Calculus Examples

$2 \cos (x) + \cos^{2} (x)$

Step 1

Write $2 \cos (x) + \cos^{2} (x)$ as a function.

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

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 $2 \cos (x) + \cos^{2} (x)$ with respect to $x$ is $\frac{d}{d x} [2 \cos (x)] + \frac{d}{d x} [\cos^{2} (x)]$ .

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

Step 2.1.2

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

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Step 2.1.2.1

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

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

Step 2.1.2.2

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

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

Step 2.1.2.3

Multiply $- 1$ by $2$ .

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

Step 2.1.3

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

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Step 2.1.3.1

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 $g (x) = \cos (x)$ .

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Step 2.1.3.1.1

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

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

Step 2.1.3.1.2

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

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

Step 2.1.3.1.3

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

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

Step 2.1.3.2

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

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

Step 2.1.3.3

Multiply $- 1$ by $2$ .

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

Step 2.1.4

Reorder terms.

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

Step 2.2

Find the second derivative.

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Step 2.2.1

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

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

Step 2.2.2

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

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Step 2.2.2.1

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

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

Step 2.2.2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \cos (x)$ and $g (x) = \sin (x)$ .

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

Step 2.2.2.3

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

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

Step 2.2.2.4

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

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

Step 2.2.2.5

Raise $\cos (x)$ to the power of $1$ .

$- 2 (\cos^{1} (x) \cos (x) + \sin (x) (- \sin (x))) + \frac{d}{d x} [- 2 \sin (x)]$

Step 2.2.2.6

Raise $\cos (x)$ to the power of $1$ .

$- 2 (\cos^{1} (x) \cos^{1} (x) + \sin (x) (- \sin (x))) + \frac{d}{d x} [- 2 \sin (x)]$

Step 2.2.2.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 2 ({\cos (x)}^{1 + 1} + \sin (x) (- \sin (x))) + \frac{d}{d x} [- 2 \sin (x)]$

Step 2.2.2.8

Add $1$ and $1$ .

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

Step 2.2.2.9

Raise $\sin (x)$ to the power of $1$ .

$- 2 (\cos^{2} (x) - (\sin^{1} (x) \sin (x))) + \frac{d}{d x} [- 2 \sin (x)]$

Step 2.2.2.10

Raise $\sin (x)$ to the power of $1$ .

$- 2 (\cos^{2} (x) - (\sin^{1} (x) \sin^{1} (x))) + \frac{d}{d x} [- 2 \sin (x)]$

Step 2.2.2.11

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 2 (\cos^{2} (x) - {\sin (x)}^{1 + 1}) + \frac{d}{d x} [- 2 \sin (x)]$

Step 2.2.2.12

Add $1$ and $1$ .

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

Step 2.2.3

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

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Step 2.2.3.1

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

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

Step 2.2.3.2

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

$- 2 (\cos^{2} (x) - \sin^{2} (x)) - 2 \cos (x)$

Step 2.2.4

Simplify.

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Step 2.2.4.1

Apply the distributive property.

$- 2 \cos^{2} (x) - 2 (- \sin^{2} (x)) - 2 \cos (x)$

Step 2.2.4.2

Multiply $- 1$ by $- 2$ .

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

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $- 2 \cos^{2} (x) + 2 \sin^{2} (x) - 2 \cos (x)$ .

$- 2 \cos^{2} (x) + 2 \sin^{2} (x) - 2 \cos (x)$

Step 3

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

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Step 3.1

Set the second derivative equal to $0$ .

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

Step 3.2

Graph each side of the equation. The solution is the x-value of the point of intersection.

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

Step 4

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

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Step 4.1

Substitute $\frac{π}{3}$ in $f (x) = 2 \cos (x) + \cos^{2} (x)$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $\frac{π}{3}$ in the expression.

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

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 $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$f (\frac{π}{3}) = 2 (\frac{1}{2}) + \cos^{2} (\frac{π}{3})$

Step 4.1.2.1.2

Cancel the common factor of $2$ .

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Step 4.1.2.1.2.1

Cancel the common factor.

$f (\frac{π}{3}) = 2 (\frac{1}{2}) + \cos^{2} (\frac{π}{3})$

Step 4.1.2.1.2.2

Rewrite the expression.

$f (\frac{π}{3}) = 1 + \cos^{2} (\frac{π}{3})$

Step 4.1.2.1.3

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$f (\frac{π}{3}) = 1 + {(\frac{1}{2})}^{2}$

Step 4.1.2.1.4

Apply the product rule to $\frac{1}{2}$ .

$f (\frac{π}{3}) = 1 + \frac{1^{2}}{2^{2}}$

Step 4.1.2.1.5

One to any power is one.

$f (\frac{π}{3}) = 1 + \frac{1}{2^{2}}$

Step 4.1.2.1.6

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

$f (\frac{π}{3}) = 1 + \frac{1}{4}$

Step 4.1.2.2

Simplify the expression.

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Step 4.1.2.2.1

Write $1$ as a fraction with a common denominator.

$f (\frac{π}{3}) = \frac{4}{4} + \frac{1}{4}$

Step 4.1.2.2.2

Combine the numerators over the common denominator.

$f (\frac{π}{3}) = \frac{4 + 1}{4}$

Step 4.1.2.2.3

Add $4$ and $1$ .

$f (\frac{π}{3}) = \frac{5}{4}$

Step 4.1.2.3

The final answer is $\frac{5}{4}$ .

$\frac{5}{4}$

Step 4.2

The point found by substituting $\frac{π}{3}$ in $f (x) = 2 \cos (x) + \cos^{2} (x)$ is $(\frac{π}{3}, \frac{5}{4})$ . This point can be an inflection point.

$(\frac{π}{3}, \frac{5}{4})$

Step 5

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

$(- \infty, \frac{π}{3}) \cup (\frac{π}{3}, \infty)$

Step 6

Substitute a value from the interval $(- \infty, \frac{π}{3})$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $0.94719755$ in the expression.

$f'' (0.94719755) = - 2 \cos^{2} (0.94719755) + 2 \sin^{2} (0.94719755) - 2 \cos (0.94719755)$

Step 6.2

The final answer is $- 2 \cos^{2} (0.94719755) + 2 \sin^{2} (0.94719755) - 2 \cos (0.94719755)$ .

$- 2 \cos^{2} (0.94719755) + 2 \sin^{2} (0.94719755) - 2 \cos (0.94719755)$

Step 6.3

At $0.94719755$ , the second derivative is $- 0.53195951$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, \frac{π}{3})$

Decreasing on $(- \infty, \frac{π}{3})$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{π}{3}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $1.14719755$ in the expression.

$f'' (1.14719755) = - 2 \cos^{2} (1.14719755) + 2 \sin^{2} (1.14719755) - 2 \cos (1.14719755)$

Step 7.2

The final answer is $- 2 \cos^{2} (1.14719755) + 2 \sin^{2} (1.14719755) - 2 \cos (1.14719755)$ .

$- 2 \cos^{2} (1.14719755) + 2 \sin^{2} (1.14719755) - 2 \cos (1.14719755)$

Step 7.3

At $1.14719755$ , the second derivative is $0.50208433$ . Since this is positive, the second derivative is increasing on the interval $(\frac{π}{3}, \infty)$ .

Increasing on $(\frac{π}{3}, \infty)$ 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 $(\frac{π}{3}, \frac{5}{4})$ .

$(\frac{π}{3}, \frac{5}{4})$

Step 9