Calculus Examples

$x + \sin (2 x) + 1$

Step 1

Write $x + \sin (2 x) + 1$ as a function.

$f (x) = x + \sin (2 x) + 1$

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

Differentiate.

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Step 2.1.1.1

By the Sum Rule, the derivative of $x + \sin (2 x) + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\sin (2 x)] + \frac{d}{d x} [1]$ .

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

Step 2.1.1.2

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

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

Step 2.1.2

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

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Step 2.1.2.1

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 2.1.2.1.1

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

$1 + \frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x] + \frac{d}{d x} [1]$

Step 2.1.2.1.2

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

$1 + \cos (u) \frac{d}{d x} [2 x] + \frac{d}{d x} [1]$

Step 2.1.2.1.3

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

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

Step 2.1.2.2

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

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

Step 2.1.2.3

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

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

Step 2.1.2.4

Multiply $2$ by $1$ .

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

Step 2.1.2.5

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

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

Step 2.1.3

Differentiate using the Constant Rule.

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Step 2.1.3.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Step 2.1.3.2

Add $1 + 2 \cos (2 x)$ and $0$ .

$f' (x) = 1 + 2 \cos (2 x)$

Step 2.2

Find the second derivative.

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Step 2.2.1

Differentiate.

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Step 2.2.1.1

By the Sum Rule, the derivative of $1 + 2 \cos (2 x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [2 \cos (2 x)]$ .

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

Step 2.2.1.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [2 \cos (2 x)]$

Step 2.2.2

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

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Step 2.2.2.1

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

$0 + 2 \frac{d}{d x} [\cos (2 x)]$

Step 2.2.2.2

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 2.2.2.2.1

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

$0 + 2 (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x])$

Step 2.2.2.2.2

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

$0 + 2 (- \sin (u) \frac{d}{d x} [2 x])$

Step 2.2.2.2.3

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

$0 + 2 (- \sin (2 x) \frac{d}{d x} [2 x])$

Step 2.2.2.3

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

$0 + 2 (- \sin (2 x) (2 \frac{d}{d x} [x]))$

Step 2.2.2.4

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

$0 + 2 (- \sin (2 x) (2 \cdot 1))$

Step 2.2.2.5

Multiply $2$ by $1$ .

$0 + 2 (- \sin (2 x) \cdot 2)$

Step 2.2.2.6

Multiply $2$ by $- 1$ .

$0 + 2 (- 2 \sin (2 x))$

Step 2.2.2.7

Multiply $- 2$ by $2$ .

$0 - 4 \sin (2 x)$

Step 2.2.3

Subtract $4 \sin (2 x)$ from $0$ .

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

Step 2.3

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

$- 4 \sin (2 x)$

Step 3

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

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Step 3.1

Set the second derivative equal to $0$ .

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

Step 3.2

Divide each term in $- 4 \sin (2 x) = 0$ by $- 4$ and simplify.

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Step 3.2.1

Divide each term in $- 4 \sin (2 x) = 0$ by $- 4$ .

$\frac{- 4 \sin (2 x)}{- 4} = \frac{0}{- 4}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $- 4$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{- 4 \sin (2 x)}{- 4} = \frac{0}{- 4}$

Step 3.2.2.1.2

Divide $\sin (2 x)$ by $1$ .

$\sin (2 x) = \frac{0}{- 4}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Divide $0$ by $- 4$ .

$\sin (2 x) = 0$

Step 3.3

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

$2 x = \arcsin (0)$

Step 3.4

Simplify the right side.

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Step 3.4.1

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

$2 x = 0$

Step 3.5

Divide each term in $2 x = 0$ by $2$ and simplify.

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Step 3.5.1

Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

Cancel the common factor of $2$ .

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Step 3.5.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Step 3.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Step 3.5.3

Simplify the right side.

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Step 3.5.3.1

Divide $0$ by $2$ .

$x = 0$

Step 3.6

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.

$2 x = π - 0$

Step 3.7

Solve for $x$ .

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Step 3.7.1

Simplify.

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Step 3.7.1.1

Multiply $- 1$ by $0$ .

$2 x = π + 0$

Step 3.7.1.2

Add $π$ and $0$ .

$2 x = π$

Step 3.7.2

Divide each term in $2 x = π$ by $2$ and simplify.

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Step 3.7.2.1

Divide each term in $2 x = π$ by $2$ .

$\frac{2 x}{2} = \frac{π}{2}$

Step 3.7.2.2

Simplify the left side.

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Step 3.7.2.2.1

Cancel the common factor of $2$ .

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Step 3.7.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Step 3.7.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{2}$

Step 3.8

Find the period of $\sin (2 x)$ .

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Step 3.8.1

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

$\frac{2 π}{| b |}$

Step 3.8.2

Replace $b$ with $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

Step 3.8.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Step 3.8.4

Cancel the common factor of $2$ .

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Step 3.8.4.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 3.8.4.2

Divide $π$ by $1$ .

$π$

Step 3.9

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

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

Step 3.10

Consolidate the answers.

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

Step 4

The point found by substituting in $f (x) = x + \sin (2 x) + 1$ is $(,)$ . This point can be an inflection point.

$(,)$

Step 5

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

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

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.

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

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $2$ by $- 0.1$ .

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

Step 6.2.2

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

$- 4 \sin (- 0.2)$

Step 6.3

At $- 0.1$ , the second derivative is $0.79467732$ . 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 $(, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

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

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

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $2$ by $0.1$ .

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

Step 7.2.2

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

$- 4 \sin (0.2)$

Step 7.3

At $0.1$ , the second derivative is $- 0.79467732$ . Since this is negative, the second derivative is decreasing on the interval $(, \infty)$

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

Step 8

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 point in this case is $(,)$ .

$(,)$

Step 9