, the function is decreasing on $(- \infty, - 5)$ .

Decreasing on $(- \infty, - 5)$ since $f' (t) < 0$

Step 7

Substitute a value from the interval $(- 5, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $t$ with $- \frac{5}{2}$ in the expression.

$f' (- \frac{5}{2}) = \frac{18 (- \frac{5}{2})}{{(- {(- \frac{5}{2})}^{2} + 25)}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify the numerator.

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Step 7.2.1.1

Multiply $- 1$ by $18$ .

$f' (- \frac{5}{2}) = \frac{- 18 (\frac{5}{2})}{{(- {(- \frac{5}{2})}^{2} + 25)}^{2}}$

Step 7.2.1.2

Combine $- 18$ and $\frac{5}{2}$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{(- {(- \frac{5}{2})}^{2} + 25)}^{2}}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 7.2.2.1.1

Apply the product rule to $- \frac{5}{2}$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{(- ({(- 1)}^{2} {(\frac{5}{2})}^{2}) + 25)}^{2}}$

Step 7.2.2.1.2

Apply the product rule to $\frac{5}{2}$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{(- ({(- 1)}^{2} (\frac{5^{2}}{2^{2}})) + 25)}^{2}}$

Step 7.2.2.2

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

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Step 7.2.2.2.1

Move ${(- 1)}^{2}$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{({(- 1)}^{2} \cdot (- 1 \frac{5^{2}}{2^{2}}) + 25)}^{2}}$

Step 7.2.2.2.2

Multiply ${(- 1)}^{2}$ by $- 1$ .

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Step 7.2.2.2.2.1

Raise $- 1$ to the power of $1$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{({(- 1)}^{2} \cdot ((- 1) (\frac{5^{2}}{2^{2}})) + 25)}^{2}}$

Step 7.2.2.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{({(- 1)}^{2 + 1} (\frac{5^{2}}{2^{2}}) + 25)}^{2}}$

Step 7.2.2.2.3

Add $2$ and $1$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{({(- 1)}^{3} (\frac{5^{2}}{2^{2}}) + 25)}^{2}}$

Step 7.2.2.3

Raise $- 1$ to the power of $3$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{(- \frac{5^{2}}{2^{2}} + 25)}^{2}}$

Step 7.2.2.4

Raise $5$ to the power of $2$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{(- \frac{25}{2^{2}} + 25)}^{2}}$

Step 7.2.2.5

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

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{(- \frac{25}{4} + 25)}^{2}}$

Step 7.2.2.6

To write $25$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{(- \frac{25}{4} + 25 \cdot \frac{4}{4})}^{2}}$

Step 7.2.2.7

Combine $25$ and $\frac{4}{4}$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{(- \frac{25}{4} + \frac{25 \cdot 4}{4})}^{2}}$

Step 7.2.2.8

Combine the numerators over the common denominator.

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{(\frac{- 25 + 25 \cdot 4}{4})}^{2}}$

Step 7.2.2.9

Simplify the numerator.

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Step 7.2.2.9.1

Multiply $25$ by $4$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{(\frac{- 25 + 100}{4})}^{2}}$

Step 7.2.2.9.2

Add $- 25$ and $100$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{{(\frac{75}{4})}^{2}}$

Step 7.2.2.10

Apply the product rule to $\frac{75}{4}$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{\frac{75^{2}}{4^{2}}}$

Step 7.2.2.11

Raise $75$ to the power of $2$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{\frac{5625}{4^{2}}}$

Step 7.2.2.12

Raise $4$ to the power of $2$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 18 \cdot 5}{2}}{\frac{5625}{16}}$

Step 7.2.3

Simplify the expression.

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Step 7.2.3.1

Multiply $- 18$ by $5$ .

$f' (- \frac{5}{2}) = \frac{\frac{- 90}{2}}{\frac{5625}{16}}$

Step 7.2.3.2

Divide $- 90$ by $2$ .

$f' (- \frac{5}{2}) = \frac{- 45}{\frac{5625}{16}}$

Step 7.2.4

Multiply the numerator by the reciprocal of the denominator.

$f' (- \frac{5}{2}) = - 45 (\frac{16}{5625})$

Step 7.2.5

Cancel the common factor of $45$ .

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Step 7.2.5.1

Factor $45$ out of $- 45$ .

$f' (- \frac{5}{2}) = 45 (- 1) (\frac{16}{5625})$

Step 7.2.5.2

Factor $45$ out of $5625$ .

$f' (- \frac{5}{2}) = 45 \cdot (- 1 \frac{16}{45 \cdot 125})$

Step 7.2.5.3

Cancel the common factor.

$f' (- \frac{5}{2}) = 45 \cdot (- 1 \frac{16}{45 \cdot 125})$

Step 7.2.5.4

Rewrite the expression.

$f' (- \frac{5}{2}) = - \frac{16}{125}$

Step 7.2.6

The final answer is $- \frac{16}{125}$ .

$- \frac{16}{125}$

Step 7.3

At $t = - \frac{5}{2}$ the derivative is $- \frac{16}{125}$ . Since this is negative, the function is decreasing on $(- 5, 0)$ .

Decreasing on $(- 5, 0)$ since $f' (t) < 0$

Step 8

Substitute a value from the interval $(0, 5)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $t$ with $\frac{5}{2}$ in the expression.

$f' (\frac{5}{2}) = \frac{18 (\frac{5}{2})}{{(- {(\frac{5}{2})}^{2} + 25)}^{2}}$

Step 8.2

Simplify the result.

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Step 8.2.1

Combine $18$ and $\frac{5}{2}$ .

