Calculus Examples

$f (x) = 3 \csc (4 x)$

Step 1

Find the first derivative of the function.

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Step 1.1

Since $3$ is constant with respect to $x$ , the derivative of $3 \csc (4 x)$ with respect to $x$ is $3 \frac{d}{d x} [\csc (4 x)]$ .

$3 \frac{d}{d x} [\csc (4 x)]$

Step 1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \csc (x)$ and $g (x) = 4 x$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $4 x$ .

$3 (\frac{d}{d u} [\csc (u)] \frac{d}{d x} [4 x])$

Step 1.2.2

The derivative of $\csc (u)$ with respect to $u$ is $- \csc (u) \cot (u)$ .

$3 (- \csc (u) \cot (u) \frac{d}{d x} [4 x])$

Step 1.2.3

Replace all occurrences of $u$ with $4 x$ .

$3 (- \csc (4 x) \cot (4 x) \frac{d}{d x} [4 x])$

Step 1.3

Differentiate.

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Step 1.3.1

Multiply $- 1$ by $3$ .

$- 3 (\csc (4 x) \cot (4 x) \frac{d}{d x} [4 x])$

Step 1.3.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$- 3 \csc (4 x) \cot (4 x) (4 \frac{d}{d x} [x])$

Step 1.3.3

Multiply $4$ by $- 3$ .

$- 12 \csc (4 x) \cot (4 x) \frac{d}{d x} [x]$

Step 1.3.4

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

$- 12 \csc (4 x) \cot (4 x) \cdot 1$

Step 1.3.5

Simplify the expression.

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Step 1.3.5.1

Multiply $- 12$ by $1$ .

$- 12 \csc (4 x) \cot (4 x)$

Step 1.3.5.2

Reorder the factors of $- 12 \csc (4 x) \cot (4 x)$ .

$- 12 \cot (4 x) \csc (4 x)$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12 \cot (4 x) \csc (4 x)$ with respect to $x$ is $- 12 \frac{d}{d x} [\cot (4 x) \csc (4 x)]$ .

$f'' (x) = - 12 \frac{d}{d x} (\cot (4 x) \csc (4 x))$

Step 2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \cot (4 x)$ and $g (x) = \csc (4 x)$ .

$f'' (x) = - 12 (\cot (4 x) \frac{d}{d x} (\csc (4 x)) + \csc (4 x) \frac{d}{d x} (\cot (4 x)))$

Step 2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \csc (x)$ and $g (x) = 4 x$ .

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Step 2.3.1

To apply the Chain Rule, set $u_{1}$ as $4 x$ .

$f'' (x) = - 12 (\cot (4 x) (\frac{d}{d u (1)} (\csc (u_{1})) \frac{d}{d x} (4 x)) + \csc (4 x) \frac{d}{d x} (\cot (4 x)))$

Step 2.3.2

The derivative of $\csc (u_{1})$ with respect to $u_{1}$ is $- \csc (u_{1}) \cot (u_{1})$ .

$f'' (x) = - 12 (\cot (4 x) (- \csc (u_{1}) \cot (u_{1}) \frac{d}{d x} (4 x)) + \csc (4 x) \frac{d}{d x} (\cot (4 x)))$

Step 2.3.3

Replace all occurrences of $u_{1}$ with $4 x$ .

$f'' (x) = - 12 (\cot (4 x) (- \csc (4 x) \cot (4 x) \frac{d}{d x} (4 x)) + \csc (4 x) \frac{d}{d x} (\cot (4 x)))$

Step 2.4

Raise $\cot (4 x)$ to the power of $1$ .

$f'' (x) = - 12 (- \csc (4 x) (\cot (4 x) \cot (4 x)) \frac{d}{d x} (4 x) + \csc (4 x) \frac{d}{d x} (\cot (4 x)))$

Step 2.5

Raise $\cot (4 x)$ to the power of $1$ .

$f'' (x) = - 12 (- \csc (4 x) (\cot (4 x) \cot (4 x)) \frac{d}{d x} (4 x) + \csc (4 x) \frac{d}{d x} (\cot (4 x)))$

Step 2.6

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

$f'' (x) = - 12 (- \csc (4 x) {\cot (4 x)}^{1 + 1} \frac{d}{d x} (4 x) + \csc (4 x) \frac{d}{d x} (\cot (4 x)))$

Step 2.7

Differentiate.

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Step 2.7.1

Add $1$ and $1$ .

