Calculus Examples

$f (x) = x + \cos (2 x)$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

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

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

Step 1.1.2

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

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

Step 1.2

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

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Step 1.2.1

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

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Step 1.2.1.1

To apply the Chain Rule, set $u$ as $2 x$ .

$1 + \frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x]$

Step 1.2.1.2

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

$1 - \sin (u) \frac{d}{d x} [2 x]$

Step 1.2.1.3

Replace all occurrences of $u$ with $2 x$ .

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

Step 1.2.2

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

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

Step 1.2.3

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

$1 - \sin (2 x) (2 \cdot 1)$

Step 1.2.4

Multiply $2$ by $1$ .

$1 - \sin (2 x) \cdot 2$

Step 1.2.5

Multiply $2$ by $- 1$ .

$1 - 2 \sin (2 x)$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $1 - 2 \sin (2 x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 2 \sin (2 x)]$ .

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

Step 2.1.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f'' (x) = 0 + \frac{d}{d x} (- 2 \sin (2 x))$

Step 2.2

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

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Step 2.2.1

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

$f'' (x) = 0 - 2 \frac{d}{d x} \sin (2 x)$

Step 2.2.2

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

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Step 2.2.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$f'' (x) = 0 - 2 (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (2 x))$

Step 2.2.2.2

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

$f'' (x) = 0 - 2 (\cos (u) \frac{d}{d x} (2 x))$

Step 2.2.2.3

Replace all occurrences of $u$ with $2 x$ .

$f'' (x) = 0 - 2 (\cos (2 x) \frac{d}{d x} (2 x))$

Step 2.2.3

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

$f'' (x) = 0 - 2 (\cos (2 x) (2 \frac{d}{d x} (x)))$

Step 2.2.4

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

$f'' (x) = 0 - 2 (\cos (2 x) (2 \cdot 1))$

Step 2.2.5

Multiply $2$ by $1$ .

$f'' (x) = 0 - 2 (\cos (2 x) \cdot 2)$

Step 2.2.6

Move $2$ to the left of $\cos (2 x)$ .

$f'' (x) = 0 - 2 (2 \cdot \cos (2 x))$

Step 2.2.7

Multiply $2$ by $- 2$ .

$f'' (x) = 0 - 4 \cos (2 x)$

Step 2.3

Subtract $4 \cos (2 x)$ from $0$ .

$f'' (x) = - 4 \cos (2 x)$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$1 - 2 \sin (2 x) = 0$

Step 4

Subtract $1$ from both sides of the equation.

$- 2 \sin (2 x) = - 1$

Step 5

Divide each term in $- 2 \sin (2 x) = - 1$ by $- 2$ and simplify.

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Step 5.1

Divide each term in $- 2 \sin (2 x) = - 1$ by $- 2$ .

$\frac{- 2 \sin (2 x)}{- 2} = \frac{- 1}{- 2}$

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $- 2$ .

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Step 5.2.1.1

Cancel the common factor.

$\frac{- 2 \sin (2 x)}{- 2} = \frac{- 1}{- 2}$

Step 5.2.1.2

Divide $\sin (2 x)$ by $1$ .

$\sin (2 x) = \frac{- 1}{- 2}$

Step 5.3

Simplify the right side.

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Step 5.3.1

Dividing two negative values results in a positive value.

$\sin (2 x) = \frac{1}{2}$

Step 6

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

$2 x = \arcsin (\frac{1}{2})$

Step 7

Simplify the right side.

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Step 7.1

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$2 x = \frac{π}{6}$

Step 8

Divide each term in $2 x = \frac{π}{6}$ by $2$ and simplify.

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Step 8.1

Divide each term in $2 x = \frac{π}{6}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{π}{6}}{2}$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $2$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{π}{6}}{2}$

Step 8.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{6}}{2}$

Step 8.3

Simplify the right side.

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Step 8.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{6} \cdot \frac{1}{2}$

Step 8.3.2

Multiply $\frac{π}{6} \cdot \frac{1}{2}$ .

