Calculus Examples

$\sin (2 x) - 2 \sin (x)$

Step 1

Find the derivative.

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Step 1.1

By the Sum Rule, the derivative of $\sin (2 x) - 2 \sin (x)$ with respect to $x$ is $\frac{d}{d x} [\sin (2 x)] + \frac{d}{d x} [- 2 \sin (x)]$ .

$\frac{d}{d x} [\sin (2 x)] + \frac{d}{d x} [- 2 \sin (x)]$

Step 1.2

Evaluate $\frac{d}{d x} [\sin (2 x)]$ .

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Step 1.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 1.2.1.1

To apply the Chain Rule, set $u$ as $2 x$ .

$\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x] + \frac{d}{d x} [- 2 \sin (x)]$

Step 1.2.1.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$\cos (u) \frac{d}{d x} [2 x] + \frac{d}{d x} [- 2 \sin (x)]$

Step 1.2.1.3

Replace all occurrences of $u$ with $2 x$ .

$\cos (2 x) \frac{d}{d x} [2 x] + \frac{d}{d x} [- 2 \sin (x)]$

Step 1.2.2

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

$\cos (2 x) (2 \frac{d}{d x} [x]) + \frac{d}{d x} [- 2 \sin (x)]$

Step 1.2.3

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

$\cos (2 x) (2 \cdot 1) + \frac{d}{d x} [- 2 \sin (x)]$

Step 1.2.4

Multiply $2$ by $1$ .

$\cos (2 x) \cdot 2 + \frac{d}{d x} [- 2 \sin (x)]$

Step 1.2.5

Move $2$ to the left of $\cos (2 x)$ .

$2 \cos (2 x) + \frac{d}{d x} [- 2 \sin (x)]$

Step 1.3

Evaluate $\frac{d}{d x} [- 2 \sin (x)]$ .

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Step 1.3.1

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

$2 \cos (2 x) - 2 \frac{d}{d x} [\sin (x)]$

Step 1.3.2

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$2 \cos (2 x) - 2 \cos (x)$

Step 2

Set the derivative equal to $0$ then solve the equation $2 \cos (2 x) - 2 \cos (x) = 0$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Use the double-angle identity to transform $\cos (2 x)$ to $2 \cos^{2} (x) - 1$ .

$2 (2 \cos^{2} (x) - 1) - 2 \cos (x) = 0$

Step 2.1.2

Apply the distributive property.

$2 (2 \cos^{2} (x)) + 2 \cdot - 1 - 2 \cos (x) = 0$

Step 2.1.3

Multiply $2$ by $2$ .

$4 \cos^{2} (x) + 2 \cdot - 1 - 2 \cos (x) = 0$

Step 2.1.4

Multiply $2$ by $- 1$ .

$4 \cos^{2} (x) - 2 - 2 \cos (x) = 0$

Step 2.2

Factor $4 \cos^{2} (x) - 2 - 2 \cos (x)$ .

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Step 2.2.1

Factor $2$ out of $4 \cos^{2} (x) - 2 - 2 \cos (x)$ .

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Step 2.2.1.1

Factor $2$ out of $4 \cos^{2} (x)$ .

$2 (2 \cos^{2} (x)) - 2 - 2 \cos (x) = 0$

Step 2.2.1.2

Factor $2$ out of $- 2$ .

$2 (2 \cos^{2} (x)) + 2 (- 1) - 2 \cos (x) = 0$

Step 2.2.1.3

Factor $2$ out of $- 2 \cos (x)$ .

$2 (2 \cos^{2} (x)) + 2 (- 1) + 2 (- \cos (x)) = 0$

Step 2.2.1.4

Factor $2$ out of $2 (2 \cos^{2} (x)) + 2 (- 1)$ .

$2 (2 \cos^{2} (x) - 1) + 2 (- \cos (x)) = 0$

Step 2.2.1.5

Factor $2$ out of $2 (2 \cos^{2} (x) - 1) + 2 (- \cos (x))$ .

$2 (2 \cos^{2} (x) - 1 - \cos (x)) = 0$

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor by grouping.

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Step 2.2.2.1.1

Reorder terms.

$2 (2 \cos^{2} (x) - \cos (x) - 1) = 0$

Step 2.2.2.1.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = - 1$ .

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Step 2.2.2.1.2.1

Factor $- 1$ out of $- \cos (x)$ .

$2 (2 \cos^{2} (x) - \cos (x) - 1) = 0$

Step 2.2.2.1.2.2

Rewrite $- 1$ as $1$ plus $- 2$

$2 (2 \cos^{2} (x) + (1 - 2) \cos (x) - 1) = 0$

Step 2.2.2.1.2.3

Apply the distributive property.

$2 (2 \cos^{2} (x) + 1 \cos (x) - 2 \cos (x) - 1) = 0$

Step 2.2.2.1.2.4

Multiply $\cos (x)$ by $1$ .

$2 (2 \cos^{2} (x) + \cos (x) - 2 \cos (x) - 1) = 0$

Step 2.2.2.1.3

Factor out the greatest common factor from each group.

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Step 2.2.2.1.3.1

Group the first two terms and the last two terms.

$2 (2 \cos^{2} (x) + \cos (x) - 2 \cos (x) - 1) = 0$

Step 2.2.2.1.3.2

Factor out the greatest common factor (GCF) from each group.

$2 (\cos (x) (2 \cos (x) + 1) - (2 \cos (x) + 1)) = 0$

Step 2.2.2.1.4

Factor the polynomial by factoring out the greatest common factor, $2 \cos (x) + 1$ .

