Calculus Examples

$6 \sin (x) + \sin (2 x)$

Step 1

Write $6 \sin (x) + \sin (2 x)$ as a function.

$f (x) = 6 \sin (x) + \sin (2 x)$

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

By the Sum Rule, the derivative of $6 \sin (x) + \sin (2 x)$ with respect to $x$ is $\frac{d}{d x} [6 \sin (x)] + \frac{d}{d x} [\sin (2 x)]$ .

$\frac{d}{d x} [6 \sin (x)] + \frac{d}{d x} [\sin (2 x)]$

Step 2.1.2

Evaluate $\frac{d}{d x} [6 \sin (x)]$ .

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Step 2.1.2.1

Since $6$ is constant with respect to $x$ , the derivative of $6 \sin (x)$ with respect to $x$ is $6 \frac{d}{d x} [\sin (x)]$ .

$6 \frac{d}{d x} [\sin (x)] + \frac{d}{d x} [\sin (2 x)]$

Step 2.1.2.2

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

$6 \cos (x) + \frac{d}{d x} [\sin (2 x)]$

Step 2.1.3

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

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Step 2.1.3.1

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 2.1.3.1.1

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

$6 \cos (x) + \frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x]$

Step 2.1.3.1.2

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

$6 \cos (x) + \cos (u) \frac{d}{d x} [2 x]$

Step 2.1.3.1.3

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

$6 \cos (x) + \cos (2 x) \frac{d}{d x} [2 x]$

Step 2.1.3.2

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

$6 \cos (x) + \cos (2 x) (2 \frac{d}{d x} [x])$

Step 2.1.3.3

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

$6 \cos (x) + \cos (2 x) (2 \cdot 1)$

Step 2.1.3.4

Multiply $2$ by $1$ .

$6 \cos (x) + \cos (2 x) \cdot 2$

Step 2.1.3.5

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

$f' (x) = 6 \cos (x) + 2 \cos (2 x)$

Step 2.2

Find the second derivative.

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Step 2.2.1

By the Sum Rule, the derivative of $6 \cos (x) + 2 \cos (2 x)$ with respect to $x$ is $\frac{d}{d x} [6 \cos (x)] + \frac{d}{d x} [2 \cos (2 x)]$ .

$\frac{d}{d x} [6 \cos (x)] + \frac{d}{d x} [2 \cos (2 x)]$

Step 2.2.2

Evaluate $\frac{d}{d x} [6 \cos (x)]$ .

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Step 2.2.2.1

Since $6$ is constant with respect to $x$ , the derivative of $6 \cos (x)$ with respect to $x$ is $6 \frac{d}{d x} [\cos (x)]$ .

$6 \frac{d}{d x} [\cos (x)] + \frac{d}{d x} [2 \cos (2 x)]$

Step 2.2.2.2

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

$6 (- \sin (x)) + \frac{d}{d x} [2 \cos (2 x)]$

Step 2.2.2.3

Multiply $- 1$ by $6$ .

$- 6 \sin (x) + \frac{d}{d x} [2 \cos (2 x)]$

Step 2.2.3

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

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Step 2.2.3.1

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

$- 6 \sin (x) + 2 \frac{d}{d x} [\cos (2 x)]$

Step 2.2.3.2

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

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Step 2.2.3.2.1

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

$- 6 \sin (x) + 2 (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x])$

Step 2.2.3.2.2

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

$- 6 \sin (x) + 2 (- \sin (u) \frac{d}{d x} [2 x])$

Step 2.2.3.2.3

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

$- 6 \sin (x) + 2 (- \sin (2 x) \frac{d}{d x} [2 x])$

Step 2.2.3.3

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

$- 6 \sin (x) + 2 (- \sin (2 x) (2 \frac{d}{d x} [x]))$

Step 2.2.3.4

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

$- 6 \sin (x) + 2 (- \sin (2 x) (2 \cdot 1))$

Step 2.2.3.5

Multiply $2$ by $1$ .

