Calculus Examples

$2 \cos (θ) + \sin^{2} (θ)$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

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

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

Step 1.1.2

Evaluate $\frac{d}{d θ} [2 \cos (θ)]$ .

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Multiply $- 1$ by $2$ .

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

Step 1.1.3

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

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Step 1.1.3.1

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

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Step 1.1.3.1.1

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

$- 2 \sin (θ) + \frac{d}{d u} [u^{2}] \frac{d}{d θ} [\sin (θ)]$

Step 1.1.3.1.2

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

$- 2 \sin (θ) + 2 u \frac{d}{d θ} [\sin (θ)]$

Step 1.1.3.1.3

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

$- 2 \sin (θ) + 2 \sin (θ) \frac{d}{d θ} [\sin (θ)]$

Step 1.1.3.2

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

$- 2 \sin (θ) + 2 \sin (θ) \cos (θ)$

Step 1.1.4

Simplify.

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Step 1.1.4.1

Reorder terms.

$2 \cos (θ) \sin (θ) - 2 \sin (θ)$

Step 1.1.4.2

Simplify each term.

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Step 1.1.4.2.1

Reorder $2 \cos (θ)$ and $\sin (θ)$ .

$\sin (θ) (2 \cos (θ)) - 2 \sin (θ)$

Step 1.1.4.2.2

Reorder $\sin (θ)$ and $2$ .

$2 \cdot \sin (θ) \cos (θ) - 2 \sin (θ)$

Step 1.1.4.2.3

Apply the sine double-angle identity.

$f' (θ) = \sin (2 θ) - 2 \sin (θ)$

Step 1.2

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

$\sin (2 θ) - 2 \sin (θ)$

Step 2

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

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Step 2.1

Set the first derivative equal to $0$ .

$\sin (2 θ) - 2 \sin (θ) = 0$

Step 2.2

Apply the sine double-angle identity.

$2 \sin (θ) \cos (θ) - 2 \sin (θ) = 0$

Step 2.3

Factor $2 \sin (θ)$ out of $2 \sin (θ) \cos (θ) - 2 \sin (θ)$ .

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Step 2.3.1

Factor $2 \sin (θ)$ out of $- 2 \sin (θ)$ .

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

Step 2.3.2

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

$2 \sin (θ) (\cos (θ) - 1) = 0$

Step 2.4

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

$\sin (θ) = 0$

$\cos (θ) - 1 = 0$

Step 2.5

Set $\sin (θ)$ equal to $0$ and solve for $θ$ .

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Step 2.5.1

Set $\sin (θ)$ equal to $0$ .

$\sin (θ) = 0$

Step 2.5.2

Solve $\sin (θ) = 0$ for $θ$ .

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Step 2.5.2.1

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

$θ = \arcsin (0)$

Step 2.5.2.2

Simplify the right side.

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Step 2.5.2.2.1

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

$θ = 0$

Step 2.5.2.3

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.

$θ = π - 0$

Step 2.5.2.4

Subtract $0$ from $π$ .

$θ = π$

Step 2.5.2.5

Find the period of $\sin (θ)$ .

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Step 2.5.2.5.1

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

$\frac{2 π}{| b |}$

Step 2.5.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 2.5.2.5.3

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

$\frac{2 π}{1}$

Step 2.5.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.5.2.6

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

$θ = 2 π n, π + 2 π n$ , for any integer $n$

Step 2.6

Set $\cos (θ) - 1$ equal to $0$ and solve for $θ$ .

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Step 2.6.1

Set $\cos (θ) - 1$ equal to $0$ .

$\cos (θ) - 1 = 0$

Step 2.6.2

Solve $\cos (θ) - 1 = 0$ for $θ$ .

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Step 2.6.2.1

Add $1$ to both sides of the equation.

$\cos (θ) = 1$

Step 2.6.2.2

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

$θ = \arccos (1)$

Step 2.6.2.3

Simplify the right side.

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Step 2.6.2.3.1

The exact value of $\arccos (1)$ is $0$ .

$θ = 0$

Step 2.6.2.4

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.

$θ = 2 π - 0$

Step 2.6.2.5

Subtract $0$ from $2 π$ .

$θ = 2 π$

Step 2.6.2.6

Find the period of $\cos (θ)$ .

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Step 2.6.2.6.1

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

$\frac{2 π}{| b |}$

Step 2.6.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 2.6.2.6.3

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

$\frac{2 π}{1}$

Step 2.6.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.6.2.7

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

$θ = 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 2.7

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

$θ = 2 π n, π + 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 2.8

Consolidate the answers.

$θ = π n$ , for any integer $n$

Step 3

Find the values where the derivative is undefined.

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Step 3.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 4

Evaluate $2 \cos (θ) + \sin^{2} (θ)$ at each $θ$ value where the derivative is $0$ or undefined.

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Step 4.1

Evaluate at $θ = 0$ .

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Step 4.1.1

Substitute $0$ for $θ$ .

$2 \cos (0) + \sin^{2} (0)$

Step 4.1.2

Simplify.

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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 $\cos (0)$ is $1$ .

$2 \cdot 1 + \sin^{2} (0)$

Step 4.1.2.1.2

Multiply $2$ by $1$ .

$2 + \sin^{2} (0)$

Step 4.1.2.1.3

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

$2 + 0^{2}$

Step 4.1.2.1.4

Raising $0$ to any positive power yields $0$ .

$2 + 0$

Step 4.1.2.2

Add $2$ and $0$ .

$2$

Step 4.2

Evaluate at $θ = π$ .

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Step 4.2.1

Substitute $π$ for $θ$ .

$2 \cos (π) + \sin^{2} (π)$

Step 4.2.2

Simplify.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

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 second quadrant.

$2 (- \cos (0)) + \sin^{2} (π)$

Step 4.2.2.1.2

The exact value of $\cos (0)$ is $1$ .

$2 (- 1 \cdot 1) + \sin^{2} (π)$

Step 4.2.2.1.3

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

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Step 4.2.2.1.3.1

Multiply $- 1$ by $1$ .

$2 \cdot - 1 + \sin^{2} (π)$

Step 4.2.2.1.3.2

Multiply $2$ by $- 1$ .

$- 2 + \sin^{2} (π)$

Step 4.2.2.1.4

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

$- 2 + \sin^{2} (0)$

Step 4.2.2.1.5

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

$- 2 + 0^{2}$

Step 4.2.2.1.6

Raising $0$ to any positive power yields $0$ .

$- 2 + 0$

Step 4.2.2.2

Add $- 2$ and $0$ .

$- 2$

Step 4.3

List all of the points.

$(0 + 2 π n, 2), (π + 2 π n, - 2)$ , for any integer $n$

Step 5