Calculus Examples

$g (t) = \frac{5 t}{t^{2} + 1}$

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

Since $5$ is constant with respect to $t$ , the derivative of $\frac{5 t}{t^{2} + 1}$ with respect to $t$ is $5 \frac{d}{d t} [\frac{t}{t^{2} + 1}]$ .

$5 \frac{d}{d t} [\frac{t}{t^{2} + 1}]$

Step 1.1.2

Differentiate using the Quotient Rule which states that $\frac{d}{d t} [\frac{f (t)}{g (t)}]$ is $\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}$ where $f (t) = t$ and $g (t) = t^{2} + 1$ .

$5 \frac{(t^{2} + 1) \frac{d}{d t} [t] - t \frac{d}{d t} [t^{2} + 1]}{{(t^{2} + 1)}^{2}}$

Step 1.1.3

Differentiate.

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Step 1.1.3.1

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

$5 \frac{(t^{2} + 1) \cdot 1 - t \frac{d}{d t} [t^{2} + 1]}{{(t^{2} + 1)}^{2}}$

Step 1.1.3.2

Multiply $t^{2} + 1$ by $1$ .

$5 \frac{t^{2} + 1 - t \frac{d}{d t} [t^{2} + 1]}{{(t^{2} + 1)}^{2}}$

Step 1.1.3.3

By the Sum Rule, the derivative of $t^{2} + 1$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$ .

$5 \frac{t^{2} + 1 - t (\frac{d}{d t} [t^{2}] + \frac{d}{d t} [1])}{{(t^{2} + 1)}^{2}}$

Step 1.1.3.4

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

$5 \frac{t^{2} + 1 - t (2 t + \frac{d}{d t} [1])}{{(t^{2} + 1)}^{2}}$

Step 1.1.3.5

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

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

Step 1.1.3.6

Simplify the expression.

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Step 1.1.3.6.1

Add $2 t$ and $0$ .

$5 \frac{t^{2} + 1 - t (2 t)}{{(t^{2} + 1)}^{2}}$

Step 1.1.3.6.2

Multiply $2$ by $- 1$ .

$5 \frac{t^{2} + 1 - 2 t \cdot t}{{(t^{2} + 1)}^{2}}$

Step 1.1.4

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

$5 \frac{t^{2} + 1 - 2 (t^{1} t)}{{(t^{2} + 1)}^{2}}$

Step 1.1.5

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

$5 \frac{t^{2} + 1 - 2 (t^{1} t^{1})}{{(t^{2} + 1)}^{2}}$

Step 1.1.6

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

$5 \frac{t^{2} + 1 - 2 t^{1 + 1}}{{(t^{2} + 1)}^{2}}$

Step 1.1.7

Add $1$ and $1$ .

$5 \frac{t^{2} + 1 - 2 t^{2}}{{(t^{2} + 1)}^{2}}$

Step 1.1.8

Subtract $2 t^{2}$ from $t^{2}$ .

$5 \frac{- t^{2} + 1}{{(t^{2} + 1)}^{2}}$

Step 1.1.9

Combine $5$ and $\frac{- t^{2} + 1}{{(t^{2} + 1)}^{2}}$ .

$\frac{5 (- t^{2} + 1)}{{(t^{2} + 1)}^{2}}$

Step 1.1.10

Simplify.

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Step 1.1.10.1

Apply the distributive property.

$\frac{5 (- t^{2}) + 5 \cdot 1}{{(t^{2} + 1)}^{2}}$

Step 1.1.10.2

Simplify each term.

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Step 1.1.10.2.1

Multiply $- 1$ by $5$ .

$\frac{- 5 t^{2} + 5 \cdot 1}{{(t^{2} + 1)}^{2}}$

Step 1.1.10.2.2

Multiply $5$ by $1$ .

$f' (t) = \frac{- 5 t^{2} + 5}{{(t^{2} + 1)}^{2}}$

Step 1.2

The first derivative of $g (t)$ with respect to $t$ is $\frac{- 5 t^{2} + 5}{{(t^{2} + 1)}^{2}}$ .

