Calculus Examples

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

Step 1

Write $y = 2 x + \sin (4 x)$ as a function.

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

Step 2

Find the first derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $2 x + \sin (4 x)$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [\sin (4 x)]$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $2$ by $1$ .

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

Step 2.3

Evaluate $\frac{d}{d x} [\sin (4 x)]$ .

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Step 2.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = 4 x$ .

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Step 2.3.1.1

To apply the Chain Rule, set $u$ as $4 x$ .

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

Step 2.3.1.2

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

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

Step 2.3.1.3

Replace all occurrences of $u$ with $4 x$ .

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

Step 2.3.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

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

Step 2.3.3

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

$2 + \cos (4 x) (4 \cdot 1)$

Step 2.3.4

Multiply $4$ by $1$ .

$2 + \cos (4 x) \cdot 4$

Step 2.3.5

Move $4$ to the left of $\cos (4 x)$ .

$2 + 4 \cos (4 x)$

Step 3

Find the second derivative of the function.

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Step 3.1

Differentiate.

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Step 3.1.1

By the Sum Rule, the derivative of $2 + 4 \cos (4 x)$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [4 \cos (4 x)]$ .

$f'' (x) = \frac{d}{d x} (2) + \frac{d}{d x} (4 \cos (4 x))$

Step 3.1.2

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$f'' (x) = 0 + \frac{d}{d x} (4 \cos (4 x))$

Step 3.2

Evaluate $\frac{d}{d x} [4 \cos (4 x)]$ .

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Step 3.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 \cos (4 x)$ with respect to $x$ is $4 \frac{d}{d x} [\cos (4 x)]$ .

$f'' (x) = 0 + 4 \frac{d}{d x} (\cos (4 x))$

Step 3.2.2

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 $g (x) = 4 x$ .

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Step 3.2.2.1

To apply the Chain Rule, set $u$ as $4 x$ .

$f'' (x) = 0 + 4 (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (4 x))$

Step 3.2.2.2

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

$f'' (x) = 0 + 4 (- \sin (u) \frac{d}{d x} (4 x))$

Step 3.2.2.3

Replace all occurrences of $u$ with $4 x$ .

$f'' (x) = 0 + 4 (- \sin (4 x) \frac{d}{d x} (4 x))$

Step 3.2.3

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$f'' (x) = 0 + 4 (- \sin (4 x) (4 \frac{d}{d x} (x)))$

Step 3.2.4

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

$f'' (x) = 0 + 4 (- \sin (4 x) (4 \cdot 1))$

Step 3.2.5

Multiply $4$ by $1$ .

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

Step 3.2.6

Multiply $4$ by $- 1$ .

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

Step 3.2.7

Multiply $- 4$ by $4$ .

$f'' (x) = 0 - 16 \sin (4 x)$

Step 3.3

Subtract $16 \sin (4 x)$ from $0$ .

$f'' (x) = - 16 \sin (4 x)$

Step 4

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

$2 + 4 \cos (4 x) = 0$

Step 5

Subtract $2$ from both sides of the equation.

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

Step 6

Divide each term in $4 \cos (4 x) = - 2$ by $4$ and simplify.

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Step 6.1

Divide each term in $4 \cos (4 x) = - 2$ by $4$ .

$\frac{4 \cos (4 x)}{4} = \frac{- 2}{4}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $4$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{4 \cos (4 x)}{4} = \frac{- 2}{4}$

Step 6.2.1.2

Divide $\cos (4 x)$ by $1$ .

$\cos (4 x) = \frac{- 2}{4}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Cancel the common factor of $- 2$ and $4$ .

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Step 6.3.1.1

Factor $2$ out of $- 2$ .

$\cos (4 x) = \frac{2 (- 1)}{4}$

Step 6.3.1.2

Cancel the common factors.

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Step 6.3.1.2.1

Factor $2$ out of $4$ .

$\cos (4 x) = \frac{2 \cdot - 1}{2 \cdot 2}$

Step 6.3.1.2.2

Cancel the common factor.

$\cos (4 x) = \frac{2 \cdot - 1}{2 \cdot 2}$

Step 6.3.1.2.3

Rewrite the expression.

$\cos (4 x) = \frac{- 1}{2}$

Step 6.3.2

Move the negative in front of the fraction.

