Calculus Examples

$y = 4 x + \sin (8 x)$

Write $y = 4 x + \sin (8 x)$ as a function.

$f (x) = 4 x + \sin (8 x)$

Find the first derivative of the function.

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By the Sum Rule, the derivative of $4 x + \sin (8 x)$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [\sin (8 x)]$ .

$\frac{d}{d x} [4 x] + \frac{d}{d x} [\sin (8 x)]$

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

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

$4 \frac{d}{d x} [x] + \frac{d}{d x} [\sin (8 x)]$

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

$4 \cdot 1 + \frac{d}{d x} [\sin (8 x)]$

Multiply $4$ by $1$ .

$4 + \frac{d}{d x} [\sin (8 x)]$

Evaluate $\frac{d}{d x} [\sin (8 x)]$ .

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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 $g (x) = 8 x$ .

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To apply the Chain Rule, set $u$ as $8 x$ .

$4 + \frac{d}{d u} [\sin (u)] \frac{d}{d x} [8 x]$

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

$4 + \cos (u) \frac{d}{d x} [8 x]$

Replace all occurrences of $u$ with $8 x$ .

$4 + \cos (8 x) \frac{d}{d x} [8 x]$

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$4 + \cos (8 x) (8 \frac{d}{d x} [x])$

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

$4 + \cos (8 x) (8 \cdot 1)$

Multiply $8$ by $1$ .

$4 + \cos (8 x) \cdot 8$

Move $8$ to the left of $\cos (8 x)$ .

$4 + 8 \cos (8 x)$

Find the second derivative of the function.

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Differentiate.

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By the Sum Rule, the derivative of $4 + 8 \cos (8 x)$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [8 \cos (8 x)]$ .

$f'' (x) = \frac{d}{d x} (4) + \frac{d}{d x} (8 \cos (8 x))$

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

$f'' (x) = 0 + \frac{d}{d x} (8 \cos (8 x))$

Evaluate $\frac{d}{d x} [8 \cos (8 x)]$ .

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Since $8$ is constant with respect to $x$ , the derivative of $8 \cos (8 x)$ with respect to $x$ is $8 \frac{d}{d x} [\cos (8 x)]$ .

$f'' (x) = 0 + 8 \frac{d}{d x} (\cos (8 x))$

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 $g (x) = 8 x$ .

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To apply the Chain Rule, set $u$ as $8 x$ .

$f'' (x) = 0 + 8 (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (8 x))$

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

$f'' (x) = 0 + 8 (- \sin (u) \frac{d}{d x} (8 x))$

Replace all occurrences of $u$ with $8 x$ .

$f'' (x) = 0 + 8 (- \sin (8 x) \frac{d}{d x} (8 x))$

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$f'' (x) = 0 + 8 (- \sin (8 x) (8 \frac{d}{d x} (x)))$

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

$f'' (x) = 0 + 8 (- \sin (8 x) (8 \cdot 1))$

Multiply $8$ by $1$ .

$f'' (x) = 0 + 8 (- \sin (8 x) \cdot 8)$

Multiply $8$ by $- 1$ .

$f'' (x) = 0 + 8 (- 8 \sin (8 x))$

Multiply $- 8$ by $8$ .

$f'' (x) = 0 - 64 \sin (8 x)$

Subtract $64 \sin (8 x)$ from $0$ .

$f'' (x) = - 64 \sin (8 x)$

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

$4 + 8 \cos (8 x) = 0$

Subtract $4$ from both sides of the equation.

$8 \cos (8 x) = - 4$

Divide each term in $8 \cos (8 x) = - 4$ by $8$ and simplify.

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Divide each term in $8 \cos (8 x) = - 4$ by $8$ .

$\frac{8 \cos (8 x)}{8} = \frac{- 4}{8}$

Simplify the left side.

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Cancel the common factor of $8$ .

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Cancel the common factor.

$\frac{8 \cos (8 x)}{8} = \frac{- 4}{8}$

Divide $\cos (8 x)$ by $1$ .

$\cos (8 x) = \frac{- 4}{8}$

Simplify the right side.

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Cancel the common factor of $- 4$ and $8$ .

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Factor $4$ out of $- 4$ .

$\cos (8 x) = \frac{4 (- 1)}{8}$

Cancel the common factors.

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Factor $4$ out of $8$ .