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Step 8.3.2.1

Multiply $\frac{π}{6}$ by $\frac{1}{2}$ .

$x = \frac{π}{6 \cdot 2}$

Step 8.3.2.2

Multiply $6$ by $2$ .

$x = \frac{π}{12}$

Step 9

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.

$2 x = π - \frac{π}{6}$

Step 10

Solve for $x$ .

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Step 10.1

Simplify.

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Step 10.1.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$2 x = π \cdot \frac{6}{6} - \frac{π}{6}$

Step 10.1.2

Combine $π$ and $\frac{6}{6}$ .

$2 x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Step 10.1.3

Combine the numerators over the common denominator.

$2 x = \frac{π \cdot 6 - π}{6}$

Step 10.1.4

Subtract $π$ from $π \cdot 6$ .

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Step 10.1.4.1

Reorder $π$ and $6$ .

$2 x = \frac{6 \cdot π - π}{6}$

Step 10.1.4.2

Subtract $π$ from $6 \cdot π$ .

$2 x = \frac{5 \cdot π}{6}$

Step 10.2

Divide each term in $2 x = \frac{5 \cdot π}{6}$ by $2$ and simplify.

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Step 10.2.1

Divide each term in $2 x = \frac{5 \cdot π}{6}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{5 \cdot π}{6}}{2}$

Step 10.2.2

Simplify the left side.

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Step 10.2.2.1

Cancel the common factor of $2$ .

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Step 10.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{5 \cdot π}{6}}{2}$

Step 10.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{5 \cdot π}{6}}{2}$

Step 10.2.3

Simplify the right side.

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Step 10.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{5 \cdot π}{6} \cdot \frac{1}{2}$

Step 10.2.3.2

Multiply $\frac{5 π}{6} \cdot \frac{1}{2}$ .

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Step 10.2.3.2.1

Multiply $\frac{5 π}{6}$ by $\frac{1}{2}$ .

$x = \frac{5 π}{6 \cdot 2}$

Step 10.2.3.2.2

Multiply $6$ by $2$ .

$x = \frac{5 π}{12}$

Step 11

The solution to the equation $2 x = \frac{π}{6}$ .

$x = \frac{π}{12}, \frac{5 π}{12}$

Step 12

Evaluate the second derivative at $x = \frac{π}{12}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 4 \cos (2 (\frac{π}{12}))$

Step 13

Evaluate the second derivative.

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Step 13.1

Cancel the common factor of $2$ .

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Step 13.1.1

Factor $2$ out of $12$ .

$- 4 \cos (2 \frac{π}{2 (6)})$

Step 13.1.2

Cancel the common factor.

$- 4 \cos (2 \frac{π}{2 \cdot 6})$

Step 13.1.3

Rewrite the expression.

$- 4 \cos (\frac{π}{6})$

Step 13.2

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$- 4 \frac{\sqrt{3}}{2}$

Step 13.3

Cancel the common factor of $2$ .

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Step 13.3.1

Factor $2$ out of $- 4$ .

$2 (- 2) \frac{\sqrt{3}}{2}$

Step 13.3.2

Cancel the common factor.

$2 \cdot - 2 \frac{\sqrt{3}}{2}$

Step 13.3.3

Rewrite the expression.

$- 2 \sqrt{3}$

Step 14

$x = \frac{π}{12}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{12}$ is a local maximum

Step 15

Find the y-value when $x = \frac{π}{12}$ .

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Step 15.1

Replace the variable $x$ with $\frac{π}{12}$ in the expression.

$f (\frac{π}{12}) = (\frac{π}{12}) + \cos (2 (\frac{π}{12}))$

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify each term.

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Step 15.2.1.1

Cancel the common factor of $2$ .

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Step 15.2.1.1.1

Factor $2$ out of $12$ .

$f (\frac{π}{12}) = \frac{π}{12} + \cos (2 (\frac{π}{2 (6)}))$

Step 15.2.1.1.2

Cancel the common factor.