$- 6 \sin (x) + 2 (- \sin (2 x) \cdot 2)$

Step 2.2.3.6

Multiply $2$ by $- 1$ .

$- 6 \sin (x) + 2 (- 2 \sin (2 x))$

Step 2.2.3.7

Multiply $- 2$ by $2$ .

$f'' (x) = - 6 \sin (x) - 4 \sin (2 x)$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $- 6 \sin (x) - 4 \sin (2 x)$ .

$- 6 \sin (x) - 4 \sin (2 x)$

Step 3

Set the second derivative equal to $0$ then solve the equation $- 6 \sin (x) - 4 \sin (2 x) = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$- 6 \sin (x) - 4 \sin (2 x) = 0$

Step 3.2

Simplify each term.

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Step 3.2.1

Apply the sine double-angle identity.

$- 6 \sin (x) - 4 (2 \sin (x) \cos (x)) = 0$

Step 3.2.2

Multiply $2$ by $- 4$ .

$- 6 \sin (x) - 8 \sin (x) \cos (x) = 0$

Step 3.3

Factor $2 \sin (x)$ out of $- 6 \sin (x) - 8 \sin (x) \cos (x)$ .

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Step 3.3.1

Factor $2 \sin (x)$ out of $- 6 \sin (x)$ .

$2 \sin (x) \cdot - 3 - 8 \sin (x) \cos (x) = 0$

Step 3.3.2

Factor $2 \sin (x)$ out of $- 8 \sin (x) \cos (x)$ .

$2 \sin (x) \cdot - 3 + 2 \sin (x) (- 4 \cos (x)) = 0$

Step 3.3.3

Factor $2 \sin (x)$ out of $2 \sin (x) \cdot - 3 + 2 \sin (x) (- 4 \cos (x))$ .

$2 \sin (x) (- 3 - 4 \cos (x)) = 0$

Step 3.4

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

$\sin (x) = 0$

$- 3 - 4 \cos (x) = 0$

Step 3.5

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

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Step 3.5.1

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

$\sin (x) = 0$

Step 3.5.2

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

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Step 3.5.2.1

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

$x = \arcsin (0)$

Step 3.5.2.2

Simplify the right side.

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Step 3.5.2.2.1

The exact value of $\arcsin (0)$ is $0$ .

$x = 0$

Step 3.5.2.3

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

$x = π - 0$

Step 3.5.2.4

Subtract $0$ from $π$ .

$x = π$

Step 3.5.2.5

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

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Step 3.5.2.5.1

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

$\frac{2 π}{| b |}$

Step 3.5.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 3.5.2.5.3

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

$\frac{2 π}{1}$

Step 3.5.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.5.2.6

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

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

Step 3.6

Set $- 3 - 4 \cos (x)$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $- 3 - 4 \cos (x)$ equal to $0$ .

$- 3 - 4 \cos (x) = 0$

Step 3.6.2

Solve $- 3 - 4 \cos (x) = 0$ for $x$ .

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Step 3.6.2.1

Add $3$ to both sides of the equation.

$- 4 \cos (x) = 3$

Step 3.6.2.2

Divide each term in $- 4 \cos (x) = 3$ by $- 4$ and simplify.

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Step 3.6.2.2.1

Divide each term in $- 4 \cos (x) = 3$ by $- 4$ .

$\frac{- 4 \cos (x)}{- 4} = \frac{3}{- 4}$

Step 3.6.2.2.2

Simplify the left side.

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Step 3.6.2.2.2.1

Cancel the common factor of $- 4$ .

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Step 3.6.2.2.2.1.1

Cancel the common factor.

$\frac{- 4 \cos (x)}{- 4} = \frac{3}{- 4}$

Step 3.6.2.2.2.1.2

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

$\cos (x) = \frac{3}{- 4}$

Step 3.6.2.2.3

Simplify the right side.

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Step 3.6.2.2.3.1

Move the negative in front of the fraction.