$\frac{- 5 t^{2} + 5}{{(t^{2} + 1)}^{2}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{- 5 t^{2} + 5}{{(t^{2} + 1)}^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

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

Step 2.2

Set the numerator equal to zero.

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

Step 2.3

Solve the equation for $t$ .

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Step 2.3.1

Subtract $5$ from both sides of the equation.

$- 5 t^{2} = - 5$

Step 2.3.2

Divide each term in $- 5 t^{2} = - 5$ by $- 5$ and simplify.

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Step 2.3.2.1

Divide each term in $- 5 t^{2} = - 5$ by $- 5$ .

$\frac{- 5 t^{2}}{- 5} = \frac{- 5}{- 5}$

Step 2.3.2.2

Simplify the left side.

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Step 2.3.2.2.1

Cancel the common factor of $- 5$ .

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Step 2.3.2.2.1.1

Cancel the common factor.

$\frac{- 5 t^{2}}{- 5} = \frac{- 5}{- 5}$

Step 2.3.2.2.1.2

Divide $t^{2}$ by $1$ .

$t^{2} = \frac{- 5}{- 5}$

Step 2.3.2.3

Simplify the right side.

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Step 2.3.2.3.1

Divide $- 5$ by $- 5$ .

$t^{2} = 1$

Step 2.3.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$t = \pm \sqrt{1}$

Step 2.3.4

Any root of $1$ is $1$ .

$t = \pm 1$

Step 2.3.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.3.5.1

First, use the positive value of the $\pm$ to find the first solution.

$t = 1$

Step 2.3.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$t = - 1$

Step 2.3.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$t = 1, - 1$

Step 3

The values which make the derivative equal to $0$ are $1, - 1$ .

$1, - 1$

Step 4

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

$(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 1)$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $t$ with $- 2$ in the expression.

$g' (- 2) = \frac{- 5 {(- 2)}^{2} + 5}{{({(- 2)}^{2} + 1)}^{2}}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify the numerator.

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Step 5.2.1.1

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

$g' (- 2) = \frac{- 5 \cdot 4 + 5}{{({(- 2)}^{2} + 1)}^{2}}$

Step 5.2.1.2

Multiply $- 5$ by $4$ .

$g' (- 2) = \frac{- 20 + 5}{{({(- 2)}^{2} + 1)}^{2}}$

Step 5.2.1.3

Add $- 20$ and $5$ .

$g' (- 2) = \frac{- 15}{{({(- 2)}^{2} + 1)}^{2}}$

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

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

$g' (- 2) = \frac{- 15}{{(4 + 1)}^{2}}$

Step 5.2.2.2

Add $4$ and $1$ .

$g' (- 2) = \frac{- 15}{5^{2}}$

Step 5.2.2.3

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

$g' (- 2) = \frac{- 15}{25}$

Step 5.2.3

Reduce the expression by cancelling the common factors.

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Step 5.2.3.1

Cancel the common factor of $- 15$ and $25$ .

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Step 5.2.3.1.1

Factor $5$ out of $- 15$ .

$g' (- 2) = \frac{5 (- 3)}{25}$

Step 5.2.3.1.2

Cancel the common factors.

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Step 5.2.3.1.2.1

Factor $5$ out of $25$ .

$g' (- 2) = \frac{5 \cdot - 3}{5 \cdot 5}$

Step 5.2.3.1.2.2

Cancel the common factor.

$g' (- 2) = \frac{5 \cdot - 3}{5 \cdot 5}$

Step 5.2.3.1.2.3

Rewrite the expression.

$g' (- 2) = \frac{- 3}{5}$

Step 5.2.3.2

Move the negative in front of the fraction.

$g' (- 2) = - \frac{3}{5}$

Step 5.2.4

The final answer is $- \frac{3}{5}$ .