$f'' (x) = - 12 (- \csc (4 x) \cot^{2} (4 x) \frac{d}{d x} (4 x) + \csc (4 x) \frac{d}{d x} (\cot (4 x)))$

Step 2.7.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$f'' (x) = - 12 (- \csc (4 x) \cot^{2} (4 x) (4 \frac{d}{d x} (x)) + \csc (4 x) \frac{d}{d x} (\cot (4 x)))$

Step 2.7.3

Multiply $4$ by $- 1$ .

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) \frac{d}{d x} x + \csc (4 x) \frac{d}{d x} (\cot (4 x)))$

Step 2.7.4

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

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) \cdot 1 + \csc (4 x) \frac{d}{d x} (\cot (4 x)))$

Step 2.7.5

Multiply $- 4$ by $1$ .

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) + \csc (4 x) \frac{d}{d x} (\cot (4 x)))$

Step 2.8

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cot (x)$ and $g (x) = 4 x$ .

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Step 2.8.1

To apply the Chain Rule, set $u_{2}$ as $4 x$ .

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) + \csc (4 x) (\frac{d}{d u (2)} (\cot (u_{2})) \frac{d}{d x} (4 x)))$

Step 2.8.2

The derivative of $\cot (u_{2})$ with respect to $u_{2}$ is $- \csc^{2} (u_{2})$ .

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) + \csc (4 x) (- \csc^{2} (u_{2}) \frac{d}{d x} (4 x)))$

Step 2.8.3

Replace all occurrences of $u_{2}$ with $4 x$ .

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) + \csc (4 x) (- \csc^{2} (4 x) \frac{d}{d x} (4 x)))$

Step 2.9

Raise $\csc (4 x)$ to the power of $1$ .

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) - \csc (4 x) \csc^{2} (4 x) \frac{d}{d x} (4 x))$

Step 2.10

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

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) - {\csc (4 x)}^{1 + 2} \frac{d}{d x} (4 x))$

Step 2.11

Add $1$ and $2$ .

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) - \csc^{3} (4 x) \frac{d}{d x} (4 x))$

Step 2.12

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) - \csc^{3} (4 x) (4 \frac{d}{d x} (x)))$

Step 2.13

Multiply $4$ by $- 1$ .

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) - 4 \csc^{3} (4 x) \frac{d}{d x} x)$

Step 2.14

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

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) - 4 \csc^{3} (4 x) \cdot 1)$

Step 2.15

Multiply $- 4$ by $1$ .

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x) - 4 \csc^{3} (4 x))$

Step 2.16

Simplify.

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Step 2.16.1

Apply the distributive property.

$f'' (x) = - 12 (- 4 \csc (4 x) \cot^{2} (4 x)) - 12 (- 4 \csc^{3} (4 x))$

Step 2.16.2

Combine terms.

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Step 2.16.2.1

Multiply $- 4$ by $- 12$ .

$f'' (x) = 48 (\csc (4 x) \cot^{2} (4 x)) - 12 (- 4 \csc^{3} (4 x))$

Step 2.16.2.2

Multiply $- 4$ by $- 12$ .

$f'' (x) = 48 \csc (4 x) \cot^{2} (4 x) + 48 \csc^{3} (4 x)$

Step 2.16.3

Reorder terms.

$f'' (x) = 48 \cot^{2} (4 x) \csc (4 x) + 48 \csc^{3} (4 x)$

Step 3

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

$- 12 \cot (4 x) \csc (4 x) = 0$

Step 4

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

$\cot (4 x) = 0$

$\csc (4 x) = 0$

Step 5

Set $\cot (4 x)$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $\cot (4 x)$ equal to $0$ .

$\cot (4 x) = 0$

Step 5.2

Solve $\cot (4 x) = 0$ for $x$ .

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Step 5.2.1

Take the inverse cotangent of both sides of the equation to extract $x$ from inside the cotangent.

$4 x = arccot (0)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

The exact value of $arccot (0)$ is $\frac{π}{2}$ .

$4 x = \frac{π}{2}$

Step 5.2.3

Divide each term in $4 x = \frac{π}{2}$ by $4$ and simplify.

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Step 5.2.3.1

Divide each term in $4 x = \frac{π}{2}$ by $4$ .

$\frac{4 x}{4} = \frac{\frac{π}{2}}{4}$

Step 5.2.3.2

Simplify the left side.

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Step 5.2.3.2.1

Cancel the common factor of $4$ .

