Calculus Examples

$\sin (x) - \cos (x)$

Step 1

Write $\sin (x) - \cos (x)$ as a function.

$f (x) = \sin (x) - \cos (x)$

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 $\sin (x) - \cos (x)$ with respect to $x$ is $\frac{d}{d x} [\sin (x)] + \frac{d}{d x} [- \cos (x)]$ .

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

Step 2.1.2

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

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

Step 2.1.3

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

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Step 2.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos (x)$ with respect to $x$ is $- \frac{d}{d x} [\cos (x)]$ .

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

Step 2.1.3.2

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

$\cos (x) - - \sin (x)$

Step 2.1.3.3

Multiply $- 1$ by $- 1$ .

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

Step 2.1.3.4

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

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

Step 2.2

Find the second derivative.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

$f'' (x) = - \sin (x) + \cos (x)$

Step 2.3

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

$- \sin (x) + \cos (x)$

Step 3

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

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Step 3.1

Set the second derivative equal to $0$ .

$- \sin (x) + \cos (x) = 0$

Step 3.2

Divide each term in the equation by $\cos (x)$ .

$\frac{- \sin (x)}{\cos (x)} + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 3.3

Separate fractions.

$\frac{- 1}{1} \cdot \frac{\sin (x)}{\cos (x)} + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 3.4

Convert from $\frac{\sin (x)}{\cos (x)}$ to $\tan (x)$ .

$\frac{- 1}{1} \cdot \tan (x) + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 3.5

Divide $- 1$ by $1$ .

$- \tan (x) + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 3.6

Cancel the common factor of $\cos (x)$ .

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Step 3.6.1

Cancel the common factor.

$- \tan (x) + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 3.6.2

Rewrite the expression.

$- \tan (x) + 1 = \frac{0}{\cos (x)}$

Step 3.7

Separate fractions.

$- \tan (x) + 1 = \frac{0}{1} \cdot \frac{1}{\cos (x)}$

Step 3.8

Convert from $\frac{1}{\cos (x)}$ to $\sec (x)$ .

$- \tan (x) + 1 = \frac{0}{1} \cdot \sec (x)$

Step 3.9

Divide $0$ by $1$ .

$- \tan (x) + 1 = 0 \sec (x)$

Step 3.10

Multiply $0$ by $\sec (x)$ .

$- \tan (x) + 1 = 0$

Step 3.11

Subtract $1$ from both sides of the equation.

$- \tan (x) = - 1$

Step 3.12

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

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Step 3.12.1

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

$\frac{- \tan (x)}{- 1} = \frac{- 1}{- 1}$

Step 3.12.2

Simplify the left side.

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Step 3.12.2.1

Dividing two negative values results in a positive value.

$\frac{\tan (x)}{1} = \frac{- 1}{- 1}$

Step 3.12.2.2

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

$\tan (x) = \frac{- 1}{- 1}$

Step 3.12.3

Simplify the right side.

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Step 3.12.3.1

Divide $- 1$ by $- 1$ .

$\tan (x) = 1$

Step 3.13

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

$x = \arctan (1)$

Step 3.14

Simplify the right side.

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Step 3.14.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$x = \frac{π}{4}$

Step 3.15

The tangent 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.

$x = π + \frac{π}{4}$

Step 3.16

Simplify $π + \frac{π}{4}$ .

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Step 3.16.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 3.16.2

Combine fractions.

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Step 3.16.2.1

Combine $π$ and $\frac{4}{4}$ .

$x = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 3.16.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Step 3.16.3

Simplify the numerator.

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Step 3.16.3.1

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Step 3.16.3.2

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

Step 3.17

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

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Step 3.17.1

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

$\frac{π}{| b |}$

Step 3.17.2

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

$\frac{π}{| 1 |}$

Step 3.17.3

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

$\frac{π}{1}$

Step 3.17.4

Divide $π$ by $1$ .

$π$

Step 3.18

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

$x = \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer $n$

Step 4

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

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Step 4.1

Substitute $\frac{π}{4}$ in $f (x) = \sin (x) - \cos (x)$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $\frac{π}{4}$ in the expression.

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

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{π}{4}) = \frac{\sqrt{2}}{2} - \cos (\frac{π}{4})$

Step 4.1.2.1.2

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

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

Step 4.1.2.2

Simplify terms.

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Step 4.1.2.2.1

Combine the numerators over the common denominator.

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

Step 4.1.2.2.2

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

$f (\frac{π}{4}) = \frac{0}{2}$

Step 4.1.2.2.3

Divide $0$ by $2$ .

$f (\frac{π}{4}) = 0$

Step 4.1.2.3

The final answer is $0$ .