$\cos (8 x) = \frac{4 \cdot - 1}{4 \cdot 2}$

Cancel the common factor.

$\cos (8 x) = \frac{4 \cdot - 1}{4 \cdot 2}$

Rewrite the expression.

$\cos (8 x) = \frac{- 1}{2}$

Move the negative in front of the fraction.

$\cos (8 x) = - \frac{1}{2}$

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

$8 x = \arccos (- \frac{1}{2})$

Simplify the right side.

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

$8 x = \frac{2 π}{3}$

Divide each term in $8 x = \frac{2 π}{3}$ by $8$ and simplify.

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Divide each term in $8 x = \frac{2 π}{3}$ by $8$ .

$\frac{8 x}{8} = \frac{\frac{2 π}{3}}{8}$

Simplify the left side.

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Cancel the common factor of $8$ .

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Cancel the common factor.

$\frac{8 x}{8} = \frac{\frac{2 π}{3}}{8}$

Divide $x$ by $1$ .

$x = \frac{\frac{2 π}{3}}{8}$

Simplify the right side.

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Multiply the numerator by the reciprocal of the denominator.

$x = \frac{2 π}{3} \cdot \frac{1}{8}$

Cancel the common factor of $2$ .

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Factor $2$ out of $2 π$ .

$x = \frac{2 (π)}{3} \cdot \frac{1}{8}$

Factor $2$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

$x = \frac{π}{3} \cdot \frac{1}{4}$

Multiply $\frac{π}{3}$ by $\frac{1}{4}$ .

$x = \frac{π}{3 \cdot 4}$

Multiply $3$ by $4$ .

$x = \frac{π}{12}$

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.

$8 x = 2 π - \frac{2 π}{3}$

Solve for $x$ .

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Simplify.

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

$8 x = 2 π \cdot \frac{3}{3} - \frac{2 π}{3}$

Combine $2 π$ and $\frac{3}{3}$ .

$8 x = \frac{2 π \cdot 3}{3} - \frac{2 π}{3}$

Combine the numerators over the common denominator.

$8 x = \frac{2 π \cdot 3 - 2 π}{3}$

Multiply $3$ by $2$ .

$8 x = \frac{6 π - 2 π}{3}$

Subtract $2 π$ from $6 π$ .

$8 x = \frac{4 π}{3}$

Divide each term in $8 x = \frac{4 π}{3}$ by $8$ and simplify.

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Divide each term in $8 x = \frac{4 π}{3}$ by $8$ .

$\frac{8 x}{8} = \frac{\frac{4 π}{3}}{8}$

Simplify the left side.

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Cancel the common factor of $8$ .

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Cancel the common factor.

$\frac{8 x}{8} = \frac{\frac{4 π}{3}}{8}$

Divide $x$ by $1$ .

$x = \frac{\frac{4 π}{3}}{8}$

Simplify the right side.

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Multiply the numerator by the reciprocal of the denominator.

$x = \frac{4 π}{3} \cdot \frac{1}{8}$

Cancel the common factor of $4$ .

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Factor $4$ out of $4 π$ .

$x = \frac{4 (π)}{3} \cdot \frac{1}{8}$

Factor $4$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

$x = \frac{π}{3} \cdot \frac{1}{2}$

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

$x = \frac{π}{3 \cdot 2}$

Multiply $3$ by $2$ .

$x = \frac{π}{6}$

The solution to the equation $8 x = \frac{2 π}{3}$ .

$x = \frac{π}{12}, \frac{π}{6}$

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.

$- 64 \sin (8 (\frac{π}{12}))$

Evaluate the second derivative.

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Cancel the common factor of $4$ .

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Factor $4$ out of $8$ .

$- 64 \sin (4 (2) \frac{π}{12})$

Factor $4$ out of $12$ .

$- 64 \sin (4 \cdot 2 \frac{π}{4 \cdot 3})$

Cancel the common factor.

$- 64 \sin (4 \cdot 2 \frac{π}{4 \cdot 3})$

Rewrite the expression.

$- 64 \sin (2 \frac{π}{3})$

Combine $2$ and $\frac{π}{3}$ .

$- 64 \sin (\frac{2 π}{3})$

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

$- 64 \sin (\frac{π}{3})$

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

$- 64 \frac{\sqrt{3}}{2}$

Cancel the common factor of $2$ .

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Factor $2$ out of $- 64$ .

