Calculus Examples

$7 \cos (2 x) + 7 x$

Step 1

Write $7 \cos (2 x) + 7 x$ as a function.

$f (x) = 7 \cos (2 x) + 7 x$

Step 2

Find the first derivative of the function.

Tap for more steps...

Step 2.1

By the Sum Rule, the derivative of $7 \cos (2 x) + 7 x$ with respect to $x$ is $\frac{d}{d x} [7 \cos (2 x)] + \frac{d}{d x} [7 x]$ .

$\frac{d}{d x} [7 \cos (2 x)] + \frac{d}{d x} [7 x]$

Step 2.2

Evaluate $\frac{d}{d x} [7 \cos (2 x)]$ .

Tap for more steps...

Step 2.2.1

Since $7$ is constant with respect to $x$ , the derivative of $7 \cos (2 x)$ with respect to $x$ is $7 \frac{d}{d x} [\cos (2 x)]$ .

$7 \frac{d}{d x} [\cos (2 x)] + \frac{d}{d x} [7 x]$

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 $g (x) = 2 x$ .

Tap for more steps...

Step 2.2.2.1

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

$7 (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x]) + \frac{d}{d x} [7 x]$

Step 2.2.2.2

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

$7 (- \sin (u) \frac{d}{d x} [2 x]) + \frac{d}{d x} [7 x]$

Step 2.2.2.3

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

$7 (- \sin (2 x) \frac{d}{d x} [2 x]) + \frac{d}{d x} [7 x]$

Step 2.2.3

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

$7 (- \sin (2 x) (2 \frac{d}{d x} [x])) + \frac{d}{d x} [7 x]$

Step 2.2.4

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

$7 (- \sin (2 x) (2 \cdot 1)) + \frac{d}{d x} [7 x]$

Step 2.2.5

Multiply $2$ by $1$ .

$7 (- \sin (2 x) \cdot 2) + \frac{d}{d x} [7 x]$

Step 2.2.6

Multiply $2$ by $- 1$ .

$7 (- 2 \sin (2 x)) + \frac{d}{d x} [7 x]$

Step 2.2.7

Multiply $- 2$ by $7$ .

$- 14 \sin (2 x) + \frac{d}{d x} [7 x]$

Step 2.3

Evaluate $\frac{d}{d x} [7 x]$ .

Tap for more steps...

Step 2.3.1

Since $7$ is constant with respect to $x$ , the derivative of $7 x$ with respect to $x$ is $7 \frac{d}{d x} [x]$ .

$- 14 \sin (2 x) + 7 \frac{d}{d x} [x]$

Step 2.3.2

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

$- 14 \sin (2 x) + 7 \cdot 1$

Step 2.3.3

Multiply $7$ by $1$ .

$- 14 \sin (2 x) + 7$

Step 2.4

Reorder terms.

$7 - 14 \sin (2 x)$

Step 3

Find the second derivative of the function.

Tap for more steps...

Step 3.1

Differentiate.

Tap for more steps...

Step 3.1.1

By the Sum Rule, the derivative of $7 - 14 \sin (2 x)$ with respect to $x$ is $\frac{d}{d x} [7] + \frac{d}{d x} [- 14 \sin (2 x)]$ .

$f'' (x) = \frac{d}{d x} (7) + \frac{d}{d x} (- 14 \sin (2 x))$

Step 3.1.2

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$f'' (x) = 0 + \frac{d}{d x} (- 14 \sin (2 x))$

Step 3.2

Evaluate $\frac{d}{d x} [- 14 \sin (2 x)]$ .

Tap for more steps...

Step 3.2.1

Since $- 14$ is constant with respect to $x$ , the derivative of $- 14 \sin (2 x)$ with respect to $x$ is $- 14 \frac{d}{d x} [\sin (2 x)]$ .

$f'' (x) = 0 - 14 \frac{d}{d x} \sin (2 x)$

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

Tap for more steps...