$f (\frac{π}{12}) = \frac{π}{12} + \cos (2 (\frac{π}{2 \cdot 6}))$

Step 15.2.1.1.3

Rewrite the expression.

$f (\frac{π}{12}) = \frac{π}{12} + \cos (\frac{π}{6})$

Step 15.2.1.2

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{12}) = \frac{π}{12} + \frac{\sqrt{3}}{2}$

Step 15.2.2

The final answer is $\frac{π}{12} + \frac{\sqrt{3}}{2}$ .

$y = \frac{π}{12} + \frac{\sqrt{3}}{2}$

Step 16

Evaluate the second derivative at $x = \frac{5 π}{12}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 4 \cos (2 (\frac{5 π}{12}))$

Step 17

Evaluate the second derivative.

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Step 17.1

Cancel the common factor of $2$ .

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Step 17.1.1

Factor $2$ out of $12$ .

$- 4 \cos (2 \frac{5 π}{2 (6)})$

Step 17.1.2

Cancel the common factor.

$- 4 \cos (2 \frac{5 π}{2 \cdot 6})$

Step 17.1.3

Rewrite the expression.

$- 4 \cos (\frac{5 π}{6})$

Step 17.2

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.

$- 4 (- \cos (\frac{π}{6}))$

Step 17.3

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$- 4 (- \frac{\sqrt{3}}{2})$

Step 17.4

Cancel the common factor of $2$ .

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Step 17.4.1

Move the leading negative in $- \frac{\sqrt{3}}{2}$ into the numerator.

$- 4 \frac{- \sqrt{3}}{2}$

Step 17.4.2

Factor $2$ out of $- 4$ .

$2 (- 2) \frac{- \sqrt{3}}{2}$

Step 17.4.3

Cancel the common factor.

$2 \cdot - 2 \frac{- \sqrt{3}}{2}$

Step 17.4.4

Rewrite the expression.

$- 2 (- \sqrt{3})$

Step 17.5

Multiply $- 1$ by $- 2$ .

$2 \sqrt{3}$

Step 18

$x = \frac{5 π}{12}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{5 π}{12}$ is a local minimum

Step 19

Find the y-value when $x = \frac{5 π}{12}$ .

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Step 19.1

Replace the variable $x$ with $\frac{5 π}{12}$ in the expression.

$f (\frac{5 π}{12}) = (\frac{5 π}{12}) + \cos (2 (\frac{5 π}{12}))$

Step 19.2

Simplify the result.

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Step 19.2.1

Simplify each term.

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Step 19.2.1.1

Cancel the common factor of $2$ .

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Step 19.2.1.1.1

Factor $2$ out of $12$ .

$f (\frac{5 π}{12}) = \frac{5 π}{12} + \cos (2 (\frac{5 π}{2 (6)}))$

Step 19.2.1.1.2

Cancel the common factor.

$f (\frac{5 π}{12}) = \frac{5 π}{12} + \cos (2 (\frac{5 π}{2 \cdot 6}))$

Step 19.2.1.1.3

Rewrite the expression.

$f (\frac{5 π}{12}) = \frac{5 π}{12} + \cos (\frac{5 π}{6})$

Step 19.2.1.2

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.

$f (\frac{5 π}{12}) = \frac{5 π}{12} - \cos (\frac{π}{6})$

Step 19.2.1.3

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{5 π}{12}) = \frac{5 π}{12} - \frac{\sqrt{3}}{2}$

Step 19.2.2

The final answer is $\frac{5 π}{12} - \frac{\sqrt{3}}{2}$ .

$y = \frac{5 π}{12} - \frac{\sqrt{3}}{2}$

Step 20

These are the local extrema for $f (x) = x + \cos (2 x)$ .

$(\frac{π}{12}, \frac{π}{12} + \frac{\sqrt{3}}{2})$ is a local maxima

$(\frac{5 π}{12}, \frac{5 π}{12} - \frac{\sqrt{3}}{2})$ is a local minima

Step 21