Calculus Examples

$y = 2 (\frac{1}{\sin (x)})$

Step 1

Write $y = 2 (\frac{1}{\sin (x)})$ as a function.

$f (x) = 2 (\frac{1}{\sin (x)})$

Step 2

Find the first derivative of the function.

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Step 2.1

Convert from $\frac{1}{\sin (x)}$ to $\csc (x)$ .

$\frac{d}{d x} [2 \csc (x)]$

Step 2.2

Since $2$ is constant with respect to $x$ , the derivative of $2 \csc (x)$ with respect to $x$ is $2 \frac{d}{d x} [\csc (x)]$ .

$2 \frac{d}{d x} [\csc (x)]$

Step 2.3

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

$2 (- \csc (x) \cot (x))$

Step 2.4

Simplify the expression.

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Step 2.4.1

Multiply $- 1$ by $2$ .

$- 2 (\csc (x) \cot (x))$

Step 2.4.2

Reorder the factors of $- 2 \csc (x) \cot (x)$ .

$- 2 \cot (x) \csc (x)$

Step 3

Find the second derivative of the function.

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Step 3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 \cot (x) \csc (x)$ with respect to $x$ is $- 2 \frac{d}{d x} [\cot (x) \csc (x)]$ .

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

Step 3.2

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 $f (x) = \cot (x)$ and $g (x) = \csc (x)$ .

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

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

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

Step 3.7

Add $1$ and $1$ .

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

Step 3.8

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

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

Step 3.9

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

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

Step 3.10

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

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

Step 3.11

Add $1$ and $2$ .

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

Step 3.12

Simplify.

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Step 3.12.1

Apply the distributive property.

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

Step 3.12.2

Combine terms.

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Step 3.12.2.1

Multiply $- 1$ by $- 2$ .

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

Step 3.12.2.2

Multiply $- 1$ by $- 2$ .

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

Step 3.12.3

Reorder terms.

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

Step 4

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

$- 2 \cot (x) \csc (x) = 0$

Step 5

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

$\cot (x) = 0$

$\csc (x) = 0$

Step 6

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

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Step 6.1

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

$\cot (x) = 0$

Step 6.2

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

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Step 6.2.1

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

$x = arccot (0)$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

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

$x = \frac{π}{2}$

Step 6.2.3

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.

$x = π + \frac{π}{2}$

Step 6.2.4

Simplify $π + \frac{π}{2}$ .

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Step 6.2.4.1

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

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

Step 6.2.4.2

Combine fractions.

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Step 6.2.4.2.1

Combine $π$ and $\frac{2}{2}$ .

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

Step 6.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 + π}{2}$

Step 6.2.4.3

Simplify the numerator.

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Step 6.2.4.3.1

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π + π}{2}$

Step 6.2.4.3.2

Add $2 π$ and $π$ .

$x = \frac{3 π}{2}$

Step 6.2.5

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

$x = \frac{π}{2}, \frac{3 π}{2}$

Step 7

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

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Step 7.1

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

$\csc (x) = 0$

Step 7.2

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 8

The final solution is all the values that make $- 2 \cot (x) \csc (x) = 0$ true.

$x = \frac{π}{2}, \frac{3 π}{2}$

Step 9

Evaluate the second derivative at $x = \frac{π}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

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

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

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

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

Step 10.1.2

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

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

Step 10.1.3

Multiply $2$ by $0$ .

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

Step 10.1.4

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

$0 \cdot 1 + 2 \csc^{3} (\frac{π}{2})$

Step 10.1.5

Multiply $0$ by $1$ .

$0 + 2 \csc^{3} (\frac{π}{2})$

Step 10.1.6

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

$0 + 2 \cdot 1^{3}$

Step 10.1.7

One to any power is one.

$0 + 2 \cdot 1$

Step 10.1.8

Multiply $2$ by $1$ .

$0 + 2$

Step 10.2

Add $0$ and $2$ .

