Calculus Examples

$f (x) = x + \sin (2 x) + 4$

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 $x + \sin (2 x) + 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\sin (2 x)] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [\sin (2 x)] + \frac{d}{d x} [4]$

Step 1.1.2

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

$1 + \frac{d}{d x} [\sin (2 x)] + \frac{d}{d x} [4]$

Step 1.2

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

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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 $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 1.2.1.1

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

$1 + \frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x] + \frac{d}{d x} [4]$

Step 1.2.1.2

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

$1 + \cos (u) \frac{d}{d x} [2 x] + \frac{d}{d x} [4]$

Step 1.2.1.3

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

$1 + \cos (2 x) \frac{d}{d x} [2 x] + \frac{d}{d x} [4]$

Step 1.2.2

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

$1 + \cos (2 x) (2 \frac{d}{d x} [x]) + \frac{d}{d x} [4]$

Step 1.2.3

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

$1 + \cos (2 x) (2 \cdot 1) + \frac{d}{d x} [4]$

Step 1.2.4

Multiply $2$ by $1$ .

$1 + \cos (2 x) \cdot 2 + \frac{d}{d x} [4]$

Step 1.2.5

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

$1 + 2 \cos (2 x) + \frac{d}{d x} [4]$

Step 1.3

Differentiate using the Constant Rule.

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Step 1.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

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

Step 1.3.2

Add $1 + 2 \cos (2 x)$ and $0$ .

$1 + 2 \cos (2 x)$

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 $1 + 2 \cos (2 x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [2 \cos (2 x)]$ .

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

Step 2.1.2

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

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

Step 2.2

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

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Step 2.2.1

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

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

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 $f (x) = \cos (x)$ and $g (x) = 2 x$ .

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Step 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $2$ by $1$ .

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

Step 2.2.6

Multiply $2$ by $- 1$ .

$f'' (x) = 0 + 2 (- 2 \sin (2 x))$

Step 2.2.7

Multiply $- 2$ by $2$ .

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

Step 2.3

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

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

Step 3

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

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

Step 4

Subtract $1$ from both sides of the equation.

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

Step 5

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

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Step 5.1

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

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

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.

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

Step 5.2.1.2

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

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

Step 5.3

Simplify the right side.

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Step 5.3.1

Move the negative in front of the fraction.

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

Step 6

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

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

Step 7

Simplify the right side.

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Step 7.1

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

$2 x = \frac{2 π}{3}$

Step 8

Divide each term in $2 x = \frac{2 π}{3}$ by $2$ and simplify.

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Step 8.1

Divide each term in $2 x = \frac{2 π}{3}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{2 π}{3}}{2}$

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.

$\frac{2 x}{2} = \frac{\frac{2 π}{3}}{2}$

Step 8.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{2 π}{3}}{2}$

Step 8.3

Simplify the right side.

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Step 8.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{2 π}{3} \cdot \frac{1}{2}$

Step 8.3.2

Cancel the common factor of $2$ .

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Step 8.3.2.1

Factor $2$ out of $2 π$ .

$x = \frac{2 (π)}{3} \cdot \frac{1}{2}$

Step 8.3.2.2

Cancel the common factor.

$x = \frac{2 π}{3} \cdot \frac{1}{2}$

Step 8.3.2.3

Rewrite the expression.

$x = \frac{π}{3}$

Step 9

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$2 x = 2 π - \frac{2 π}{3}$

Step 10

Solve for $x$ .

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Step 10.1

Simplify.

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Step 10.1.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$2 x = 2 π \cdot \frac{3}{3} - \frac{2 π}{3}$

Step 10.1.2

Combine $2 π$ and $\frac{3}{3}$ .

$2 x = \frac{2 π \cdot 3}{3} - \frac{2 π}{3}$

Step 10.1.3

Combine the numerators over the common denominator.

$2 x = \frac{2 π \cdot 3 - 2 π}{3}$

Step 10.1.4

Multiply $3$ by $2$ .

$2 x = \frac{6 π - 2 π}{3}$

Step 10.1.5

Subtract $2 π$ from $6 π$ .

$2 x = \frac{4 π}{3}$

Step 10.2

Divide each term in $2 x = \frac{4 π}{3}$ by $2$ and simplify.

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Step 10.2.1

Divide each term in $2 x = \frac{4 π}{3}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{4 π}{3}}{2}$

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.

$\frac{2 x}{2} = \frac{\frac{4 π}{3}}{2}$

Step 10.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{4 π}{3}}{2}$

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.

$x = \frac{4 π}{3} \cdot \frac{1}{2}$

Step 10.2.3.2

Cancel the common factor of $2$ .

