Calculus Examples

$y = - \cos (4 x - π)$

Step 1

Write $y = - \cos (4 x - π)$ as a function.

$f (x) = - \cos (4 x - π)$

Step 2

Find the first derivative of the function.

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Step 2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos (4 x - π)$ with respect to $x$ is $- \frac{d}{d x} [\cos (4 x - π)]$ .

$- \frac{d}{d x} [\cos (4 x - π)]$

Step 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 $g (x) = 4 x - π$ .

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Differentiate.

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Step 2.3.1

Multiply $- 1$ by $- 1$ .

$1 (\sin (4 x - π) \frac{d}{d x} [4 x - π])$

Step 2.3.2

Multiply $\sin (4 x - π)$ by $1$ .

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

Multiply $4$ by $1$ .

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

Step 2.3.7

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

$\sin (4 x - π) (4 + 0)$

Step 2.3.8

Simplify the expression.

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Step 2.3.8.1

Add $4$ and $0$ .

$\sin (4 x - π) \cdot 4$

Step 2.3.8.2

Move $4$ to the left of $\sin (4 x - π)$ .

$4 \sin (4 x - π)$

Step 3

Find the second derivative of the function.

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Step 3.1

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

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

Step 3.2

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 $g (x) = 4 x - π$ .

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Step 3.2.1

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

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

Step 3.2.2

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

$f'' (x) = 4 (\cos (u) \frac{d}{d x} (4 x - π))$

Step 3.2.3

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

$f'' (x) = 4 (\cos (4 x - π) \frac{d}{d x} (4 x - π))$

Step 3.3

Differentiate.

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

$f'' (x) = 4 \cos (4 x - π) (4 \cdot 1 + \frac{d}{d x} (- π))$

Step 3.3.4

Multiply $4$ by $1$ .

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

Step 3.3.5

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

$f'' (x) = 4 \cos (4 x - π) (4 + 0)$

Step 3.3.6

Simplify the expression.

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Step 3.3.6.1

Add $4$ and $0$ .

$f'' (x) = 4 \cos (4 x - π) \cdot 4$

Step 3.3.6.2

Multiply $4$ by $4$ .

$f'' (x) = 16 \cos (4 x - π)$

Step 4

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

$4 \sin (4 x - π) = 0$

Step 5

Divide each term in $4 \sin (4 x - π) = 0$ by $4$ and simplify.

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Step 5.1

Divide each term in $4 \sin (4 x - π) = 0$ by $4$ .

$\frac{4 \sin (4 x - π)}{4} = \frac{0}{4}$

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $4$ .

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Step 5.2.1.1

Cancel the common factor.

$\frac{4 \sin (4 x - π)}{4} = \frac{0}{4}$

Step 5.2.1.2

Divide $\sin (4 x - π)$ by $1$ .

$\sin (4 x - π) = \frac{0}{4}$

Step 5.3

Simplify the right side.

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Step 5.3.1

Divide $0$ by $4$ .

$\sin (4 x - π) = 0$

Step 6

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

$4 x - π = \arcsin (0)$

Step 7

Simplify the right side.

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Step 7.1

The exact value of $\arcsin (0)$ is $0$ .

$4 x - π = 0$

Step 8

Add $π$ to both sides of the equation.

$4 x = π$

Step 9

Divide each term in $4 x = π$ by $4$ and simplify.

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Step 9.1

Divide each term in $4 x = π$ by $4$ .

$\frac{4 x}{4} = \frac{π}{4}$

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.

$\frac{4 x}{4} = \frac{π}{4}$

Step 9.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{4}$

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.

$4 x - π = π - 0$

Step 11

Solve for $x$ .

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Step 11.1

Subtract $0$ from $π$ .

$4 x - π = π$

Step 11.2

Move all terms not containing $x$ to the right side of the equation.

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Step 11.2.1

Add $π$ to both sides of the equation.

$4 x = π + π$

Step 11.2.2

Add $π$ and $π$ .

$4 x = 2 π$

Step 11.3

Divide each term in $4 x = 2 π$ by $4$ and simplify.

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Step 11.3.1

Divide each term in $4 x = 2 π$ by $4$ .

$\frac{4 x}{4} = \frac{2 π}{4}$

Step 11.3.2

Simplify the left side.

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Step 11.3.2.1

Cancel the common factor of $4$ .

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Step 11.3.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{2 π}{4}$

Step 11.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{2 π}{4}$

Step 11.3.3

Simplify the right side.

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Step 11.3.3.1

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

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Step 11.3.3.1.1

Factor $2$ out of $2 π$ .

$x = \frac{2 (π)}{4}$

Step 11.3.3.1.2

Cancel the common factors.

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Step 11.3.3.1.2.1

Factor $2$ out of $4$ .