$2$

Step 11

$x = \frac{π}{2}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{π}{2}$ is a local minimum

Step 12

Find the y-value when $x = \frac{π}{2}$ .

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Step 12.1

Replace the variable $x$ with $\frac{π}{2}$ in the expression.

$f (\frac{π}{2}) = 2 (\frac{1}{\sin (\frac{π}{2})})$

Step 12.2

Simplify the result.

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Step 12.2.1

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

$f (\frac{π}{2}) = 2 (\frac{1}{1})$

Step 12.2.2

Cancel the common factor of $1$ .

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Step 12.2.2.1

Cancel the common factor.

$f (\frac{π}{2}) = 2 (\frac{1}{1})$

Step 12.2.2.2

Rewrite the expression.

$f (\frac{π}{2}) = 2 \cdot 1$

Step 12.2.3

Multiply $2$ by $1$ .

$f (\frac{π}{2}) = 2$

Step 12.2.4

The final answer is $2$ .

$y = 2$

Step 13

Evaluate the second derivative at $x = \frac{3 π}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

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

Step 14

Evaluate the second derivative.

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Step 14.1

Simplify each term.

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Step 14.1.1

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.

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

Step 14.1.2

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

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

Step 14.1.3

Multiply $- 1$ by $0$ .

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

Step 14.1.4

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

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

Step 14.1.5

Multiply $2$ by $0$ .

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

Step 14.1.6

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.

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

Step 14.1.7

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

$0 (- 1 \cdot 1) + 2 \csc^{3} (\frac{3 π}{2})$

Step 14.1.8

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

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Step 14.1.8.1

Multiply $- 1$ by $1$ .

$0 \cdot - 1 + 2 \csc^{3} (\frac{3 π}{2})$

Step 14.1.8.2

Multiply $0$ by $- 1$ .

$0 + 2 \csc^{3} (\frac{3 π}{2})$

Step 14.1.9

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.

$0 + 2 {(- \csc (\frac{π}{2}))}^{3}$

Step 14.1.10

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

$0 + 2 {(- 1 \cdot 1)}^{3}$

Step 14.1.11

Multiply $- 1$ by $1$ .

$0 + 2 {(- 1)}^{3}$

Step 14.1.12

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

$0 + 2 \cdot - 1$

Step 14.1.13

Multiply $2$ by $- 1$ .

$0 - 2$

Step 14.2

Subtract $2$ from $0$ .

$- 2$

Step 15

$x = \frac{3 π}{2}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{3 π}{2}$ is a local maximum

Step 16

Find the y-value when $x = \frac{3 π}{2}$ .

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Step 16.1

Replace the variable $x$ with $\frac{3 π}{2}$ in the expression.

$f (\frac{3 π}{2}) = 2 (\frac{1}{\sin (\frac{3 π}{2})})$

Step 16.2

Simplify the result.

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Step 16.2.1

Simplify the denominator.

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Step 16.2.1.1

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

$f (\frac{3 π}{2}) = 2 (\frac{1}{- \sin (\frac{π}{2})})$

Step 16.2.1.2

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

$f (\frac{3 π}{2}) = 2 (\frac{1}{- 1 \cdot 1})$

Step 16.2.1.3

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{2}) = 2 (\frac{1}{- 1})$

Step 16.2.2

Simplify the expression.

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Step 16.2.2.1

Divide $1$ by $- 1$ .

$f (\frac{3 π}{2}) = 2 \cdot - 1$

Step 16.2.2.2

Multiply $2$ by $- 1$ .

$f (\frac{3 π}{2}) = - 2$

Step 16.2.3

The final answer is $- 2$ .

$y = - 2$

Step 17

These are the local extrema for $f (x) = 2 (\frac{1}{\sin (x)})$ .

$(\frac{π}{2}, 2)$ is a local minima

$(\frac{3 π}{2}, - 2)$ is a local maxima

Step 18