$- \frac{3}{5}$

Step 5.3

At $t = - 2$ the derivative is $- \frac{3}{5}$ . Since this is negative, the function is decreasing on $(- \infty, - 1)$ .

Decreasing on $(- \infty, - 1)$ since $g' (t) < 0$

Step 6

Substitute a value from the interval $(- 1, 1)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $t$ with $0$ in the expression.

$g' (0) = \frac{- 5 {(0)}^{2} + 5}{{({(0)}^{2} + 1)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the numerator.

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Step 6.2.1.1

Raising $0$ to any positive power yields $0$ .

$g' (0) = \frac{- 5 \cdot 0 + 5}{{(0^{2} + 1)}^{2}}$

Step 6.2.1.2

Multiply $- 5$ by $0$ .

$g' (0) = \frac{0 + 5}{{(0^{2} + 1)}^{2}}$

Step 6.2.1.3

Add $0$ and $5$ .

$g' (0) = \frac{5}{{(0^{2} + 1)}^{2}}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Raising $0$ to any positive power yields $0$ .

$g' (0) = \frac{5}{{(0 + 1)}^{2}}$

Step 6.2.2.2

Add $0$ and $1$ .

$g' (0) = \frac{5}{1^{2}}$

Step 6.2.2.3

One to any power is one.

$g' (0) = \frac{5}{1}$

Step 6.2.3

Divide $5$ by $1$ .

$g' (0) = 5$

Step 6.2.4

The final answer is $5$ .

$5$

Step 6.3

At $t = 0$ the derivative is $5$ . Since this is positive, the function is increasing on $(- 1, 1)$ .

Increasing on $(- 1, 1)$ since $g' (t) > 0$

Step 7

Substitute a value from the interval $(1, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

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

$g' (2) = \frac{- 5 {(2)}^{2} + 5}{{({(2)}^{2} + 1)}^{2}}$

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

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

$g' (2) = \frac{- 5 \cdot 4 + 5}{{(2^{2} + 1)}^{2}}$

Step 7.2.1.2

Multiply $- 5$ by $4$ .

$g' (2) = \frac{- 20 + 5}{{(2^{2} + 1)}^{2}}$

Step 7.2.1.3

Add $- 20$ and $5$ .

$g' (2) = \frac{- 15}{{(2^{2} + 1)}^{2}}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

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

$g' (2) = \frac{- 15}{{(4 + 1)}^{2}}$

Step 7.2.2.2

Add $4$ and $1$ .

$g' (2) = \frac{- 15}{5^{2}}$

Step 7.2.2.3

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

$g' (2) = \frac{- 15}{25}$

Step 7.2.3

Reduce the expression by cancelling the common factors.

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Step 7.2.3.1

Cancel the common factor of $- 15$ and $25$ .

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Step 7.2.3.1.1

Factor $5$ out of $- 15$ .

$g' (2) = \frac{5 (- 3)}{25}$

Step 7.2.3.1.2

Cancel the common factors.

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Step 7.2.3.1.2.1

Factor $5$ out of $25$ .

$g' (2) = \frac{5 \cdot - 3}{5 \cdot 5}$

Step 7.2.3.1.2.2

Cancel the common factor.

$g' (2) = \frac{5 \cdot - 3}{5 \cdot 5}$

Step 7.2.3.1.2.3

Rewrite the expression.

$g' (2) = \frac{- 3}{5}$

Step 7.2.3.2

Move the negative in front of the fraction.

$g' (2) = - \frac{3}{5}$

Step 7.2.4

The final answer is $- \frac{3}{5}$ .

$- \frac{3}{5}$

Step 7.3

At $t = 2$ the derivative is $- \frac{3}{5}$ . Since this is negative, the function is decreasing on $(1, \infty)$ .

Decreasing on $(1, \infty)$ since $g' (t) < 0$

Step 8

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

Increasing on: $(- 1, 1)$

Decreasing on: $(- \infty, - 1), (1, \infty)$

Step 9