Calculus Examples

$p (t) = 561 + (36 t^{0.6} - 108)$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of $561 + 36 t^{0.6} - 108$ with respect to $t$ is $\frac{d}{d t} [561] + \frac{d}{d t} [36 t^{0.6}] + \frac{d}{d t} [- 108]$ .

$\frac{d}{d t} [561] + \frac{d}{d t} [36 t^{0.6}] + \frac{d}{d t} [- 108]$

Step 1.1.2

Since $561$ is constant with respect to $t$ , the derivative of $561$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [36 t^{0.6}] + \frac{d}{d t} [- 108]$

Step 1.2

Evaluate $\frac{d}{d t} [36 t^{0.6}]$ .

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Step 1.2.1

Since $36$ is constant with respect to $t$ , the derivative of $36 t^{0.6}$ with respect to $t$ is $36 \frac{d}{d t} [t^{0.6}]$ .

$0 + 36 \frac{d}{d t} [t^{0.6}] + \frac{d}{d t} [- 108]$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 0.6$ .

$0 + 36 (0.6 t^{- 0.4}) + \frac{d}{d t} [- 108]$

Step 1.2.3

Multiply $0.6$ by $36$ .

$0 + 21.6 t^{- 0.4} + \frac{d}{d t} [- 108]$

Step 1.3

Since $- 108$ is constant with respect to $t$ , the derivative of $- 108$ with respect to $t$ is $0$ .

$0 + 21.6 t^{- 0.4} + 0$

Step 1.4

Simplify.

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Step 1.4.1

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

$0 + 21.6 \frac{1}{t^{0.4}} + 0$

Step 1.4.2

Combine terms.

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Step 1.4.2.1

Combine $21.6$ and $\frac{1}{t^{0.4}}$ .

$0 + \frac{21.6}{t^{0.4}} + 0$

Step 1.4.2.2

Add $0$ and $\frac{21.6}{t^{0.4}}$ .

$\frac{21.6}{t^{0.4}} + 0$

Step 1.4.2.3

Add $\frac{21.6}{t^{0.4}}$ and $0$ .

$\frac{21.6}{t^{0.4}}$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $21.6$ is constant with respect to $t$ , the derivative of $\frac{21.6}{t^{0.4}}$ with respect to $t$ is $21.6 \frac{d}{d t} [\frac{1}{t^{0.4}}]$ .

$p'' (t) = 21.6 \frac{d}{d t} (\frac{1}{t^{0.4}})$

Step 2.2

Apply basic rules of exponents.

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Step 2.2.1

Rewrite $\frac{1}{t^{0.4}}$ as ${(t^{0.4})}^{- 1}$ .

$p'' (t) = 21.6 \frac{d}{d t} ({(t^{0.4})}^{- 1})$

Step 2.2.2

Multiply the exponents in ${(t^{0.4})}^{- 1}$ .

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Step 2.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$p'' (t) = 21.6 \frac{d}{d t} (t^{0.4 \cdot - 1})$

Step 2.2.2.2

Multiply $0.4$ by $- 1$ .

$p'' (t) = 21.6 \frac{d}{d t} (t^{- 0.4})$

Step 2.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = - 0.4$ .

$p'' (t) = 21.6 (- 0.4 t^{- 1.4})$

Step 2.4

Multiply $- 0.4$ by $21.6$ .

$p'' (t) = - 8.64 t^{- 1.4}$

Step 2.5

Simplify.

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Step 2.5.1

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

$p'' (t) = - 8.64 \frac{1}{t^{1.4}}$

Step 2.5.2

Combine terms.

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Step 2.5.2.1

Combine $- 8.64$ and $\frac{1}{t^{1.4}}$ .

$p'' (t) = \frac{- 8.64}{t^{1.4}}$

Step 2.5.2.2

Move the negative in front of the fraction.

$p'' (t) = - \frac{8.64}{t^{1.4}}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$\frac{21.6}{t^{0.4}} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

Differentiate.

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Step 4.1.1.1

By the Sum Rule, the derivative of $561 + 36 t^{0.6} - 108$ with respect to $t$ is $\frac{d}{d t} [561] + \frac{d}{d t} [36 t^{0.6}] + \frac{d}{d t} [- 108]$ .

$\frac{d}{d t} [561] + \frac{d}{d t} [36 t^{0.6}] + \frac{d}{d t} [- 108]$

Step 4.1.1.2

Since $561$ is constant with respect to $t$ , the derivative of $561$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [36 t^{0.6}] + \frac{d}{d t} [- 108]$

Step 4.1.2

Evaluate $\frac{d}{d t} [36 t^{0.6}]$ .

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Step 4.1.2.1

Since $36$ is constant with respect to $t$ , the derivative of $36 t^{0.6}$ with respect to $t$ is $36 \frac{d}{d t} [t^{0.6}]$ .

$0 + 36 \frac{d}{d t} [t^{0.6}] + \frac{d}{d t} [- 108]$

Step 4.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 0.6$ .

