Calculus Examples

$y = - \sin (x)$

Step 1

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

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

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

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

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

Step 2.1.2

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

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

Step 2.2

Find the second derivative.

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Step 2.2.1

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

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

Step 2.2.2

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

$- - \sin (x)$

Step 2.2.3

Multiply.

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Step 2.2.3.1

Multiply $- 1$ by $- 1$ .

$1 \sin (x)$

Step 2.2.3.2

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

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

Step 2.3

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

$\sin (x)$

Step 3

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

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Step 3.1

Set the second derivative equal to $0$ .

$\sin (x) = 0$

Step 3.2

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

$x = \arcsin (0)$

Step 3.3

Simplify the right side.

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Step 3.3.1

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

$x = 0$

Step 3.4

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

Subtract $0$ from $π$ .

$x = π$

Step 3.6

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

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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 $\sin (x)$ 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 3.8

Consolidate the answers.

$x = π n$ , for any integer $n$

Step 4

The point found by substituting in $f (x) = - \sin (x)$ 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.

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

Step 6.2

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

$\sin (- 0.1)$

Step 6.3

At $- 0.1$ , the second derivative is $- 0.09983341$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty,)$

Decreasing 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.

$f'' (0.1) = \sin (0.1)$

Step 7.2

The final answer is $\sin (0.1)$ .

$\sin (0.1)$

Step 7.3

At $0.1$ , the second derivative is $0.09983341$ . Since this is positive, the second derivative is increasing on the interval $(, \infty)$ .

Increasing 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