Calculus Examples

$f (x) = 2 \sin (x) + \sin^{2} (x)$

Step 1

Find the derivative.

Tap for more steps...

Step 1.1

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

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

Step 1.2

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

Tap for more steps...

Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

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

Tap for more steps...

Step 1.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \sin (x)$ .

Tap for more steps...

Step 1.3.1.1

To apply the Chain Rule, set $u$ as $\sin (x)$ .

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

Step 1.3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

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

Step 1.3.1.3

Replace all occurrences of $u$ with $\sin (x)$ .

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

Step 1.3.2

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

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

Step 1.4

Simplify.

Tap for more steps...

Step 1.4.1

Reorder terms.

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

Step 1.4.2

Simplify each term.

Tap for more steps...

Step 1.4.2.1

Reorder $2 \cos (x)$ and $\sin (x)$ .

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

Step 1.4.2.2

Reorder $\sin (x)$ and $2$ .

$2 \cdot \sin (x) \cos (x) + 2 \cos (x)$

Step 1.4.2.3

Apply the sine double-angle identity.

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

Step 2

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

Tap for more steps...

Step 2.1

Apply the sine double-angle identity.

$2 \sin (x) \cos (x) + 2 \cos (x) = 0$

Step 2.2

Factor $2 \cos (x)$ out of $2 \sin (x) \cos (x) + 2 \cos (x)$ .

Tap for more steps...

Step 2.2.1

Factor $2 \cos (x)$ out of $2 \sin (x) \cos (x)$ .

$2 \cos (x) \sin (x) + 2 \cos (x) = 0$

Step 2.2.2

Factor $2 \cos (x)$ out of $2 \cos (x)$ .

$2 \cos (x) \sin (x) + 2 \cos (x) \cdot 1 = 0$

Step 2.2.3

Factor $2 \cos (x)$ out of $2 \cos (x) \sin (x) + 2 \cos (x) \cdot 1$ .

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

Step 2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\cos (x) = 0$

$\sin (x) + 1 = 0$

Step 2.4

Set $\cos (x)$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.4.1

Set $\cos (x)$ equal to $0$ .

$\cos (x) = 0$

Step 2.4.2

Solve $\cos (x) = 0$ for $x$ .

Tap for more steps...

Step 2.4.2.1

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

$x = \arccos (0)$

Step 2.4.2.2

Simplify the right side.

Tap for more steps...

Step 2.4.2.2.1

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

$x = \frac{π}{2}$

Step 2.4.2.3

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.

$x = 2 π - \frac{π}{2}$

Step 2.4.2.4

Simplify $2 π - \frac{π}{2}$ .

Tap for more steps...

Step 2.4.2.4.1

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

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

Step 2.4.2.4.2

Combine fractions.

Tap for more steps...

Step 2.4.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 2.4.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 2.4.2.4.3

Simplify the numerator.

Tap for more steps...

Step 2.4.2.4.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 2.4.2.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 2.4.2.5

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

Tap for more steps...

Step 2.4.2.5.1

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

$\frac{2 π}{| b |}$

Step 2.4.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 2.4.2.5.3

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

$\frac{2 π}{1}$

Step 2.4.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.4.2.6

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 $n$

Step 2.5

Set $\sin (x) + 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.5.1

Set $\sin (x) + 1$ equal to $0$ .

$\sin (x) + 1 = 0$

Step 2.5.2

Solve $\sin (x) + 1 = 0$ for $x$ .

Tap for more steps...

Step 2.5.2.1

Subtract $1$ from both sides of the equation.

$\sin (x) = - 1$

Step 2.5.2.2

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

$x = \arcsin (- 1)$

Step 2.5.2.3

Simplify the right side.

Tap for more steps...

Step 2.5.2.3.1

The exact value of $\arcsin (- 1)$ is $- \frac{π}{2}$ .

$x = - \frac{π}{2}$

Step 2.5.2.4

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{2} + π$

Step 2.5.2.5

Simplify the expression to find the second solution.

Tap for more steps...

Step 2.5.2.5.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 2.5.2.5.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 2.5.2.6

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

Tap for more steps...

Step 2.5.2.6.1

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

$\frac{2 π}{| b |}$

Step 2.5.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 2.5.2.6.3

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

$\frac{2 π}{1}$

Step 2.5.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.5.2.7

Add $2 π$ to every negative angle to get positive angles.

Tap for more steps...

Step 2.5.2.7.1

Add $2 π$ to $- \frac{π}{2}$ to find the positive angle.

$- \frac{π}{2} + 2 π$

Step 2.5.2.7.2

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

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 2.5.2.7.3

Combine fractions.

Tap for more steps...

Step 2.5.2.7.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 2.5.2.7.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 2.5.2.7.4

Simplify the numerator.

Tap for more steps...

Step 2.5.2.7.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 2.5.2.7.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 2.5.2.7.5

List the new angles.

$x = \frac{3 π}{2}$

Step 2.5.2.8

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

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

Step 2.6

The final solution is all the values that make $2 \cos (x) (\sin (x) + 1) = 0$ true.

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

Step 2.7

Consolidate the answers.

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

Step 3

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

Tap for more steps...

