Calculus Examples

$2 \cos (θ) + \cos^{2} (θ)$

Step 1

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

$f (θ) = 2 \cos (θ) + \cos^{2} (θ)$

Step 2

Find the second derivative.

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Find the first derivative.

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

$f' (θ) = \frac{d}{d θ} (2 \cos (θ)) + \frac{d}{d θ} (\cos^{2} (θ))$

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

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

$f' (θ) = 2 \frac{d}{d θ} (\cos (θ)) + \frac{d}{d θ} (\cos^{2} (θ))$

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

$f' (θ) = 2 (- \sin (θ)) + \frac{d}{d θ} (\cos^{2} (θ))$

Multiply $- 1$ by $2$ .

$f' (θ) = - 2 \sin (θ) + \frac{d}{d θ} (\cos^{2} (θ))$

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

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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 $g (θ) = \cos (θ)$ .

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

$f' (θ) = - 2 \sin (θ) + \frac{d}{d u} (u^{2}) \frac{d}{d θ} (\cos (θ))$

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

$f' (θ) = - 2 \sin (θ) + 2 u \frac{d}{d θ} (\cos (θ))$

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

$f' (θ) = - 2 \sin (θ) + 2 \cos (θ) \frac{d}{d θ} (\cos (θ))$

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

$f' (θ) = - 2 \sin (θ) + 2 \cos (θ) (- \sin (θ))$

Multiply $- 1$ by $2$ .

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

Reorder terms.

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

Find the second derivative.

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

$f'' (θ) = \frac{d}{d θ} (- 2 \cos (θ) \sin (θ)) + \frac{d}{d θ} (- 2 \sin (θ))$

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

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

$f'' (θ) = - 2 \frac{d}{d θ} (\cos (θ) \sin (θ)) + \frac{d}{d θ} (- 2 \sin (θ))$

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

$f'' (θ) = - 2 (\cos (θ) \frac{d}{d θ} (\sin (θ)) + \sin (θ) \frac{d}{d θ} (\cos (θ))) + \frac{d}{d θ} (- 2 \sin (θ))$

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

$f'' (θ) = - 2 (\cos (θ) \cos (θ) + \sin (θ) \frac{d}{d θ} (\cos (θ))) + \frac{d}{d θ} (- 2 \sin (θ))$

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

$f'' (θ) = - 2 (\cos (θ) \cos (θ) + \sin (θ) (- \sin (θ))) + \frac{d}{d θ} (- 2 \sin (θ))$

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

$f'' (θ) = - 2 (\cos (θ) \cos (θ) + \sin (θ) (- \sin (θ))) + \frac{d}{d θ} (- 2 \sin (θ))$

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

$f'' (θ) = - 2 (\cos (θ) \cos (θ) + \sin (θ) (- \sin (θ))) + \frac{d}{d θ} (- 2 \sin (θ))$

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

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

Add $1$ and $1$ .

$f'' (θ) = - 2 (\cos^{2} (θ) + \sin (θ) (- \sin (θ))) + \frac{d}{d θ} (- 2 \sin (θ))$

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

$f'' (θ) = - 2 (\cos^{2} (θ) - (\sin (θ) \sin (θ))) + \frac{d}{d θ} (- 2 \sin (θ))$

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

$f'' (θ) = - 2 (\cos^{2} (θ) - (\sin (θ) \sin (θ))) + \frac{d}{d θ} (- 2 \sin (θ))$

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

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

Add $1$ and $1$ .

$f'' (θ) = - 2 (\cos^{2} (θ) - \sin^{2} (θ)) + \frac{d}{d θ} (- 2 \sin (θ))$

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

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

$f'' (θ) = - 2 (\cos^{2} (θ) - \sin^{2} (θ)) - 2 \frac{d}{d θ} \sin (θ)$

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

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

Simplify.

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Apply the distributive property.

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

Multiply $- 1$ by $- 2$ .

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

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

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

Step 3

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

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Set the second derivative equal to $0$ .

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

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

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

Step 4

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

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

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

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

Simplify the result.

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Simplify each term.

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

One to any power is one.

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

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

$f (\frac{π}{3}) = 1 + \frac{1}{4}$

Simplify the expression.

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

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

Combine the numerators over the common denominator.

$f (\frac{π}{3}) = \frac{4 + 1}{4}$

Add $4$ and $1$ .

$f (\frac{π}{3}) = \frac{5}{4}$

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

$\frac{5}{4}$

The point found by substituting $\frac{π}{3}$ in $f (θ) = 2 \cos (θ) + \cos^{2} (θ)$ 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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Replace the variable $θ$ with $0.94719755$ in the expression.

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

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

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

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 $f'' (θ) < 0$

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

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

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

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

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 $f'' (θ) > 0$

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