$\cos (x) = - \frac{3}{4}$

Step 3.6.2.3

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

$x = \arccos (- \frac{3}{4})$

Step 3.6.2.4

Simplify the right side.

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Step 3.6.2.4.1

Evaluate $\arccos (- \frac{3}{4})$ .

$x = 2.4188584$

Step 3.6.2.5

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.

$x = 2 (3.14159265) - 2.4188584$

Step 3.6.2.6

Solve for $x$ .

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Step 3.6.2.6.1

Remove parentheses.

$x = 2 (3.14159265) - 2.4188584$

Step 3.6.2.6.2

Simplify $2 (3.14159265) - 2.4188584$ .

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Step 3.6.2.6.2.1

Multiply $2$ by $3.14159265$ .

$x = 6.2831853 - 2.4188584$

Step 3.6.2.6.2.2

Subtract $2.4188584$ from $6.2831853$ .

$x = 3.8643269$

Step 3.6.2.7

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

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Step 3.6.2.7.1

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

$\frac{2 π}{| b |}$

Step 3.6.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 3.6.2.7.3

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

$\frac{2 π}{1}$

Step 3.6.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.6.2.8

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

$x = 2.4188584 + 2 π n, 3.8643269 + 2 π n$ , for any integer $n$

Step 3.7

The final solution is all the values that make $2 \sin (x) (- 3 - 4 \cos (x)) = 0$ true.

$x = 2 π n, π + 2 π n, 2.4188584 + 2 π n, 3.8643269 + 2 π n$ , for any integer $n$

Step 3.8

Consolidate $2 π n$ and $π + 2 π n$ to $π n$ .

$x = π n, 2.4188584 + 2 π n, 3.8643269 + 2 π n$ , for any integer $n$

Step 4

Find the points where the second derivative is $0$ .

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Step 4.1

The point found by substituting in $f (x) = 6 \sin (x) + \sin (2 x)$ is $(,)$ . This point can be an inflection point.

$(,)$

Step 4.2

Substitute $2.4188584$ in $f (x) = 6 \sin (x) + \sin (2 x)$ to find the value of $y$ .

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Step 4.2.1

Replace the variable $x$ with $2.4188584$ in the expression.

$f (2.4188584) = 6 \sin (2.4188584) + \sin (2 (2.4188584))$

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Multiply $2$ by $2.4188584$ .

$f (2.4188584) = 6 \sin (2.4188584) + \sin (4.83771681)$

Step 4.2.2.2

The final answer is $6 \sin (2.4188584) + \sin (4.83771681)$ .

$6 \sin (2.4188584) + \sin (4.83771681)$

Step 4.3

The point found by substituting $2.4188584$ in $f (x) = 6 \sin (x) + \sin (2 x)$ is $(2.4188584, 6 \sin (2.4188584) + \sin (4.83771681))$ . This point can be an inflection point.

$(2.4188584, 6 \sin (2.4188584) + \sin (4.83771681))$

Step 4.4

Substitute $3.8643269$ in $f (x) = 6 \sin (x) + \sin (2 x)$ to find the value of $y$ .

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Step 4.4.1

Replace the variable $x$ with $3.8643269$ in the expression.

$f (3.8643269) = 6 \sin (3.8643269) + \sin (2 (3.8643269))$

Step 4.4.2

Simplify the result.

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Step 4.4.2.1

Multiply $2$ by $3.8643269$ .

$f (3.8643269) = 6 \sin (3.8643269) + \sin (7.7286538)$

Step 4.4.2.2

The final answer is $6 \sin (3.8643269) + \sin (7.7286538)$ .

$6 \sin (3.8643269) + \sin (7.7286538)$

Step 4.5

The point found by substituting $3.8643269$ in $f (x) = 6 \sin (x) + \sin (2 x)$ is $(3.8643269, 6 \sin (3.8643269) + \sin (7.7286538))$ . This point can be an inflection point.

$(3.8643269, 6 \sin (3.8643269) + \sin (7.7286538))$

Step 4.6

Determine the points that could be inflection points.

