Calculus Examples

$1 - 2 \sin (x)$

Step 1

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

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

Step 2

Find the first derivative.

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Find the first derivative.

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Differentiate.

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

$f' (x) = \frac{d}{d x} (1) + \frac{d}{d x} (- 2 \sin (x))$

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

$f' (x) = 0 + \frac{d}{d x} (- 2 \sin (x))$

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

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

$f' (x) = 0 - 2 \frac{d}{d x} \sin (x)$

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

$f' (x) = 0 - 2 \cos (x)$

Subtract $2 \cos (x)$ from $0$ .

$f' (x) = - 2 \cos (x)$

The first derivative of $f (x)$ with respect to $x$ is $- 2 \cos (x)$ .

$- 2 \cos (x)$

Step 3

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

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Set the first derivative equal to $0$ .

$- 2 \cos (x) = 0$

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

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

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

Simplify the left side.

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Cancel the common factor of $- 2$ .

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Cancel the common factor.

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

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

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

Simplify the right side.

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Divide $0$ by $- 2$ .

$\cos (x) = 0$

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

$x = \arccos (0)$

Simplify the right side.

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The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

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.

$x = 2 π - \frac{π}{2}$

Simplify $2 π - \frac{π}{2}$ .

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

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

Combine fractions.

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Combine $2 π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

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Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

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

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

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

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

Consolidate the answers.

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

Step 4

The values which make the derivative equal to $0$ are $\frac{π}{2} + π n$ .

$\frac{π}{2} + π n$

Step 5

After finding the point that makes the derivative $f' (x) = - 2 \cos (x)$ equal to $0$ or undefined, the interval to check where $f (x) = 1 - 2 \sin (x)$ is increasing and where it is decreasing is $(- \infty, \frac{π}{2} + π n) \cup (\frac{π}{2} + π n, \infty)$ .

$(- \infty, \frac{π}{2} + π n) \cup (\frac{π}{2} + π n, \infty)$

Step 6

Substitute a value from the interval $(- \infty, \frac{π}{2} + π n)$ into the derivative to determine if the function is increasing or decreasing.

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Replace the variable $x$ with $\frac{π}{2} + π n - 1$ in the expression.

$f' (\frac{π}{2} + π n - 1) = - 2 \cos (\frac{π}{2} + π n - 1)$

The final answer is $- 2 \cos (\frac{π}{2} + π n - 1)$ .

$- 2 \cos (\frac{π}{2} + π n - 1)$

Simplify.

$- 2 \cos (\frac{π}{2} + π n - 1)$

At $x = \frac{π}{2} + π n - 1$ the derivative is $- 2 \cos (\frac{π}{2} + π n - 1)$ . Since this is negative, the function is decreasing on $(- \infty, \frac{π}{2} + π n)$ .

Decreasing on $(- \infty, \frac{π}{2} + π n)$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{π}{2} + π n, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Replace the variable $x$ with $\frac{π}{2} + π n + 1$ in the expression.

$f' (\frac{π}{2} + π n + 1) = - 2 \cos (\frac{π}{2} + π n + 1)$

The final answer is $- 2 \cos (\frac{π}{2} + π n + 1)$ .

$- 2 \cos (\frac{π}{2} + π n + 1)$

Simplify.

$- 2 \cos (\frac{π}{2} + π n + 1)$

At $x = \frac{π}{2} + π n + 1$ the derivative is $- 2 \cos (\frac{π}{2} + π n + 1)$ . Since this is negative, the function is decreasing on $(\frac{π}{2} + π n, \infty)$ .

Decreasing on $(\frac{π}{2} + π n, \infty)$ since $f' (x) < 0$

Step 8

List the intervals on which the function is increasing and decreasing.

Decreasing on: $(- \infty, \frac{π}{2} + π n), (\frac{π}{2} + π n, \infty)$

Step 9