$2 (- 32) \frac{\sqrt{3}}{2}$

Cancel the common factor.

$2 \cdot - 32 \frac{\sqrt{3}}{2}$

Rewrite the expression.

$- 32 \sqrt{3}$

$x = \frac{π}{12}$ 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

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

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

$f (\frac{π}{12}) = 4 (\frac{π}{12}) + \sin (8 (\frac{π}{12}))$

Simplify the result.

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Simplify each term.

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Cancel the common factor of $4$ .

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Factor $4$ out of $12$ .

$f (\frac{π}{12}) = 4 (\frac{π}{4 (3)}) + \sin (8 (\frac{π}{12}))$

Cancel the common factor.

$f (\frac{π}{12}) = 4 (\frac{π}{4 \cdot 3}) + \sin (8 (\frac{π}{12}))$

Rewrite the expression.

$f (\frac{π}{12}) = \frac{π}{3} + \sin (8 (\frac{π}{12}))$

Cancel the common factor of $4$ .

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Factor $4$ out of $8$ .

$f (\frac{π}{12}) = \frac{π}{3} + \sin (4 (2) (\frac{π}{12}))$

Factor $4$ out of $12$ .

$f (\frac{π}{12}) = \frac{π}{3} + \sin (4 \cdot (2 (\frac{π}{4 \cdot 3})))$

Cancel the common factor.

$f (\frac{π}{12}) = \frac{π}{3} + \sin (4 \cdot (2 (\frac{π}{4 \cdot 3})))$

Rewrite the expression.

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

Combine $2$ and $\frac{π}{3}$ .

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

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

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

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

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

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

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

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.

$- 64 \sin (8 (\frac{π}{6}))$

Evaluate the second derivative.

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Cancel the common factor of $2$ .

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Factor $2$ out of $8$ .

$- 64 \sin (2 (4) \frac{π}{6})$

Factor $2$ out of $6$ .

$- 64 \sin (2 \cdot 4 \frac{π}{2 \cdot 3})$

Cancel the common factor.

$- 64 \sin (2 \cdot 4 \frac{π}{2 \cdot 3})$

Rewrite the expression.

$- 64 \sin (4 \frac{π}{3})$

Combine $4$ and $\frac{π}{3}$ .

$- 64 \sin (\frac{4 π}{3})$

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.

$- 64 (- \sin (\frac{π}{3}))$

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

$- 64 (- \frac{\sqrt{3}}{2})$

Cancel the common factor of $2$ .

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

$- 64 \frac{- \sqrt{3}}{2}$

Factor $2$ out of $- 64$ .

$2 (- 32) \frac{- \sqrt{3}}{2}$

Cancel the common factor.

$2 \cdot - 32 \frac{- \sqrt{3}}{2}$

Rewrite the expression.

$- 32 (- \sqrt{3})$

Multiply $- 1$ by $- 32$ .

$32 \sqrt{3}$

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

$x = \frac{π}{6}$ is a local minimum

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

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

$f (\frac{π}{6}) = 4 (\frac{π}{6}) + \sin (8 (\frac{π}{6}))$

Simplify the result.

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Simplify each term.

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Cancel the common factor of $2$ .

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Factor $2$ out of $4$ .

$f (\frac{π}{6}) = 2 (2) (\frac{π}{6}) + \sin (8 (\frac{π}{6}))$

Factor $2$ out of $6$ .

$f (\frac{π}{6}) = 2 \cdot (2 (\frac{π}{2 \cdot 3})) + \sin (8 (\frac{π}{6}))$

Cancel the common factor.

$f (\frac{π}{6}) = 2 \cdot (2 (\frac{π}{2 \cdot 3})) + \sin (8 (\frac{π}{6}))$

Rewrite the expression.

$f (\frac{π}{6}) = 2 (\frac{π}{3}) + \sin (8 (\frac{π}{6}))$

Combine $2$ and $\frac{π}{3}$ .

$f (\frac{π}{6}) = \frac{2 π}{3} + \sin (8 (\frac{π}{6}))$

Cancel the common factor of $2$ .

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Factor $2$ out of $8$ .

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

Factor $2$ out of $6$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Combine $4$ and $\frac{π}{3}$ .

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

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.

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

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

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

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

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

These are the local extrema for $f (x) = 4 x + \sin (8 x)$ .

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

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