$(,), (2.4188584, 6 \sin (2.4188584) + \sin (4.83771681)), (3.8643269, 6 \sin (3.8643269) + \sin (7.7286538))$

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty,) \cup (, 2.4188584) \cup (2.4188584, 3.8643269) \cup (3.8643269, \infty)$

Step 6

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

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Step 6.1

Replace the variable $x$ with $- 0.1$ in the expression.

$f'' (- 0.1) = - 6 \sin (- 0.1) - 4 \sin (2 (- 0.1))$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $2$ by $- 0.1$ .

$f'' (- 0.1) = - 6 \sin (- 0.1) - 4 \sin (- 0.2)$

Step 6.2.2

The final answer is $- 6 \sin (- 0.1) - 4 \sin (- 0.2)$ .

$- 6 \sin (- 0.1) - 4 \sin (- 0.2)$

Step 6.3

At $- 0.1$ , the second derivative is $1.39367782$ . Since this is positive, the second derivative is increasing on the interval $(- \infty,)$ .

Increasing on $(- \infty,)$ since $f'' (x) > 0$

Step 7

Substitute a value from the interval $(, 2.4188584)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $1.2094292$ in the expression.

$f'' (1.2094292) = - 6 \sin (1.2094292) - 4 \sin (2 (1.2094292))$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $2$ by $1.2094292$ .

$f'' (1.2094292) = - 6 \sin (1.2094292) - 4 \sin (2.4188584)$

Step 7.2.2

The final answer is $- 6 \sin (1.2094292) - 4 \sin (2.4188584)$ .

$- 6 \sin (1.2094292) - 4 \sin (2.4188584)$

Step 7.3

At $1.2094292$ , the second derivative is $- 8.25823739$ . Since this is negative, the second derivative is decreasing on the interval $(, 2.4188584)$

Decreasing on $(, 2.4188584)$ since $f'' (x) < 0$

Step 8

Substitute a value from the interval $(2.4188584, 3.8643269)$ into the second derivative to determine if it is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $3.14159265$ in the expression.

$f'' (3.14159265) = - 6 \sin (3.14159265) - 4 \sin (2 (3.14159265))$

Step 8.2

Simplify the result.

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Step 8.2.1

Multiply $2$ by $3.14159265$ .

$f'' (3.14159265) = - 6 \sin (3.14159265) - 4 \sin (6.2831853)$

Step 8.2.2

The final answer is $- 6 \sin (3.14159265) - 4 \sin (6.2831853)$ .

$- 6 \sin (3.14159265) - 4 \sin (6.2831853)$

Step 8.3

At $3.14159265$ , the second derivative is $2.44929359 E - 16$ . Since this is positive, the second derivative is increasing on the interval $(2.4188584, 3.8643269)$ .

Increasing on $(2.4188584, 3.8643269)$ since $f'' (x) > 0$

Step 9

Substitute a value from the interval $(3.8643269, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 9.1

Replace the variable $x$ with $3.9643269$ in the expression.

$f'' (3.9643269) = - 6 \sin (3.9643269) - 4 \sin (2 (3.9643269))$

Step 9.2

Simplify the result.

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Step 9.2.1

Multiply $2$ by $3.9643269$ .

$f'' (3.9643269) = - 6 \sin (3.9643269) - 4 \sin (7.9286538)$

Step 9.2.2

The final answer is $- 6 \sin (3.9643269) - 4 \sin (7.9286538)$ .

$- 6 \sin (3.9643269) - 4 \sin (7.9286538)$

Step 9.3

At $3.9643269$ , the second derivative is $0.40919742$ . Since this is positive, the second derivative is increasing on the interval $(3.8643269, \infty)$ .

Increasing on $(3.8643269, \infty)$ since $f'' (x) > 0$

Step 10

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection points in this case are $(,), (2.4188584, 6 \sin (2.4188584) + \sin (4.83771681))$ .

$(,), (2.4188584, 6 \sin (2.4188584) + \sin (4.83771681))$

Step 11