$\cos (4 x) = - \frac{1}{2}$

Step 7

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

$4 x = \arccos (- \frac{1}{2})$

Step 8

Simplify the right side.

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Step 8.1

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

$4 x = \frac{2 π}{3}$

Step 9

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

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Step 9.1

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

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

Step 9.2

Simplify the left side.

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Step 9.2.1

Cancel the common factor of $4$ .

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Step 9.2.1.1

Cancel the common factor.

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

Step 9.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{2 π}{3}}{4}$

Step 9.3

Simplify the right side.

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Step 9.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{2 π}{3} \cdot \frac{1}{4}$

Step 9.3.2

Cancel the common factor of $2$ .

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Step 9.3.2.1

Factor $2$ out of $2 π$ .

$x = \frac{2 (π)}{3} \cdot \frac{1}{4}$

Step 9.3.2.2

Factor $2$ out of $4$ .

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

Step 9.3.2.3

Cancel the common factor.

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

Step 9.3.2.4

Rewrite the expression.

$x = \frac{π}{3} \cdot \frac{1}{2}$

Step 9.3.3

Multiply $\frac{π}{3}$ by $\frac{1}{2}$ .

$x = \frac{π}{3 \cdot 2}$

Step 9.3.4

Multiply $3$ by $2$ .

$x = \frac{π}{6}$

Step 10

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.

$4 x = 2 π - \frac{2 π}{3}$

Step 11

Solve for $x$ .

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Step 11.1

Simplify.

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Step 11.1.1

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

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

Step 11.1.2

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 11.1.3

Combine the numerators over the common denominator.

$4 x = \frac{2 π \cdot 3 - 2 π}{3}$

Step 11.1.4

Multiply $3$ by $2$ .

$4 x = \frac{6 π - 2 π}{3}$

Step 11.1.5

Subtract $2 π$ from $6 π$ .

$4 x = \frac{4 π}{3}$

Step 11.2

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

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Step 11.2.1

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

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

Step 11.2.2

Simplify the left side.

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Step 11.2.2.1

Cancel the common factor of $4$ .

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Step 11.2.2.1.1

Cancel the common factor.

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

Step 11.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{4 π}{3}}{4}$

Step 11.2.3

Simplify the right side.

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Step 11.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{4 π}{3} \cdot \frac{1}{4}$

Step 11.2.3.2

Cancel the common factor of $4$ .

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Step 11.2.3.2.1

Factor $4$ out of $4 π$ .

$x = \frac{4 (π)}{3} \cdot \frac{1}{4}$

Step 11.2.3.2.2

Cancel the common factor.

$x = \frac{4 π}{3} \cdot \frac{1}{4}$

Step 11.2.3.2.3

Rewrite the expression.

$x = \frac{π}{3}$

Step 12

The solution to the equation $4 x = \frac{2 π}{3}$ .

$x = \frac{π}{6}, \frac{π}{3}$

Step 13

Evaluate the second derivative at $x = \frac{π}{6}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 16 \sin (4 (\frac{π}{6}))$

Step 14

Evaluate the second derivative.

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Step 14.1

Cancel the common factor of $2$ .

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Step 14.1.1

Factor $2$ out of $4$ .

$- 16 \sin (2 (2) \frac{π}{6})$

Step 14.1.2

Factor $2$ out of $6$ .

$- 16 \sin (2 \cdot 2 \frac{π}{2 \cdot 3})$

Step 14.1.3

Cancel the common factor.

$- 16 \sin (2 \cdot 2 \frac{π}{2 \cdot 3})$

Step 14.1.4

Rewrite the expression.

$- 16 \sin (2 \frac{π}{3})$

Step 14.2

Combine $2$ and $\frac{π}{3}$ .

$- 16 \sin (\frac{2 π}{3})$

Step 14.3

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

$- 16 \sin (\frac{π}{3})$

Step 14.4

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

$- 16 \frac{\sqrt{3}}{2}$

Step 14.5

Cancel the common factor of $2$ .

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Step 14.5.1

Factor $2$ out of $- 16$ .

$2 (- 8) \frac{\sqrt{3}}{2}$

Step 14.5.2

Cancel the common factor.

