Calculus Examples

$f (x) = 2 \cos (x - π)$

Step 1

Find the first derivative of the function.

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Step 1.1

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

$2 \frac{d}{d x} [\cos (x - π)]$

Step 1.2

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 $g (x) = x - π$ .

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Step 1.2.1

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

$2 (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [x - π])$

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.3

Differentiate.

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Step 1.3.1

Multiply $- 1$ by $2$ .

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

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

$- 2 \sin (x - π) (1 + 0)$

Step 1.3.5

Simplify the expression.

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Step 1.3.5.1

Add $1$ and $0$ .

$- 2 \sin (x - π) \cdot 1$

Step 1.3.5.2

Multiply $- 2$ by $1$ .

$- 2 \sin (x - π)$

Step 2

Find the second derivative of the function.

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Step 2.1

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

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

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 $f (x) = \sin (x)$ and $g (x) = x - π$ .

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Differentiate.

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Step 2.3.1

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

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

Step 2.3.2

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

$f'' (x) = - 2 \cos (x - π) (1 + \frac{d}{d x} (- π))$

Step 2.3.3

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

$f'' (x) = - 2 \cos (x - π) (1 + 0)$

Step 2.3.4

Simplify the expression.

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Step 2.3.4.1

Add $1$ and $0$ .

$f'' (x) = - 2 \cos (x - π) \cdot 1$

Step 2.3.4.2

Multiply $- 2$ by $1$ .

$f'' (x) = - 2 \cos (x - π)$

Step 3

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

$- 2 \sin (x - π) = 0$

Step 4

Divide each term in $- 2 \sin (x - π) = 0$ by $- 2$ and simplify.

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Step 4.1

Divide each term in $- 2 \sin (x - π) = 0$ by $- 2$ .

$\frac{- 2 \sin (x - π)}{- 2} = \frac{0}{- 2}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $- 2$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{- 2 \sin (x - π)}{- 2} = \frac{0}{- 2}$

Step 4.2.1.2

Divide $\sin (x - π)$ by $1$ .

$\sin (x - π) = \frac{0}{- 2}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 2$ .

$\sin (x - π) = 0$

Step 5

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

$x - π = \arcsin (0)$

Step 6

Simplify the right side.

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Step 6.1

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

$x - π = 0$

Step 7

Add $π$ to both sides of the equation.

$x = π$

Step 8

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.

$x - π = π - 0$

Step 9

Solve for $x$ .

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Step 9.1

Subtract $0$ from $π$ .

$x - π = π$

Step 9.2

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

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Step 9.2.1

Add $π$ to both sides of the equation.

$x = π + π$

Step 9.2.2

Add $π$ and $π$ .

$x = 2 π$

Step 10

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

$x = π, 2 π$

Step 11

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

$- 2 \cos ((π) - π)$

Step 12

Evaluate the second derivative.

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Step 12.1

Subtract $π$ from $π$ .

$- 2 \cos (0)$

Step 12.2

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

$- 2 \cdot 1$

Step 12.3

Multiply $- 2$ by $1$ .

$- 2$

Step 13

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

$x = π$ is a local maximum

Step 14

Find the y-value when $x = π$ .

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Step 14.1

Replace the variable $x$ with $π$ in the expression.

$f (π) = 2 \cos ((π) - π)$

Step 14.2

Simplify the result.

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Step 14.2.1

Subtract $π$ from $π$ .

$f (π) = 2 \cos (0)$

Step 14.2.2

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

$f (π) = 2 \cdot 1$

Step 14.2.3

Multiply $2$ by $1$ .

$f (π) = 2$

Step 14.2.4

The final answer is $2$ .

$y = 2$

Step 15

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

$- 2 \cos ((2 π) - π)$

Step 16

Evaluate the second derivative.

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Step 16.1

Subtract $π$ from $2 π$ .

$- 2 \cos (π)$

Step 16.2

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.

$- 2 (- \cos (0))$

Step 16.3

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

$- 2 (- 1 \cdot 1)$

Step 16.4

Multiply $- 2 (- 1 \cdot 1)$ .

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Step 16.4.1

Multiply $- 1$ by $1$ .

$- 2 \cdot - 1$

Step 16.4.2

Multiply $- 2$ by $- 1$ .

$2$

Step 17

$x = 2 π$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 2 π$ is a local minimum

Step 18

Find the y-value when $x = 2 π$ .

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Step 18.1

Replace the variable $x$ with $2 π$ in the expression.

$f (2 π) = 2 \cos ((2 π) - π)$

Step 18.2

Simplify the result.

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Step 18.2.1

Subtract $π$ from $2 π$ .

$f (2 π) = 2 \cos (π)$

Step 18.2.2

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.

$f (2 π) = 2 (- \cos (0))$

Step 18.2.3

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

$f (2 π) = 2 (- 1 \cdot 1)$

Step 18.2.4

Multiply $2 (- 1 \cdot 1)$ .

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Step 18.2.4.1

Multiply $- 1$ by $1$ .

$f (2 π) = 2 \cdot - 1$

Step 18.2.4.2

Multiply $2$ by $- 1$ .

$f (2 π) = - 2$

Step 18.2.5

The final answer is $- 2$ .

$y = - 2$

Step 19

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

$(π, 2)$ is a local maxima

$(2 π, - 2)$ is a local minima

Step 20