$0 + 36 (0.6 t^{- 0.4}) + \frac{d}{d t} [- 108]$

Step 4.1.2.3

Multiply $0.6$ by $36$ .

$0 + 21.6 t^{- 0.4} + \frac{d}{d t} [- 108]$

Step 4.1.3

Since $- 108$ is constant with respect to $t$ , the derivative of $- 108$ with respect to $t$ is $0$ .

$0 + 21.6 t^{- 0.4} + 0$

Step 4.1.4

Simplify.

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Step 4.1.4.1

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

$0 + 21.6 \frac{1}{t^{0.4}} + 0$

Step 4.1.4.2

Combine terms.

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Step 4.1.4.2.1

Combine $21.6$ and $\frac{1}{t^{0.4}}$ .

$0 + \frac{21.6}{t^{0.4}} + 0$

Step 4.1.4.2.2

Add $0$ and $\frac{21.6}{t^{0.4}}$ .

$\frac{21.6}{t^{0.4}} + 0$

Step 4.1.4.2.3

Add $\frac{21.6}{t^{0.4}}$ and $0$ .

$f' (t) = \frac{21.6}{t^{0.4}}$

Step 4.2

The first derivative of $p (t)$ with respect to $t$ is $\frac{21.6}{t^{0.4}}$ .

$\frac{21.6}{t^{0.4}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{21.6}{t^{0.4}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{21.6}{t^{0.4}} = 0$

Step 5.2

Set the numerator equal to zero.

$21.6 = 0$

Step 5.3

Since $21.6 \neq 0$ , there are no solutions.

No solution

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Convert expressions with fractional exponents to radicals.

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Step 6.1.1

Change $0.4$ into a fraction.

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Step 6.1.1.1

Multiply by $\frac{10}{10}$ to remove the decimal.

$\frac{21.6}{t^{\frac{10 \cdot 0.4}{10}}}$

Step 6.1.1.2

Multiply $10$ by $0.4$ .

$\frac{21.6}{t^{\frac{4}{10}}}$

Step 6.1.1.3

Cancel the common factor of $4$ and $10$ .

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Step 6.1.1.3.1

Factor $2$ out of $4$ .

$\frac{21.6}{t^{\frac{2 (2)}{10}}}$

Step 6.1.1.3.2

Cancel the common factors.

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Step 6.1.1.3.2.1

Factor $2$ out of $10$ .

$\frac{21.6}{t^{\frac{2 \cdot 2}{2 \cdot 5}}}$

Step 6.1.1.3.2.2

Cancel the common factor.

$\frac{21.6}{t^{\frac{2 \cdot 2}{2 \cdot 5}}}$

Step 6.1.1.3.2.3

Rewrite the expression.

$\frac{21.6}{t^{\frac{2}{5}}}$

Step 6.1.2

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\frac{21.6}{\sqrt[5]{t^{2}}}$

Step 6.2

Set the denominator in $\frac{21.6}{\sqrt[5]{t^{2}}}$ equal to $0$ to find where the expression is undefined.

$\sqrt[5]{t^{2}} = 0$

Step 6.3

Solve for $t$ .

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Step 6.3.1

To remove the radical on the left side of the equation, raise both sides of the equation to the power of $5$ .

${\sqrt[5]{t^{2}}}^{5} = 0^{5}$

Step 6.3.2

Simplify each side of the equation.

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Step 6.3.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{t^{2}}$ as $t^{\frac{2}{5}}$ .

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

Step 6.3.2.2

Simplify the left side.

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Step 6.3.2.2.1

Multiply the exponents in ${(t^{\frac{2}{5}})}^{5}$ .

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Step 6.3.2.2.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$t^{\frac{2}{5} \cdot 5} = 0^{5}$

Step 6.3.2.2.1.2

Cancel the common factor of $5$ .

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Step 6.3.2.2.1.2.1

Cancel the common factor.

$t^{\frac{2}{5} \cdot 5} = 0^{5}$

Step 6.3.2.2.1.2.2

Rewrite the expression.

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

Step 6.3.2.3

Simplify the right side.

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Step 6.3.2.3.1

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

$t^{2} = 0$

Step 6.3.3

Solve for $t$ .

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Step 6.3.3.1

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

$t = \pm \sqrt{0}$

Step 6.3.3.2

Simplify $\pm \sqrt{0}$ .

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Step 6.3.3.2.1

Rewrite $0$ as $0^{2}$ .

$t = \pm \sqrt{0^{2}}$

Step 6.3.3.2.2

Pull terms out from under the radical, assuming positive real numbers.

$t = \pm 0$

Step 6.3.3.2.3

Plus or minus $0$ is $0$ .

$t = 0$

Step 7

Critical points to evaluate.

$t = 0$

Step 8

Evaluate the second derivative at $t = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{8.64}{{(0)}^{1.4}}$

Step 9

Evaluate the second derivative.

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Step 9.1

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

$- \frac{8.64}{0}$

Step 9.2

The expression contains a division by $0$ . The expression is undefined.

Undefined

Step 10

Since the first derivative test failed, there are no local extrema.

No Local Extrema

Step 11