Step 3.2.2.1

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

$f'' (x) = 0 - 14 (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (2 x))$

Step 3.2.2.2

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

$f'' (x) = 0 - 14 (\cos (u) \frac{d}{d x} (2 x))$

Step 3.2.2.3

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

$f'' (x) = 0 - 14 (\cos (2 x) \frac{d}{d x} (2 x))$

Step 3.2.3

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

$f'' (x) = 0 - 14 (\cos (2 x) (2 \frac{d}{d x} (x)))$

Step 3.2.4

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

$f'' (x) = 0 - 14 (\cos (2 x) (2 \cdot 1))$

Step 3.2.5

Multiply $2$ by $1$ .

$f'' (x) = 0 - 14 (\cos (2 x) \cdot 2)$

Step 3.2.6

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

$f'' (x) = 0 - 14 (2 \cdot \cos (2 x))$

Step 3.2.7

Multiply $2$ by $- 14$ .

$f'' (x) = 0 - 28 \cos (2 x)$

Step 3.3

Subtract $28 \cos (2 x)$ from $0$ .

$f'' (x) = - 28 \cos (2 x)$

Step 4

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

$7 - 14 \sin (2 x) = 0$

Step 5

Subtract $7$ from both sides of the equation.

$- 14 \sin (2 x) = - 7$

Step 6

Divide each term in $- 14 \sin (2 x) = - 7$ by $- 14$ and simplify.

Tap for more steps...

Step 6.1

Divide each term in $- 14 \sin (2 x) = - 7$ by $- 14$ .

$\frac{- 14 \sin (2 x)}{- 14} = \frac{- 7}{- 14}$

Step 6.2

Simplify the left side.

Tap for more steps...

Step 6.2.1

Cancel the common factor of $- 14$ .

Tap for more steps...

Step 6.2.1.1

Cancel the common factor.

$\frac{- 14 \sin (2 x)}{- 14} = \frac{- 7}{- 14}$

Step 6.2.1.2

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

$\sin (2 x) = \frac{- 7}{- 14}$

Step 6.3

Simplify the right side.

Tap for more steps...

Step 6.3.1

Cancel the common factor of $- 7$ and $- 14$ .

Tap for more steps...

Step 6.3.1.1

Factor $- 7$ out of $- 7$ .

$\sin (2 x) = \frac{- 7 (1)}{- 14}$

Step 6.3.1.2

Cancel the common factors.

Tap for more steps...

Step 6.3.1.2.1

Factor $- 7$ out of $- 14$ .

$\sin (2 x) = \frac{- 7 \cdot 1}{- 7 \cdot 2}$

Step 6.3.1.2.2

Cancel the common factor.

$\sin (2 x) = \frac{- 7 \cdot 1}{- 7 \cdot 2}$

Step 6.3.1.2.3

Rewrite the expression.

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

Step 7

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

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

Step 8

Simplify the right side.

Tap for more steps...

Step 8.1

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

$2 x = \frac{π}{6}$

Step 9

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

Tap for more steps...

Step 9.1

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

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

Step 9.2

Simplify the left side.

Tap for more steps...

Step 9.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 9.2.1.1

Cancel the common factor.

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

Step 9.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{6}}{2}$

Step 9.3

Simplify the right side.

Tap for more steps...

Step 9.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{6} \cdot \frac{1}{2}$

Step 9.3.2

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

Tap for more steps...

Step 9.3.2.1

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

$x = \frac{π}{6 \cdot 2}$

Step 9.3.2.2

Multiply $6$ by $2$ .

$x = \frac{π}{12}$

Step 10

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 11

Solve for $x$ .

Tap for more steps...

Step 11.1

Simplify.

Tap for more steps...

Step 11.1.1

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 11.1.2

Combine $π$ and $\frac{6}{6}$ .

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

Step 11.1.3

Combine the numerators over the common denominator.

$2 x = \frac{π \cdot 6 - π}{6}$

Step 11.1.4

Subtract $π$ from $π \cdot 6$ .

Tap for more steps...

Step 11.1.4.1

Reorder $π$ and $6$ .