$2 ((2 \cos (x) + 1) (\cos (x) - 1)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$2 (2 \cos (x) + 1) (\cos (x) - 1) = 0$

Step 2.3

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

$2 \cos (x) + 1 = 0$

$\cos (x) - 1 = 0$

Step 2.4

Set $2 \cos (x) + 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $2 \cos (x) + 1$ equal to $0$ .

$2 \cos (x) + 1 = 0$

Step 2.4.2

Solve $2 \cos (x) + 1 = 0$ for $x$ .

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Step 2.4.2.1

Subtract $1$ from both sides of the equation.

$2 \cos (x) = - 1$

Step 2.4.2.2

Divide each term in $2 \cos (x) = - 1$ by $2$ and simplify.

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Step 2.4.2.2.1

Divide each term in $2 \cos (x) = - 1$ by $2$ .

$\frac{2 \cos (x)}{2} = \frac{- 1}{2}$

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

$\frac{2 \cos (x)}{2} = \frac{- 1}{2}$

Step 2.4.2.2.2.1.2

Divide $\cos (x)$ by $1$ .

$\cos (x) = \frac{- 1}{2}$

Step 2.4.2.2.3

Simplify the right side.

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Step 2.4.2.2.3.1

Move the negative in front of the fraction.

$\cos (x) = - \frac{1}{2}$

Step 2.4.2.3

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

$x = \arccos (- \frac{1}{2})$

Step 2.4.2.4

Simplify the right side.

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Step 2.4.2.4.1

The exact value of $\arccos (- \frac{1}{2})$ is $\frac{2 π}{3}$ .

$x = \frac{2 π}{3}$

Step 2.4.2.5

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$x = 2 π - \frac{2 π}{3}$

Step 2.4.2.6

Simplify $2 π - \frac{2 π}{3}$ .

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Step 2.4.2.6.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 2.4.2.6.2

Combine fractions.

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Step 2.4.2.6.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 2.4.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - 2 π}{3}$

Step 2.4.2.6.3

Simplify the numerator.

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Step 2.4.2.6.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - 2 π}{3}$

Step 2.4.2.6.3.2

Subtract $2 π$ from $6 π$ .

$x = \frac{4 π}{3}$

Step 2.4.2.7

Find the period of $\cos (x)$ .

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Step 2.4.2.7.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 2.4.2.7.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 2.4.2.7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 2.4.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.4.2.8

The period of the $\cos (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer $n$

Step 2.5

Set $\cos (x) - 1$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $\cos (x) - 1$ equal to $0$ .

$\cos (x) - 1 = 0$

Step 2.5.2

Solve $\cos (x) - 1 = 0$ for $x$ .

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Step 2.5.2.1

Add $1$ to both sides of the equation.

$\cos (x) = 1$

Step 2.5.2.2

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

$x = \arccos (1)$

Step 2.5.2.3

Simplify the right side.

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Step 2.5.2.3.1

The exact value of $\arccos (1)$ is $0$ .

$x = 0$

Step 2.5.2.4

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - 0$

Step 2.5.2.5

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 2.5.2.6

Find the period of $\cos (x)$ .

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Step 2.5.2.6.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 2.5.2.6.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 2.5.2.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 2.5.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.5.2.7

The period of the $\cos (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 2.6

The final solution is all the values that make $2 (2 \cos (x) + 1) (\cos (x) - 1) = 0$ true.

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 2.7

Consolidate the answers.

$x = \frac{2 π n}{3}$ , for any integer $n$

Step 3

Solve the original function $f (x) = \sin (2 x) - 2 \sin (x)$ at $x = \frac{2 π}{3}$ .

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Step 3.1

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

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply $2 (\frac{2 π}{3})$ .

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Step 3.2.1.1.1

Combine $2$ and $\frac{2 π}{3}$ .

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

Step 3.2.1.1.2

Multiply $2$ by $2$ .

$f (\frac{2 π}{3}) = \sin (\frac{4 π}{3}) - 2 \sin (\frac{2 π}{3})$

Step 3.2.1.2

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 third quadrant.

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

Step 3.2.1.3

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

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

Step 3.2.1.4

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

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

Step 3.2.1.5

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

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

Step 3.2.1.6

Cancel the common factor of $2$ .

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Step 3.2.1.6.1

Factor $2$ out of $- 2$ .

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

Step 3.2.1.6.2

Cancel the common factor.

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

Step 3.2.1.6.3

Rewrite the expression.

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

Step 3.2.1.7

Rewrite $- 1 \sqrt{3}$ as $- \sqrt{3}$ .

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

Step 3.2.2

To write $- \sqrt{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.2.3

Combine fractions.

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Step 3.2.3.1

Combine $- \sqrt{3}$ and $\frac{2}{2}$ .

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

Step 3.2.3.2

Combine the numerators over the common denominator.

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

Step 3.2.4

Simplify the numerator.

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Step 3.2.4.1

Multiply $2$ by $- 1$ .

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

Step 3.2.4.2

Subtract $2 \sqrt{3}$ from $- \sqrt{3}$ .

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

Step 3.2.5

Move the negative in front of the fraction.

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

Step 3.2.6

The final answer is $- \frac{3 \sqrt{3}}{2}$ .

$- \frac{3 \sqrt{3}}{2}$

Step 4

The horizontal tangent line on function $f (x) = \sin (2 x) - 2 \sin (x)$ is $y = x = \frac{2 π n}{3}, y = - \frac{3 \sqrt{3}}{2}$ .

$y = x = \frac{2 π n}{3}, y = - \frac{3 \sqrt{3}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$y = x = \frac{2 π n}{3}, - \frac{3 \sqrt{3}}{2}$

Decimal Form:

$y = x = \frac{2 π n}{3}, - 2.59807621 \dots$

Step 6