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Step 5.2.3.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{\frac{π}{2}}{4}$

Step 5.2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{4}$

Step 5.2.3.3

Simplify the right side.

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Step 5.2.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{4}$

Step 5.2.3.3.2

Multiply $\frac{π}{2} \cdot \frac{1}{4}$ .

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Step 5.2.3.3.2.1

Multiply $\frac{π}{2}$ by $\frac{1}{4}$ .

$x = \frac{π}{2 \cdot 4}$

Step 5.2.3.3.2.2

Multiply $2$ by $4$ .

$x = \frac{π}{8}$

Step 5.2.4

The cotangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$4 x = π + \frac{π}{2}$

Step 5.2.5

Solve for $x$ .

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Step 5.2.5.1

Simplify.

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Step 5.2.5.1.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$4 x = π \cdot \frac{2}{2} + \frac{π}{2}$

Step 5.2.5.1.2

Combine $π$ and $\frac{2}{2}$ .

$4 x = \frac{π \cdot 2}{2} + \frac{π}{2}$

Step 5.2.5.1.3

Combine the numerators over the common denominator.

$4 x = \frac{π \cdot 2 + π}{2}$

Step 5.2.5.1.4

Add $π \cdot 2$ and $π$ .

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Step 5.2.5.1.4.1

Reorder $π$ and $2$ .

$4 x = \frac{2 \cdot π + π}{2}$

Step 5.2.5.1.4.2

Add $2 \cdot π$ and $π$ .

$4 x = \frac{3 \cdot π}{2}$

Step 5.2.5.2

Divide each term in $4 x = \frac{3 \cdot π}{2}$ by $4$ and simplify.

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Step 5.2.5.2.1

Divide each term in $4 x = \frac{3 \cdot π}{2}$ by $4$ .

$\frac{4 x}{4} = \frac{\frac{3 \cdot π}{2}}{4}$

Step 5.2.5.2.2

Simplify the left side.

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Step 5.2.5.2.2.1

Cancel the common factor of $4$ .

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Step 5.2.5.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{\frac{3 \cdot π}{2}}{4}$

Step 5.2.5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{3 \cdot π}{2}}{4}$

Step 5.2.5.2.3

Simplify the right side.

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Step 5.2.5.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{3 \cdot π}{2} \cdot \frac{1}{4}$

Step 5.2.5.2.3.2

Multiply $\frac{3 π}{2} \cdot \frac{1}{4}$ .

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Step 5.2.5.2.3.2.1

Multiply $\frac{3 π}{2}$ by $\frac{1}{4}$ .

$x = \frac{3 π}{2 \cdot 4}$

Step 5.2.5.2.3.2.2

Multiply $2$ by $4$ .

$x = \frac{3 π}{8}$

Step 5.2.6

The solution to the equation $4 x = \frac{π}{2}$ .

$x = \frac{π}{8}, \frac{3 π}{8}$

Step 6

Set $\csc (4 x)$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $\csc (4 x)$ equal to $0$ .

$\csc (4 x) = 0$

Step 6.2

The range of cosecant is $y \leq - 1$ and $y \geq 1$ . Since $0$ does not fall in this range, there is no solution.

No solution

Step 7

The final solution is all the values that make $- 12 \cot (4 x) \csc (4 x) = 0$ true.

$x = \frac{π}{8}, \frac{3 π}{8}$

Step 8

Evaluate the second derivative at $x = \frac{π}{8}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$48 \cot^{2} (4 (\frac{π}{8})) \csc (4 (\frac{π}{8})) + 48 \csc^{3} (4 (\frac{π}{8}))$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

Cancel the common factor of $4$ .

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Step 9.1.1.1

Factor $4$ out of $8$ .

$48 \cot^{2} (4 \frac{π}{4 (2)}) \csc (4 (\frac{π}{8})) + 48 \csc^{3} (4 (\frac{π}{8}))$

Step 9.1.1.2

Cancel the common factor.

$48 \cot^{2} (4 \frac{π}{4 \cdot 2}) \csc (4 (\frac{π}{8})) + 48 \csc^{3} (4 (\frac{π}{8}))$

Step 9.1.1.3

Rewrite the expression.

$48 \cot^{2} (\frac{π}{2}) \csc (4 (\frac{π}{8})) + 48 \csc^{3} (4 (\frac{π}{8}))$

Step 9.1.2

The exact value of $\cot (\frac{π}{2})$ is $0$ .