$2 x = \frac{6 \cdot π - π}{6}$

Step 11.1.4.2

Subtract $π$ from $6 \cdot π$ .

$2 x = \frac{5 \cdot π}{6}$

Step 11.2

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

Tap for more steps...

Step 11.2.1

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 11.2.2

Simplify the left side.

Tap for more steps...

Step 11.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 11.2.2.1.1

Cancel the common factor.

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

Step 11.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{5 \cdot π}{6}}{2}$

Step 11.2.3

Simplify the right side.

Tap for more steps...

Step 11.2.3.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 11.2.3.2

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

Tap for more steps...

Step 11.2.3.2.1

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

$x = \frac{5 π}{6 \cdot 2}$

Step 11.2.3.2.2

Multiply $6$ by $2$ .

$x = \frac{5 π}{12}$

Step 12

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

$x = \frac{π}{12}, \frac{5 π}{12}$

Step 13

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.

$- 28 \cos (2 (\frac{π}{12}))$

Step 14

Evaluate the second derivative.

Tap for more steps...

Step 14.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 14.1.1

Factor $2$ out of $12$ .

$- 28 \cos (2 \frac{π}{2 (6)})$

Step 14.1.2

Cancel the common factor.

$- 28 \cos (2 \frac{π}{2 \cdot 6})$

Step 14.1.3

Rewrite the expression.

$- 28 \cos (\frac{π}{6})$

Step 14.2

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

$- 28 \frac{\sqrt{3}}{2}$

Step 14.3

Cancel the common factor of $2$ .

Tap for more steps...

Step 14.3.1

Factor $2$ out of $- 28$ .

$2 (- 14) \frac{\sqrt{3}}{2}$

Step 14.3.2

Cancel the common factor.

$2 \cdot - 14 \frac{\sqrt{3}}{2}$

Step 14.3.3

Rewrite the expression.

$- 14 \sqrt{3}$

Step 15

$x = \frac{π}{12}$ 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 16

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

Tap for more steps...

Step 16.1

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

$f (\frac{π}{12}) = 7 \cos (2 (\frac{π}{12})) + 7 (\frac{π}{12})$

Step 16.2

Simplify the result.

Tap for more steps...

Step 16.2.1

Simplify each term.

Tap for more steps...

Step 16.2.1.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 16.2.1.1.1

Factor $2$ out of $12$ .

$f (\frac{π}{12}) = 7 \cos (2 (\frac{π}{2 (6)})) + 7 (\frac{π}{12})$

Step 16.2.1.1.2

Cancel the common factor.

$f (\frac{π}{12}) = 7 \cos (2 (\frac{π}{2 \cdot 6})) + 7 (\frac{π}{12})$

Step 16.2.1.1.3

Rewrite the expression.

$f (\frac{π}{12}) = 7 \cos (\frac{π}{6}) + 7 (\frac{π}{12})$

Step 16.2.1.2

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

$f (\frac{π}{12}) = 7 (\frac{\sqrt{3}}{2}) + 7 (\frac{π}{12})$

Step 16.2.1.3

Combine $7$ and $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{12}) = \frac{7 \sqrt{3}}{2} + 7 (\frac{π}{12})$

Step 16.2.1.4

Combine $7$ and $\frac{π}{12}$ .

$f (\frac{π}{12}) = \frac{7 \sqrt{3}}{2} + \frac{7 π}{12}$

Step 16.2.2

The final answer is $\frac{7 \sqrt{3}}{2} + \frac{7 π}{12}$ .

$y = \frac{7 \sqrt{3}}{2} + \frac{7 π}{12}$

Step 17

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.

$- 28 \cos (2 (\frac{5 π}{12}))$

Step 18

Evaluate the second derivative.

Tap for more steps...

Step 18.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 18.1.1

Factor $2$ out of $12$ .

$- 28 \cos (2 \frac{5 π}{2 (6)})$

Step 18.1.2

Cancel the common factor.

$- 28 \cos (2 \frac{5 π}{2 \cdot 6})$

Step 18.1.3

Rewrite the expression.