$2 \cdot - 8 \frac{\sqrt{3}}{2}$

Step 14.5.3

Rewrite the expression.

$- 8 \sqrt{3}$

Step 15

$x = \frac{π}{6}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{6}$ is a local maximum

Step 16

Find the y-value when $x = \frac{π}{6}$ .

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Step 16.1

Replace the variable $x$ with $\frac{π}{6}$ in the expression.

$f (\frac{π}{6}) = 2 (\frac{π}{6}) + \sin (4 (\frac{π}{6}))$

Step 16.2

Simplify the result.

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Step 16.2.1

Simplify each term.

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Step 16.2.1.1

Cancel the common factor of $2$ .

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Step 16.2.1.1.1

Factor $2$ out of $6$ .

$f (\frac{π}{6}) = 2 (\frac{π}{2 (3)}) + \sin (4 (\frac{π}{6}))$

Step 16.2.1.1.2

Cancel the common factor.

$f (\frac{π}{6}) = 2 (\frac{π}{2 \cdot 3}) + \sin (4 (\frac{π}{6}))$

Step 16.2.1.1.3

Rewrite the expression.

$f (\frac{π}{6}) = \frac{π}{3} + \sin (4 (\frac{π}{6}))$

Step 16.2.1.2

Cancel the common factor of $2$ .

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Step 16.2.1.2.1

Factor $2$ out of $4$ .

$f (\frac{π}{6}) = \frac{π}{3} + \sin (2 (2) (\frac{π}{6}))$

Step 16.2.1.2.2

Factor $2$ out of $6$ .

$f (\frac{π}{6}) = \frac{π}{3} + \sin (2 \cdot (2 (\frac{π}{2 \cdot 3})))$

Step 16.2.1.2.3

Cancel the common factor.

$f (\frac{π}{6}) = \frac{π}{3} + \sin (2 \cdot (2 (\frac{π}{2 \cdot 3})))$

Step 16.2.1.2.4

Rewrite the expression.

$f (\frac{π}{6}) = \frac{π}{3} + \sin (2 (\frac{π}{3}))$

Step 16.2.1.3

Combine $2$ and $\frac{π}{3}$ .

$f (\frac{π}{6}) = \frac{π}{3} + \sin (\frac{2 π}{3})$

Step 16.2.1.4

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

$f (\frac{π}{6}) = \frac{π}{3} + \sin (\frac{π}{3})$

Step 16.2.1.5

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

$f (\frac{π}{6}) = \frac{π}{3} + \frac{\sqrt{3}}{2}$

Step 16.2.2

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

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

Step 17

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.

$- 16 \sin (4 (\frac{π}{3}))$

Step 18

Evaluate the second derivative.

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Step 18.1

Combine $4$ and $\frac{π}{3}$ .

$- 16 \sin (\frac{4 π}{3})$

Step 18.2

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.

$- 16 (- \sin (\frac{π}{3}))$

Step 18.3

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

$- 16 (- \frac{\sqrt{3}}{2})$

Step 18.4

Cancel the common factor of $2$ .

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Step 18.4.1

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

$- 16 \frac{- \sqrt{3}}{2}$

Step 18.4.2

Factor $2$ out of $- 16$ .

$2 (- 8) \frac{- \sqrt{3}}{2}$

Step 18.4.3

Cancel the common factor.

$2 \cdot - 8 \frac{- \sqrt{3}}{2}$

Step 18.4.4

Rewrite the expression.

$- 8 (- \sqrt{3})$

Step 18.5

Multiply $- 1$ by $- 8$ .

$8 \sqrt{3}$

Step 19

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

$x = \frac{π}{3}$ is a local minimum

Step 20

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

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Step 20.1

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

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

Step 20.2

Simplify the result.

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Step 20.2.1

Simplify each term.

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Step 20.2.1.1

Combine $2$ and $\frac{π}{3}$ .

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

Step 20.2.1.2

Combine $4$ and $\frac{π}{3}$ .

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

Step 20.2.1.3

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.

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

Step 20.2.1.4

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

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

Step 20.2.2

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

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

Step 21

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

$(\frac{π}{6}, \frac{π}{3} + \frac{\sqrt{3}}{2})$ is a local maxima

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

Step 22