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Step 10.2.3.2.1

Factor $2$ out of $4 π$ .

$x = \frac{2 (2 π)}{3} \cdot \frac{1}{2}$

Step 10.2.3.2.2

Cancel the common factor.

$x = \frac{2 (2 π)}{3} \cdot \frac{1}{2}$

Step 10.2.3.2.3

Rewrite the expression.

$x = \frac{2 π}{3}$

Step 11

The solution to the equation $2 x = \frac{2 π}{3}$ .

$x = \frac{π}{3}, \frac{2 π}{3}$

Step 12

Evaluate the second derivative at $x = \frac{π}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 4 \sin (2 (\frac{π}{3}))$

Step 13

Evaluate the second derivative.

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Step 13.1

Combine $2$ and $\frac{π}{3}$ .

$- 4 \sin (\frac{2 π}{3})$

Step 13.2

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

$- 4 \sin (\frac{π}{3})$

Step 13.3

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

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

Step 13.4

Cancel the common factor of $2$ .

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Step 13.4.1

Factor $2$ out of $- 4$ .

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

Step 13.4.2

Cancel the common factor.

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

Step 13.4.3

Rewrite the expression.

$- 2 \sqrt{3}$

Step 14

$x = \frac{π}{3}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{3}$ is a local maximum

Step 15

Find the y-value when $x = \frac{π}{3}$ .

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Step 15.1

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

$f (\frac{π}{3}) = (\frac{π}{3}) + \sin (2 (\frac{π}{3})) + 4$

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

Combine $2$ and $\frac{π}{3}$ .

$f (\frac{π}{3}) = \frac{π}{3} + \sin (\frac{2 π}{3}) + 4$

Step 15.2.1.2

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

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

Step 15.2.1.3

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

$f (\frac{π}{3}) = \frac{π}{3} + \frac{\sqrt{3}}{2} + 4$

Step 15.2.2

The final answer is $\frac{π}{3} + \frac{\sqrt{3}}{2} + 4$ .

$y = \frac{π}{3} + \frac{\sqrt{3}}{2} + 4$

Step 16

Evaluate the second derivative at $x = \frac{2 π}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 4 \sin (2 (\frac{2 π}{3}))$

Step 17

Evaluate the second derivative.

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Step 17.1

Multiply $2 (\frac{2 π}{3})$ .

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Step 17.1.1

Combine $2$ and $\frac{2 π}{3}$ .

$- 4 \sin (\frac{2 (2 π)}{3})$

Step 17.1.2

Multiply $2$ by $2$ .

$- 4 \sin (\frac{4 π}{3})$

Step 17.2

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 third quadrant.

$- 4 (- \sin (\frac{π}{3}))$

Step 17.3

The exact value of $\sin (\frac{π}{3})$ 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

$x = \frac{2 π}{3}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{2 π}{3}$ is a local minimum

Step 19

Find the y-value when $x = \frac{2 π}{3}$ .

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Step 19.1

Replace the variable $x$ with $\frac{2 π}{3}$ in the expression.

$f (\frac{2 π}{3}) = (\frac{2 π}{3}) + \sin (2 (\frac{2 π}{3})) + 4$

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

Multiply $2 (\frac{2 π}{3})$ .

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Step 19.2.1.1.1

Combine $2$ and $\frac{2 π}{3}$ .

$f (\frac{2 π}{3}) = \frac{2 π}{3} + \sin (\frac{2 (2 π)}{3}) + 4$

Step 19.2.1.1.2

Multiply $2$ by $2$ .

$f (\frac{2 π}{3}) = \frac{2 π}{3} + \sin (\frac{4 π}{3}) + 4$

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 sine is negative in the third quadrant.

$f (\frac{2 π}{3}) = \frac{2 π}{3} - \sin (\frac{π}{3}) + 4$

Step 19.2.1.3

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

$f (\frac{2 π}{3}) = \frac{2 π}{3} - \frac{\sqrt{3}}{2} + 4$

Step 19.2.2

The final answer is $\frac{2 π}{3} - \frac{\sqrt{3}}{2} + 4$ .

$y = \frac{2 π}{3} - \frac{\sqrt{3}}{2} + 4$

Step 20

These are the local extrema for $f (x) = x + \sin (2 x) + 4$ .

$(\frac{π}{3}, \frac{π}{3} + \frac{\sqrt{3}}{2} + 4)$ is a local maxima

$(\frac{2 π}{3}, \frac{2 π}{3} - \frac{\sqrt{3}}{2} + 4)$ is a local minima

Step 21