Calculus Examples

$y = 6 x - 4 \cos (3 x)$

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (6 x - 4 \cos (3 x))$

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Differentiate the right side of the equation.

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By the Sum Rule, the derivative of $6 x - 4 \cos (3 x)$ with respect to $x$ is $\frac{d}{d x} [6 x] + \frac{d}{d x} [- 4 \cos (3 x)]$ .

$\frac{d}{d x} [6 x] + \frac{d}{d x} [- 4 \cos (3 x)]$

Evaluate $\frac{d}{d x} [6 x]$ .

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Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$6 \frac{d}{d x} [x] + \frac{d}{d x} [- 4 \cos (3 x)]$

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

$6 \cdot 1 + \frac{d}{d x} [- 4 \cos (3 x)]$

Multiply $6$ by $1$ to get $6$ .

$6 + \frac{d}{d x} [- 4 \cos (3 x)]$

Evaluate $\frac{d}{d x} [- 4 \cos (3 x)]$ .

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Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 \cos (3 x)$ with respect to $x$ is $- 4 \frac{d}{d x} [\cos (3 x)]$ .

$6 - 4 \frac{d}{d x} [\cos (3 x)]$

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 $g (x) = 3 x$ .

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To apply the Chain Rule, set $u$ as $3 x$ .

$6 - 4 (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [3 x])$

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

$6 - 4 (- \sin (u) \frac{d}{d x} [3 x])$

Replace all occurrences of $u$ with $3 x$ .

$6 - 4 (- \sin (3 x) \frac{d}{d x} [3 x])$

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$6 - 4 (- \sin (3 x) (3 \frac{d}{d x} [x]))$

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

$6 - 4 (- \sin (3 x) (3 \cdot 1))$

Multiply $3$ by $1$ to get $3$ .

$6 - 4 (- \sin (3 x) \cdot 3)$

Multiply $3$ by $- 1$ to get $- 3$ .

$6 - 4 (- 3 \sin (3 x))$

Multiply $- 3$ by $- 4$ to get $12$ .

$6 + 12 \sin (3 x)$

Reform the equation by setting the left side equal to the right side.

$y' = 6 + 12 \sin (3 x)$

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = 6 + 12 \sin (3 x)$

Set $\frac{d y}{d x} = 0$ then solve for $x$ in terms of $y$ .

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Rewrite the equation as $6 + 12 \sin (3 x) = 0$ .

$6 + 12 \sin (3 x) = 0$

Since $6$ does not contain the variable to solve for, move it to the right side of the equation by subtracting $6$ from both sides.

$12 \sin (3 x) = - 6$

Divide each term by $12$ and simplify.

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Divide each term in $12 \sin (3 x) = - 6$ by $12$ .

$\frac{12 \sin (3 x)}{12} = - \frac{6}{12}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{12 \sin (3 x)}{12} = - \frac{6}{12}$

Divide $\sin (3 x)$ by $1$ to get $\sin (3 x)$ .

$\sin (3 x) = - \frac{6}{12}$

Reduce the expression by cancelling the common factors.

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Factor $6$ out of $12$ .

$\sin (3 x) = - \frac{6 \cdot 1}{6 \cdot 2}$

Cancel the common factor.

$\sin (3 x) = - \frac{6 \cdot 1}{6 \cdot 2}$

Rewrite the expression.

$\sin (3 x) = - \frac{1}{2}$

Simplify the expression to find the first solution.

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Take the inverse $s i n e$ of both sides of the equation to extract $x$ from inside the $s i n e$ .

$3 x = \arcsin (- \frac{1}{2})$

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

$3 x = - \frac{π}{6}$

Divide each term by $3$ and simplify.

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Divide each term in $3 x = - \frac{π}{6}$ by $3$ .

$\frac{3 x}{3} = - \frac{π}{6} \cdot \frac{1}{3}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{3 x}{3} = - \frac{π}{6} \cdot \frac{1}{3}$

Divide $x$ by $1$ to get $x$ .

$x = - \frac{π}{6} \cdot \frac{1}{3}$

Simplify the right side of the equation.

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Multiply $- \frac{π}{6}$ by $\frac{1}{3}$ to get $- \frac{π}{6} \frac{1}{3}$ .

$x = - \frac{π}{6} \cdot \frac{1}{3}$

Simplify $- \frac{π}{6} \frac{1}{3}$ .

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Multiply $\frac{1}{3}$ and $\frac{π}{6}$ to get $\frac{π}{3 \cdot 6}$ .

$x = - \frac{π}{3 \cdot 6}$

Multiply $3$ by $6$ to get $18$ .

$x = - \frac{π}{18}$

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$3 x = 2 π + \frac{π}{6} + π$

Simplify the expression to find the second solution.

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Simplify the right side.

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To write $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$3 x = \frac{2 π}{1} \cdot \frac{6}{6} + \frac{π}{6} + π$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$3 x = \frac{2 π \cdot 6}{1 \cdot 6} + \frac{π}{6} + π$

Multiply $6$ by $1$ to get $6$ .