$0$

Step 4.2

The point found by substituting $\frac{π}{4}$ in $f (x) = \sin (x) - \cos (x)$ is $(\frac{π}{4}, 0)$ . This point can be an inflection point.

$(\frac{π}{4}, 0)$

Step 4.3

Substitute $\frac{5 π}{4}$ in $f (x) = \sin (x) - \cos (x)$ to find the value of $y$ .

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Step 4.3.1

Replace the variable $x$ with $\frac{5 π}{4}$ in the expression.

$f (\frac{5 π}{4}) = \sin (\frac{5 π}{4}) - \cos (\frac{5 π}{4})$

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

Simplify each term.

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Step 4.3.2.1.1

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 third quadrant.

$f (\frac{5 π}{4}) = - \sin (\frac{π}{4}) - \cos (\frac{5 π}{4})$

Step 4.3.2.1.2

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2} - \cos (\frac{5 π}{4})$

Step 4.3.2.1.3

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

$f (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2} + \cos (\frac{π}{4})$

Step 4.3.2.1.4

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$

Step 4.3.2.1.5

Multiply $- - \frac{\sqrt{2}}{2}$ .

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Step 4.3.2.1.5.1

Multiply $- 1$ by $- 1$ .

$f (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2} + 1 (\frac{\sqrt{2}}{2})$

Step 4.3.2.1.5.2

Multiply $\frac{\sqrt{2}}{2}$ by $1$ .

$f (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$

Step 4.3.2.2

Simplify terms.

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Step 4.3.2.2.1

Combine the numerators over the common denominator.

$f (\frac{5 π}{4}) = \frac{- \sqrt{2} + \sqrt{2}}{2}$

Step 4.3.2.2.2

Add $- \sqrt{2}$ and $\sqrt{2}$ .

$f (\frac{5 π}{4}) = \frac{0}{2}$

Step 4.3.2.2.3

Divide $0$ by $2$ .

$f (\frac{5 π}{4}) = 0$

Step 4.3.2.3

The final answer is $0$ .

$0$

Step 4.4

The point found by substituting $\frac{5 π}{4}$ in $f (x) = \sin (x) - \cos (x)$ is $(\frac{5 π}{4}, 0)$ . This point can be an inflection point.

$(\frac{5 π}{4}, 0)$

Step 4.5

Determine the points that could be inflection points.

$(\frac{π}{4}, 0), (\frac{5 π}{4}, 0)$

Step 5

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

$(- \infty, \frac{π}{4}) \cup (\frac{π}{4}, \frac{5 π}{4}) \cup (\frac{5 π}{4}, \infty)$

Step 6

Substitute a value from the interval $(- \infty, \frac{π}{4})$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $0.68539816$ in the expression.

$f'' (0.68539816) = - \sin (0.68539816) + \cos (0.68539816)$

Step 6.2

The final answer is $- \sin (0.68539816) + \cos (0.68539816)$ .

$- \sin (0.68539816) + \cos (0.68539816)$

Step 6.3

At $0.68539816$ , the second derivative is $0.14118577$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, \frac{π}{4})$ .

Increasing on $(- \infty, \frac{π}{4})$ since $f'' (x) > 0$

Step 7

Substitute a value from the interval $(\frac{π}{4}, \frac{5 π}{4})$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $2.35619449$ in the expression.

$f'' (2.35619449) = - \sin (2.35619449) + \cos (2.35619449)$

Step 7.2

The final answer is $- \sin (2.35619449) + \cos (2.35619449)$ .

$- \sin (2.35619449) + \cos (2.35619449)$

Step 7.3

At $2.35619449$ , the second derivative is $- 1.41421356$ . Since this is negative, the second derivative is decreasing on the interval $(\frac{π}{4}, \frac{5 π}{4})$

Decreasing on $(\frac{π}{4}, \frac{5 π}{4})$ since $f'' (x) < 0$

Step 8

Substitute a value from the interval $(\frac{5 π}{4}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $4.02699081$ in the expression.

$f'' (4.02699081) = - \sin (4.02699081) + \cos (4.02699081)$

Step 8.2

The final answer is $- \sin (4.02699081) + \cos (4.02699081)$ .

$- \sin (4.02699081) + \cos (4.02699081)$

Step 8.3

At $4.02699081$ , the second derivative is $0.14118577$ . Since this is positive, the second derivative is increasing on the interval $(\frac{5 π}{4}, \infty)$ .

Increasing on $(\frac{5 π}{4}, \infty)$ since $f'' (x) > 0$

Step 9

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 $(\frac{π}{4}, 0), (\frac{5 π}{4}, 0)$ .

$(\frac{π}{4}, 0), (\frac{5 π}{4}, 0)$

Step 10