$48 \cdot 0^{2} \csc (4 (\frac{π}{8})) + 48 \csc^{3} (4 (\frac{π}{8}))$

Step 9.1.3

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

$48 \cdot 0 \csc (4 (\frac{π}{8})) + 48 \csc^{3} (4 (\frac{π}{8}))$

Step 9.1.4

Multiply $48$ by $0$ .

$0 \csc (4 (\frac{π}{8})) + 48 \csc^{3} (4 (\frac{π}{8}))$

Step 9.1.5

Cancel the common factor of $4$ .

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Step 9.1.5.1

Factor $4$ out of $8$ .

$0 \csc (4 \frac{π}{4 (2)}) + 48 \csc^{3} (4 (\frac{π}{8}))$

Step 9.1.5.2

Cancel the common factor.

$0 \csc (4 \frac{π}{4 \cdot 2}) + 48 \csc^{3} (4 (\frac{π}{8}))$

Step 9.1.5.3

Rewrite the expression.

$0 \csc (\frac{π}{2}) + 48 \csc^{3} (4 (\frac{π}{8}))$

Step 9.1.6

The exact value of $\csc (\frac{π}{2})$ is $1$ .

$0 \cdot 1 + 48 \csc^{3} (4 (\frac{π}{8}))$

Step 9.1.7

Multiply $0$ by $1$ .

$0 + 48 \csc^{3} (4 (\frac{π}{8}))$

Step 9.1.8

Cancel the common factor of $4$ .

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Step 9.1.8.1

Factor $4$ out of $8$ .

$0 + 48 \csc^{3} (4 \frac{π}{4 (2)})$

Step 9.1.8.2

Cancel the common factor.

$0 + 48 \csc^{3} (4 \frac{π}{4 \cdot 2})$

Step 9.1.8.3

Rewrite the expression.

$0 + 48 \csc^{3} (\frac{π}{2})$

Step 9.1.9

The exact value of $\csc (\frac{π}{2})$ is $1$ .

$0 + 48 \cdot 1^{3}$

Step 9.1.10

One to any power is one.

$0 + 48 \cdot 1$

Step 9.1.11

Multiply $48$ by $1$ .

$0 + 48$

Step 9.2

Add $0$ and $48$ .

$48$

Step 10

$x = \frac{π}{8}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{π}{8}$ is a local minimum

Step 11

Find the y-value when $x = \frac{π}{8}$ .

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Step 11.1

Replace the variable $x$ with $\frac{π}{8}$ in the expression.

$f (\frac{π}{8}) = 3 \csc (4 (\frac{π}{8}))$

Step 11.2

Simplify the result.

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Step 11.2.1

Cancel the common factor of $4$ .

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Step 11.2.1.1

Factor $4$ out of $8$ .

$f (\frac{π}{8}) = 3 \csc (4 (\frac{π}{4 (2)}))$

Step 11.2.1.2

Cancel the common factor.

$f (\frac{π}{8}) = 3 \csc (4 (\frac{π}{4 \cdot 2}))$

Step 11.2.1.3

Rewrite the expression.

$f (\frac{π}{8}) = 3 \csc (\frac{π}{2})$

Step 11.2.2

The exact value of $\csc (\frac{π}{2})$ is $1$ .

$f (\frac{π}{8}) = 3 \cdot 1$

Step 11.2.3

Multiply $3$ by $1$ .

$f (\frac{π}{8}) = 3$

Step 11.2.4

The final answer is $3$ .

$y = 3$

Step 12

Evaluate the second derivative at $x = \frac{3 π}{8}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$48 \cot^{2} (4 (\frac{3 π}{8})) \csc (4 (\frac{3 π}{8})) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify each term.

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Step 13.1.1

Cancel the common factor of $4$ .

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Step 13.1.1.1

Factor $4$ out of $8$ .

$48 \cot^{2} (4 \frac{3 π}{4 (2)}) \csc (4 (\frac{3 π}{8})) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.1.2

Cancel the common factor.

$48 \cot^{2} (4 \frac{3 π}{4 \cdot 2}) \csc (4 (\frac{3 π}{8})) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.1.3

Rewrite the expression.

$48 \cot^{2} (\frac{3 π}{2}) \csc (4 (\frac{3 π}{8})) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cotangent is negative in the fourth quadrant.