$3 x = \frac{2 π \cdot 6}{6} + \frac{π}{6} + π$

Combine the numerators over the common denominator.

$3 x = \frac{2 π \cdot 6 + π}{6} + π$

Simplify each term.

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Simplify the numerator.

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Factor $π$ out of $2 π \cdot 6 + π$ .

$3 x = \frac{π (2 \cdot 6 + 1)}{6} + π$

Multiply $2$ by $6$ to get $12$ .

$3 x = \frac{π (12 + 1)}{6} + π$

Add $12$ and $1$ to get $13$ .

$3 x = \frac{π \cdot 13}{6} + π$

Move $13$ to the left of the expression $π \cdot 13$ .

$3 x = \frac{13 \cdot π}{6} + π$

Multiply $13$ by $π$ to get $13 π$ .

$3 x = \frac{13 π}{6} + π$

To write $\frac{π}{1}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$3 x = \frac{13 π}{6} + \frac{π}{1} \cdot \frac{6}{6}$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$3 x = \frac{13 π}{6} + \frac{π \cdot 6}{1 \cdot 6}$

Multiply $6$ by $1$ to get $6$ .

$3 x = \frac{13 π}{6} + \frac{π \cdot 6}{6}$

Combine the numerators over the common denominator.

$3 x = \frac{13 π + π \cdot 6}{6}$

Simplify the numerator.

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Move $6$ to the left of the expression $π \cdot 6$ .

$3 x = \frac{13 π + 6 \cdot π}{6}$

Multiply $6$ by $π$ to get $6 π$ .

$3 x = \frac{13 π + 6 π}{6}$

Add $13 π$ and $6 π$ to get $19 π$ .

$3 x = \frac{19 π}{6}$

Divide each term by $3$ and simplify.

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Divide each term in $3 x = \frac{19 π}{6}$ by $3$ .

$\frac{3 x}{3} = \frac{19 π}{6} \cdot \frac{1}{3}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{3 x}{3} = \frac{19 π}{6} \cdot \frac{1}{3}$

Divide $x$ by $1$ to get $x$ .

$x = \frac{19 π}{6} \cdot \frac{1}{3}$

Simplify the right side of the equation.

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Multiply $\frac{19 π}{6}$ by $\frac{1}{3}$ to get $\frac{19 π}{6} \frac{1}{3}$ .

$x = \frac{19 π}{6} \cdot \frac{1}{3}$

Simplify $\frac{19 π}{6} \frac{1}{3}$ .

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Multiply $\frac{19 π}{6}$ and $\frac{1}{3}$ to get $\frac{19 π}{6 \cdot 3}$ .

$x = \frac{19 π}{6 \cdot 3}$

Multiply $6$ by $3$ to get $18$ .

$x = \frac{19 π}{18}$

Find the period.

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The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $3$ in the formula for period.

$\frac{2 π}{| 3 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$\frac{2 π}{3}$

Add $2 π$ to every negative angle to get positive angles.

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Add $2 π$ to $- \frac{π}{18}$ to find the positive angle.

$- \frac{π}{18} + 2 π$

To write $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{18}{18}$ .

$\frac{2 π}{1} \frac{18}{18} - \frac{π}{18}$

Write each expression with a common denominator of $18$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$\frac{2 π \cdot 18}{1 \cdot 18} - \frac{π}{18}$

Multiply $18$ by $1$ to get $18$ .

$\frac{2 π \cdot 18}{18} - \frac{π}{18}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 18 - π}{18}$

Simplify the numerator.

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Factor $π$ out of $2 π \cdot 18 - π$ .

$\frac{π (2 \cdot 18 - 1)}{18}$

Multiply $2$ by $18$ to get $36$ .

$\frac{π (36 - 1)}{18}$

Subtract $1$ from $36$ to get $35$ .

$\frac{π \cdot 35}{18}$

Simplify the expression.

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Move $35$ to the left of the expression $π \cdot 35$ .

$\frac{35 \cdot π}{18}$

Multiply $35$ by $π$ to get $35 π$ .

$\frac{35 π}{18}$

List the new angles.

$3 x = \frac{35 π}{18}$

The period of the $\sin (3 x)$ function is $\frac{2 π}{3}$ so values will repeat every $\frac{2 π}{3}$ radians in both directions.

$x = \frac{19 π}{18} \pm \frac{2 π n}{3}, \frac{35 π}{18} \pm \frac{2 π n}{3}$

Find the points where $\frac{d y}{d x} = 0$ .

$(\frac{19 π}{18} \pm \frac{2 π n}{3}, \frac{19 π}{18} \pm \frac{2 π n}{3})$

$(\frac{19 π}{18} \pm \frac{2 π n}{3}, \frac{35 π}{18} \pm \frac{2 π n}{3})$

$(\frac{35 π}{18} \pm \frac{2 π n}{3}, \frac{19 π}{18} \pm \frac{2 π n}{3})$

$(\frac{35 π}{18} \pm \frac{2 π n}{3}, \frac{35 π}{18} \pm \frac{2 π n}{3})$

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Calculus Examples

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