Calculus Examples

$y = x + \sin (x)$

Step 1

Set $y$ as a function of $x$ .

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

Step 2

Find the derivative.

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Step 2.1

Differentiate.

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.2

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

$1 + \cos (x)$

Step 3

Set the derivative equal to $0$ then solve the equation $1 + \cos (x) = 0$ .

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Step 3.1

Subtract $1$ from both sides of the equation.

$\cos (x) = - 1$

Step 3.2

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

$x = \arccos (- 1)$

Step 3.3

Simplify the right side.

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Step 3.3.1

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

$x = π$

Step 3.4

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.

$x = 2 π - π$

Step 3.5

Subtract $π$ from $2 π$ .

$x = π$

Step 3.6

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

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Step 3.6.1

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

$\frac{2 π}{| b |}$

Step 3.6.2

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

$\frac{2 π}{| 1 |}$

Step 3.6.3

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

$\frac{2 π}{1}$

Step 3.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.7

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

$x = π + 2 π n$ , for any integer $n$

Step 4

Solve the original function $f (x) = x + \sin (x)$ at $x = π$ .

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Step 4.1

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

$f (π) = (π) + \sin (π)$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

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

$f (π) = π + \sin (0)$

Step 4.2.1.2

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

$f (π) = π + 0$

Step 4.2.2

Add $π$ and $0$ .

$f (π) = π$

Step 4.2.3

The final answer is $π$ .

$π$

Step 5

Solve the original function $f (x) = x + \sin (x)$ at $x = π + 2 π$ .

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Step 5.1

Replace the variable $x$ with $π + 2 π$ in the expression.

$f (π + 2 π) = (π + 2 π) + \sin (π + 2 π)$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Add $π$ and $2 π$ .

$f (π + 2 π) = π + 2 π + \sin (3 π)$

Step 5.2.1.2

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$f (π + 2 π) = π + 2 π + \sin (π)$

Step 5.2.1.3

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

$f (π + 2 π) = π + 2 π + \sin (0)$

Step 5.2.1.4

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

$f (π + 2 π) = π + 2 π + 0$

Step 5.2.2

Simplify by adding terms.

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Step 5.2.2.1

Add $π$ and $2 π$ .

$f (π + 2 π) = 3 π + 0$

Step 5.2.2.2

Add $3 π$ and $0$ .

$f (π + 2 π) = 3 π$

Step 5.2.3

The final answer is $3 π$ .

$3 π$

Step 6

The horizontal tangent line on function $f (x) = x + \sin (x)$ is $y = π, y = 3 π$ .

$y = π, y = 3 π$

Step 7