$x = \frac{2 π}{2 \cdot 2}$

Step 11.3.3.1.2.2

Cancel the common factor.

$x = \frac{2 π}{2 \cdot 2}$

Step 11.3.3.1.2.3

Rewrite the expression.

$x = \frac{π}{2}$

Step 12

The solution to the equation $4 x - π = 0$ .

$x = \frac{π}{4}, \frac{π}{2}$

Step 13

Evaluate the second derivative at $x = \frac{π}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$16 \cos (4 (\frac{π}{4}) - π)$

Step 14

Evaluate the second derivative.

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Step 14.1

Cancel the common factor of $4$ .

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Step 14.1.1

Cancel the common factor.

$16 \cos (4 \frac{π}{4} - π)$

Step 14.1.2

Rewrite the expression.

$16 \cos (π - π)$

Step 14.2

Subtract $π$ from $π$ .

$16 \cos (0)$

Step 14.3

The exact value of $\cos (0)$ is $1$ .

$16 \cdot 1$

Step 14.4

Multiply $16$ by $1$ .

$16$

Step 15

$x = \frac{π}{4}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{π}{4}$ is a local minimum

Step 16

Find the y-value when $x = \frac{π}{4}$ .

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Step 16.1

Replace the variable $x$ with $\frac{π}{4}$ in the expression.

$f (\frac{π}{4}) = - \cos (4 (\frac{π}{4}) - π)$

Step 16.2

Simplify the result.

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Step 16.2.1

Cancel the common factor of $4$ .

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Step 16.2.1.1

Cancel the common factor.

$f (\frac{π}{4}) = - \cos (4 (\frac{π}{4}) - π)$

Step 16.2.1.2

Rewrite the expression.

$f (\frac{π}{4}) = - \cos (π - π)$

Step 16.2.2

Subtract $π$ from $π$ .

$f (\frac{π}{4}) = - \cos (0)$

Step 16.2.3

The exact value of $\cos (0)$ is $1$ .

$f (\frac{π}{4}) = - 1 \cdot 1$

Step 16.2.4

Multiply $- 1$ by $1$ .

$f (\frac{π}{4}) = - 1$

Step 16.2.5

The final answer is $- 1$ .

$y = - 1$

Step 17

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

$16 \cos (4 (\frac{π}{2}) - π)$

Step 18

Evaluate the second derivative.

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Step 18.1

Cancel the common factor of $2$ .

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Step 18.1.1

Factor $2$ out of $4$ .

$16 \cos (2 (2) \frac{π}{2} - π)$

Step 18.1.2

Cancel the common factor.

$16 \cos (2 \cdot 2 \frac{π}{2} - π)$

Step 18.1.3

Rewrite the expression.

$16 \cos (2 π - π)$

Step 18.2

Subtract $π$ from $2 π$ .

$16 \cos (π)$

Step 18.3

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.

$16 (- \cos (0))$

Step 18.4

The exact value of $\cos (0)$ is $1$ .

$16 (- 1 \cdot 1)$

Step 18.5

Multiply $16 (- 1 \cdot 1)$ .

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Step 18.5.1

Multiply $- 1$ by $1$ .

$16 \cdot - 1$

Step 18.5.2

Multiply $16$ by $- 1$ .

$- 16$

Step 19

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

$x = \frac{π}{2}$ is a local maximum

Step 20

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

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Step 20.1

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

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

Step 20.2

Simplify the result.

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Step 20.2.1

Cancel the common factor of $2$ .

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Step 20.2.1.1

Factor $2$ out of $4$ .

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

Step 20.2.1.2

Cancel the common factor.

$f (\frac{π}{2}) = - \cos (2 \cdot (2 (\frac{π}{2})) - π)$

Step 20.2.1.3

Rewrite the expression.

$f (\frac{π}{2}) = - \cos (2 π - π)$

Step 20.2.2

Subtract $π$ from $2 π$ .

$f (\frac{π}{2}) = - \cos (π)$

Step 20.2.3

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.

$f (\frac{π}{2}) = \cos (0)$

Step 20.2.4

The exact value of $\cos (0)$ is $1$ .

$f (\frac{π}{2}) = - (- 1 \cdot 1)$

Step 20.2.5

Multiply $- (- 1 \cdot 1)$ .

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Step 20.2.5.1

Multiply $- 1$ by $1$ .

$f (\frac{π}{2}) = 1$

Step 20.2.5.2

Multiply $- 1$ by $- 1$ .

$f (\frac{π}{2}) = 1$

Step 20.2.6

The final answer is $1$ .

$y = 1$

Step 21

These are the local extrema for $f (x) = - \cos (4 x - π)$ .

$(\frac{π}{4}, - 1)$ is a local minima

$(\frac{π}{2}, 1)$ is a local maxima

Step 22