$48 {(- \cot (\frac{π}{2}))}^{2} \csc (4 (\frac{3 π}{8})) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.3

The exact value of $\cot (\frac{π}{2})$ is $0$ .

$48 {(- 0)}^{2} \csc (4 (\frac{3 π}{8})) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.4

Multiply $- 1$ by $0$ .

$48 \cdot 0^{2} \csc (4 (\frac{3 π}{8})) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.5

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

$48 \cdot 0 \csc (4 (\frac{3 π}{8})) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.6

Multiply $48$ by $0$ .

$0 \csc (4 (\frac{3 π}{8})) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.7

Cancel the common factor of $4$ .

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Step 13.1.7.1

Factor $4$ out of $8$ .

$0 \csc (4 \frac{3 π}{4 (2)}) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.7.2

Cancel the common factor.

$0 \csc (4 \frac{3 π}{4 \cdot 2}) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.7.3

Rewrite the expression.

$0 \csc (\frac{3 π}{2}) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.8

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosecant is negative in the fourth quadrant.

$0 (- \csc (\frac{π}{2})) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.9

The exact value of $\csc (\frac{π}{2})$ is $1$ .

$0 (- 1 \cdot 1) + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.10

Multiply $0 (- 1 \cdot 1)$ .

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Step 13.1.10.1

Multiply $- 1$ by $1$ .

$0 \cdot - 1 + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.10.2

Multiply $0$ by $- 1$ .

$0 + 48 \csc^{3} (4 (\frac{3 π}{8}))$

Step 13.1.11

Cancel the common factor of $4$ .

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Step 13.1.11.1

Factor $4$ out of $8$ .

$0 + 48 \csc^{3} (4 \frac{3 π}{4 (2)})$

Step 13.1.11.2

Cancel the common factor.

$0 + 48 \csc^{3} (4 \frac{3 π}{4 \cdot 2})$

Step 13.1.11.3

Rewrite the expression.

$0 + 48 \csc^{3} (\frac{3 π}{2})$

Step 13.1.12

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosecant is negative in the fourth quadrant.

$0 + 48 {(- \csc (\frac{π}{2}))}^{3}$

Step 13.1.13

The exact value of $\csc (\frac{π}{2})$ is $1$ .

$0 + 48 {(- 1 \cdot 1)}^{3}$

Step 13.1.14

Multiply $- 1$ by $1$ .

$0 + 48 {(- 1)}^{3}$

Step 13.1.15

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

$0 + 48 \cdot - 1$

Step 13.1.16

Multiply $48$ by $- 1$ .

$0 - 48$

Step 13.2

Subtract $48$ from $0$ .

$- 48$

Step 14

$x = \frac{3 π}{8}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{3 π}{8}$ is a local maximum

Step 15

Find the y-value when $x = \frac{3 π}{8}$ .

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Step 15.1

Replace the variable $x$ with $\frac{3 π}{8}$ in the expression.

$f (\frac{3 π}{8}) = 3 \csc (4 (\frac{3 π}{8}))$

Step 15.2

Simplify the result.

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Step 15.2.1

Cancel the common factor of $4$ .

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Step 15.2.1.1

Factor $4$ out of $8$ .

$f (\frac{3 π}{8}) = 3 \csc (4 (\frac{3 π}{4 (2)}))$

Step 15.2.1.2

Cancel the common factor.

$f (\frac{3 π}{8}) = 3 \csc (4 (\frac{3 π}{4 \cdot 2}))$

Step 15.2.1.3

Rewrite the expression.

$f (\frac{3 π}{8}) = 3 \csc (\frac{3 π}{2})$

Step 15.2.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosecant is negative in the fourth quadrant.

$f (\frac{3 π}{8}) = 3 (- \csc (\frac{π}{2}))$

Step 15.2.3

The exact value of $\csc (\frac{π}{2})$ is $1$ .

$f (\frac{3 π}{8}) = 3 (- 1 \cdot 1)$

Step 15.2.4

Multiply $3 (- 1 \cdot 1)$ .

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Step 15.2.4.1

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{8}) = 3 \cdot - 1$

Step 15.2.4.2

Multiply $3$ by $- 1$ .

$f (\frac{3 π}{8}) = - 3$

Step 15.2.5

The final answer is $- 3$ .

$y = - 3$

Step 16

These are the local extrema for $f (x) = 3 \csc (4 x)$ .

$(\frac{π}{8}, 3)$ is a local minima

$(\frac{3 π}{8}, - 3)$ is a local maxima

Step 17