Step 3.1

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

$f (\frac{π}{2}) = 2 \sin (\frac{π}{2}) + \sin^{2} (\frac{π}{2})$

Step 3.2

Simplify the result.

Tap for more steps...

Step 3.2.1

Simplify each term.

Tap for more steps...

Step 3.2.1.1

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{π}{2}) = 2 \cdot 1 + \sin^{2} (\frac{π}{2})$

Step 3.2.1.2

Multiply $2$ by $1$ .

$f (\frac{π}{2}) = 2 + \sin^{2} (\frac{π}{2})$

Step 3.2.1.3

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{π}{2}) = 2 + 1^{2}$

Step 3.2.1.4

One to any power is one.

$f (\frac{π}{2}) = 2 + 1$

Step 3.2.2

Add $2$ and $1$ .

$f (\frac{π}{2}) = 3$

Step 3.2.3

The final answer is $3$ .

$3$

Step 4

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

Tap for more steps...

Step 4.1

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

$f (\frac{π}{2} + π) = 2 \sin (\frac{π}{2} + π) + \sin^{2} (\frac{π}{2} + π)$

Step 4.2

Simplify the result.

Tap for more steps...

Step 4.2.1

Simplify each term.

Tap for more steps...

Step 4.2.1.1

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

$f (\frac{π}{2} + π) = 2 \sin (\frac{π}{2} + π \cdot \frac{2}{2}) + \sin^{2} (\frac{π}{2} + π)$

Step 4.2.1.2

Combine $π$ and $\frac{2}{2}$ .

$f (\frac{π}{2} + π) = 2 \sin (\frac{π}{2} + \frac{π \cdot 2}{2}) + \sin^{2} (\frac{π}{2} + π)$

Step 4.2.1.3

Combine the numerators over the common denominator.

$f (\frac{π}{2} + π) = 2 \sin (\frac{π + π \cdot 2}{2}) + \sin^{2} (\frac{π}{2} + π)$

Step 4.2.1.4

Simplify the numerator.

Tap for more steps...

Step 4.2.1.4.1

Move $2$ to the left of $π$ .

$f (\frac{π}{2} + π) = 2 \sin (\frac{π + 2 \cdot π}{2}) + \sin^{2} (\frac{π}{2} + π)$

Step 4.2.1.4.2

Add $π$ and $2 π$ .

$f (\frac{π}{2} + π) = 2 \sin (\frac{3 π}{2}) + \sin^{2} (\frac{π}{2} + π)$

Step 4.2.1.5

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 fourth quadrant.

$f (\frac{π}{2} + π) = 2 (- \sin (\frac{π}{2})) + \sin^{2} (\frac{π}{2} + π)$

Step 4.2.1.6

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{π}{2} + π) = 2 (- 1 \cdot 1) + \sin^{2} (\frac{π}{2} + π)$

Step 4.2.1.7

Multiply $2 (- 1 \cdot 1)$ .

Tap for more steps...

Step 4.2.1.7.1

Multiply $- 1$ by $1$ .

$f (\frac{π}{2} + π) = 2 \cdot - 1 + \sin^{2} (\frac{π}{2} + π)$

Step 4.2.1.7.2

Multiply $2$ by $- 1$ .

$f (\frac{π}{2} + π) = - 2 + \sin^{2} (\frac{π}{2} + π)$

Step 4.2.1.8

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

$f (\frac{π}{2} + π) = - 2 + \sin^{2} (\frac{π}{2} + π \cdot \frac{2}{2})$

Step 4.2.1.9

Combine $π$ and $\frac{2}{2}$ .

$f (\frac{π}{2} + π) = - 2 + \sin^{2} (\frac{π}{2} + \frac{π \cdot 2}{2})$

Step 4.2.1.10

Combine the numerators over the common denominator.

$f (\frac{π}{2} + π) = - 2 + \sin^{2} (\frac{π + π \cdot 2}{2})$

Step 4.2.1.11

Simplify the numerator.

Tap for more steps...

Step 4.2.1.11.1

Move $2$ to the left of $π$ .

$f (\frac{π}{2} + π) = - 2 + \sin^{2} (\frac{π + 2 \cdot π}{2})$

Step 4.2.1.11.2

Add $π$ and $2 π$ .

$f (\frac{π}{2} + π) = - 2 + \sin^{2} (\frac{3 π}{2})$

Step 4.2.1.12

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 fourth quadrant.

$f (\frac{π}{2} + π) = - 2 + {(- \sin (\frac{π}{2}))}^{2}$

Step 4.2.1.13

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{π}{2} + π) = - 2 + {(- 1 \cdot 1)}^{2}$

Step 4.2.1.14

Multiply $- 1$ by $1$ .

$f (\frac{π}{2} + π) = - 2 + {(- 1)}^{2}$

Step 4.2.1.15

Raise $- 1$ to the power of $2$ .

$f (\frac{π}{2} + π) = - 2 + 1$

Step 4.2.2

Add $- 2$ and $1$ .

$f (\frac{π}{2} + π) = - 1$

Step 4.2.3

The final answer is $- 1$ .

$- 1$

Step 5

The horizontal tangent line on function $f (x) = 2 \sin (x) + \sin^{2} (x)$ is $y = 3, y = - 1$ .

$y = 3, y = - 1$

Step 6