$- 28 \cos (\frac{5 π}{6})$

Step 18.2

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.

$- 28 (- \cos (\frac{π}{6}))$

Step 18.3

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

$- 28 (- \frac{\sqrt{3}}{2})$

Step 18.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 18.4.1

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

$- 28 \frac{- \sqrt{3}}{2}$

Step 18.4.2

Factor $2$ out of $- 28$ .

$2 (- 14) \frac{- \sqrt{3}}{2}$

Step 18.4.3

Cancel the common factor.

$2 \cdot - 14 \frac{- \sqrt{3}}{2}$

Step 18.4.4

Rewrite the expression.

$- 14 (- \sqrt{3})$

Step 18.5

Multiply $- 1$ by $- 14$ .

$14 \sqrt{3}$

Step 19

$x = \frac{5 π}{12}$ 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 20

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

Tap for more steps...

Step 20.1

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

$f (\frac{5 π}{12}) = 7 \cos (2 (\frac{5 π}{12})) + 7 (\frac{5 π}{12})$

Step 20.2

Simplify the result.

Tap for more steps...

Step 20.2.1

Simplify each term.

Tap for more steps...

Step 20.2.1.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 20.2.1.1.1

Factor $2$ out of $12$ .

$f (\frac{5 π}{12}) = 7 \cos (2 (\frac{5 π}{2 (6)})) + 7 (\frac{5 π}{12})$

Step 20.2.1.1.2

Cancel the common factor.

$f (\frac{5 π}{12}) = 7 \cos (2 (\frac{5 π}{2 \cdot 6})) + 7 (\frac{5 π}{12})$

Step 20.2.1.1.3

Rewrite the expression.

$f (\frac{5 π}{12}) = 7 \cos (\frac{5 π}{6}) + 7 (\frac{5 π}{12})$

Step 20.2.1.2

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.

$f (\frac{5 π}{12}) = 7 (- \cos (\frac{π}{6})) + 7 (\frac{5 π}{12})$

Step 20.2.1.3

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

$f (\frac{5 π}{12}) = 7 (- \frac{\sqrt{3}}{2}) + 7 (\frac{5 π}{12})$

Step 20.2.1.4

Multiply $7 (- \frac{\sqrt{3}}{2})$ .

Tap for more steps...

Step 20.2.1.4.1

Multiply $- 1$ by $7$ .

$f (\frac{5 π}{12}) = - 7 \frac{\sqrt{3}}{2} + 7 (\frac{5 π}{12})$

Step 20.2.1.4.2

Combine $- 7$ and $\frac{\sqrt{3}}{2}$ .

$f (\frac{5 π}{12}) = \frac{- 7 \sqrt{3}}{2} + 7 (\frac{5 π}{12})$

Step 20.2.1.5

Move the negative in front of the fraction.

$f (\frac{5 π}{12}) = - \frac{7 \sqrt{3}}{2} + 7 (\frac{5 π}{12})$

Step 20.2.1.6

Multiply $7 (\frac{5 π}{12})$ .

Tap for more steps...

Step 20.2.1.6.1

Combine $7$ and $\frac{5 π}{12}$ .

$f (\frac{5 π}{12}) = - \frac{7 \sqrt{3}}{2} + \frac{7 (5 π)}{12}$

Step 20.2.1.6.2

Multiply $5$ by $7$ .

$f (\frac{5 π}{12}) = - \frac{7 \sqrt{3}}{2} + \frac{35 π}{12}$

Step 20.2.2

The final answer is $- \frac{7 \sqrt{3}}{2} + \frac{35 π}{12}$ .

$y = - \frac{7 \sqrt{3}}{2} + \frac{35 π}{12}$

Step 21

These are the local extrema for $f (x) = 7 \cos (2 x) + 7 x$ .

$(\frac{π}{12}, \frac{7 \sqrt{3}}{2} + \frac{7 π}{12})$ is a local maxima

$(\frac{5 π}{12}, - \frac{7 \sqrt{3}}{2} + \frac{35 π}{12})$ is a local minima

Step 22