$f' (\frac{5}{2}) = \frac{\frac{18 \cdot 5}{2}}{{(- {(\frac{5}{2})}^{2} + 25)}^{2}}$

Step 8.2.2

Simplify the denominator.

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Step 8.2.2.1

Apply the product rule to $\frac{5}{2}$ .

$f' (\frac{5}{2}) = \frac{\frac{18 \cdot 5}{2}}{{(- \frac{5^{2}}{2^{2}} + 25)}^{2}}$

Step 8.2.2.2

Raise $5$ to the power of $2$ .

$f' (\frac{5}{2}) = \frac{\frac{18 \cdot 5}{2}}{{(- \frac{25}{2^{2}} + 25)}^{2}}$

Step 8.2.2.3

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

$f' (\frac{5}{2}) = \frac{\frac{18 \cdot 5}{2}}{{(- \frac{25}{4} + 25)}^{2}}$

Step 8.2.2.4

To write $25$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f' (\frac{5}{2}) = \frac{\frac{18 \cdot 5}{2}}{{(- \frac{25}{4} + 25 \cdot \frac{4}{4})}^{2}}$

Step 8.2.2.5

Combine $25$ and $\frac{4}{4}$ .

$f' (\frac{5}{2}) = \frac{\frac{18 \cdot 5}{2}}{{(- \frac{25}{4} + \frac{25 \cdot 4}{4})}^{2}}$

Step 8.2.2.6

Combine the numerators over the common denominator.

$f' (\frac{5}{2}) = \frac{\frac{18 \cdot 5}{2}}{{(\frac{- 25 + 25 \cdot 4}{4})}^{2}}$

Step 8.2.2.7

Simplify the numerator.

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Step 8.2.2.7.1

Multiply $25$ by $4$ .

$f' (\frac{5}{2}) = \frac{\frac{18 \cdot 5}{2}}{{(\frac{- 25 + 100}{4})}^{2}}$

Step 8.2.2.7.2

Add $- 25$ and $100$ .

$f' (\frac{5}{2}) = \frac{\frac{18 \cdot 5}{2}}{{(\frac{75}{4})}^{2}}$

Step 8.2.2.8

Apply the product rule to $\frac{75}{4}$ .

$f' (\frac{5}{2}) = \frac{\frac{18 \cdot 5}{2}}{\frac{75^{2}}{4^{2}}}$

Step 8.2.2.9

Raise $75$ to the power of $2$ .

$f' (\frac{5}{2}) = \frac{\frac{18 \cdot 5}{2}}{\frac{5625}{4^{2}}}$

Step 8.2.2.10

Raise $4$ to the power of $2$ .

$f' (\frac{5}{2}) = \frac{\frac{18 \cdot 5}{2}}{\frac{5625}{16}}$

Step 8.2.3

Simplify the expression.

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Step 8.2.3.1

Multiply $18$ by $5$ .

$f' (\frac{5}{2}) = \frac{\frac{90}{2}}{\frac{5625}{16}}$

Step 8.2.3.2

Divide $90$ by $2$ .

$f' (\frac{5}{2}) = \frac{45}{\frac{5625}{16}}$

Step 8.2.4

Multiply the numerator by the reciprocal of the denominator.

$f' (\frac{5}{2}) = 45 (\frac{16}{5625})$

Step 8.2.5

Cancel the common factor of $45$ .

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Step 8.2.5.1

Factor $45$ out of $5625$ .

$f' (\frac{5}{2}) = 45 (\frac{16}{45 (125)})$

Step 8.2.5.2

Cancel the common factor.

$f' (\frac{5}{2}) = 45 (\frac{16}{45 \cdot 125})$

Step 8.2.5.3

Rewrite the expression.

$f' (\frac{5}{2}) = \frac{16}{125}$

Step 8.2.6

The final answer is $\frac{16}{125}$ .

$\frac{16}{125}$

Step 8.3

At $t = \frac{5}{2}$ the derivative is $\frac{16}{125}$ . Since this is positive, the function is increasing on $(0, 5)$ .

Increasing on $(0, 5)$ since $f' (t) > 0$

Step 9

Substitute a value from the interval $(5, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 9.1

Replace the variable $t$ with $6$ in the expression.

$f' (6) = \frac{18 (6)}{{(- {(6)}^{2} + 25)}^{2}}$

Step 9.2

Simplify the result.

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Step 9.2.1

Multiply $18$ by $6$ .

$f' (6) = \frac{108}{{(- 6^{2} + 25)}^{2}}$

Step 9.2.2

Simplify the denominator.

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Step 9.2.2.1

Raise $6$ to the power of $2$ .

$f' (6) = \frac{108}{{(- 1 \cdot 36 + 25)}^{2}}$

Step 9.2.2.2

Multiply $- 1$ by $36$ .

$f' (6) = \frac{108}{{(- 36 + 25)}^{2}}$

Step 9.2.2.3

Add $- 36$ and $25$ .

$f' (6) = \frac{108}{{(- 11)}^{2}}$

Step 9.2.2.4

Raise $- 11$ to the power of $2$ .

$f' (6) = \frac{108}{121}$

Step 9.2.3

The final answer is $\frac{108}{121}$ .

$\frac{108}{121}$

Step 9.3

At $t = 6$ the derivative is $\frac{108}{121}$ . Since this is positive, the function is increasing on $(5, \infty)$ .

Increasing on $(5, \infty)$ since $f' (t) > 0$

Step 10

List the intervals on which the function is increasing and decreasing.

Increasing on: $(0, 5), (5, \infty)$

Decreasing on: $(- \infty, - 5), (- 5, 0)$

Step 11