Calculus Examples

$y = 4 \sec (x)$

Step 1

Write $y = 4 \sec (x)$ as a function.

$f (x) = 4 \sec (x)$

Step 2

Find the first derivative of the function.

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Step 2.1

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

$4 \frac{d}{d x} [\sec (x)]$

Step 2.2

The derivative of $\sec (x)$ with respect to $x$ is $\sec (x) \tan (x)$ .

$4 \sec (x) \tan (x)$

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 $4 \sec (x) \tan (x)$ with respect to $x$ is $4 \frac{d}{d x} [\sec (x) \tan (x)]$ .

$f'' (x) = 4 \frac{d}{d x} (\sec (x) \tan (x))$

Step 3.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \sec (x)$ and $g (x) = \tan (x)$ .

$f'' (x) = 4 (\sec (x) \frac{d}{d x} (\tan (x)) + \tan (x) \frac{d}{d x} (\sec (x)))$

Step 3.3

The derivative of $\tan (x)$ with respect to $x$ is $\sec^{2} (x)$ .

$f'' (x) = 4 (\sec (x) \sec^{2} (x) + \tan (x) \frac{d}{d x} (\sec (x)))$

Step 3.4

Multiply $\sec (x)$ by $\sec^{2} (x)$ by adding the exponents.

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Step 3.4.1

Multiply $\sec (x)$ by $\sec^{2} (x)$ .

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Step 3.4.1.1

Raise $\sec (x)$ to the power of $1$ .

$f'' (x) = 4 (\sec (x) \sec^{2} (x) + \tan (x) \frac{d}{d x} (\sec (x)))$

Step 3.4.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = 4 ({\sec (x)}^{1 + 2} + \tan (x) \frac{d}{d x} (\sec (x)))$

Step 3.4.2

Add $1$ and $2$ .

$f'' (x) = 4 (\sec^{3} (x) + \tan (x) \frac{d}{d x} (\sec (x)))$

Step 3.5

The derivative of $\sec (x)$ with respect to $x$ is $\sec (x) \tan (x)$ .

$f'' (x) = 4 (\sec^{3} (x) + \tan (x) (\sec (x) \tan (x)))$

Step 3.6

Raise $\tan (x)$ to the power of $1$ .

$f'' (x) = 4 (\sec^{3} (x) + \tan (x) \tan (x) \sec (x))$

Step 3.7

Raise $\tan (x)$ to the power of $1$ .

$f'' (x) = 4 (\sec^{3} (x) + \tan (x) \tan (x) \sec (x))$

Step 3.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = 4 (\sec^{3} (x) + {\tan (x)}^{1 + 1} \sec (x))$

Step 3.9

Add $1$ and $1$ .

$f'' (x) = 4 (\sec^{3} (x) + \tan^{2} (x) \sec (x))$

Step 3.10

Simplify.

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Step 3.10.1

Apply the distributive property.

$f'' (x) = 4 \sec^{3} (x) + 4 (\tan^{2} (x) \sec (x))$

Step 3.10.2

Reorder terms.

$f'' (x) = 4 \tan^{2} (x) \sec (x) + 4 \sec^{3} (x)$

Step 4

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

$4 \sec (x) \tan (x) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\sec (x) = 0$

$\tan (x) = 0$

Step 6

Set $\sec (x)$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $\sec (x)$ equal to $0$ .

$\sec (x) = 0$

Step 6.2

The range of secant is $y \leq - 1$ and $y \geq 1$ . Since $0$ does not fall in this range, there is no solution.

No solution

Step 7

Set $\tan (x)$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $\tan (x)$ equal to $0$ .

$\tan (x) = 0$

Step 7.2

Solve $\tan (x) = 0$ for $x$ .

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Step 7.2.1

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

$x = \arctan (0)$

Step 7.2.2

Simplify the right side.

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Step 7.2.2.1

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

$x = 0$

Step 7.2.3

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = π + 0$

Step 7.2.4

Add $π$ and $0$ .

$x = π$

Step 7.2.5

The solution to the equation $x = 0$ .

$x = 0, π$

Step 8

The final solution is all the values that make $4 \sec (x) \tan (x) = 0$ true.

$x = 0, π$

Step 9

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$4 \tan^{2} (0) \sec (0) + 4 \sec^{3} (0)$

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

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

$4 \cdot 0^{2} \sec (0) + 4 \sec^{3} (0)$

Step 10.1.2

Raising $0$ to any positive power yields $0$ .

$4 \cdot 0 \sec (0) + 4 \sec^{3} (0)$

Step 10.1.3

Multiply $4$ by $0$ .

$0 \sec (0) + 4 \sec^{3} (0)$

Step 10.1.4

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

$0 \cdot 1 + 4 \sec^{3} (0)$

Step 10.1.5

Multiply $0$ by $1$ .

$0 + 4 \sec^{3} (0)$

Step 10.1.6

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

$0 + 4 \cdot 1^{3}$

Step 10.1.7

One to any power is one.

$0 + 4 \cdot 1$

Step 10.1.8

Multiply $4$ by $1$ .

$0 + 4$

Step 10.2

Add $0$ and $4$ .

$4$

Step 11

$x = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 0$ is a local minimum

Step 12

Find the y-value when $x = 0$ .

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Step 12.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = 4 \sec (0)$

Step 12.2

Simplify the result.

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Step 12.2.1

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

$f (0) = 4 \cdot 1$

Step 12.2.2

Multiply $4$ by $1$ .

$f (0) = 4$

Step 12.2.3

The final answer is $4$ .

$y = 4$

Step 13

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.

$4 \tan^{2} (π) \sec (π) + 4 \sec^{3} (π)$

Step 14

Evaluate the second derivative.

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Step 14.1

Simplify each term.

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Step 14.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the second quadrant.

$4 {(- \tan (0))}^{2} \sec (π) + 4 \sec^{3} (π)$

Step 14.1.2

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

$4 {(- 0)}^{2} \sec (π) + 4 \sec^{3} (π)$

Step 14.1.3

Multiply $- 1$ by $0$ .

$4 \cdot 0^{2} \sec (π) + 4 \sec^{3} (π)$

Step 14.1.4

Raising $0$ to any positive power yields $0$ .

$4 \cdot 0 \sec (π) + 4 \sec^{3} (π)$

Step 14.1.5

Multiply $4$ by $0$ .

$0 \sec (π) + 4 \sec^{3} (π)$

Step 14.1.6

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the second quadrant.

$0 (- \sec (0)) + 4 \sec^{3} (π)$

Step 14.1.7

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

$0 (- 1 \cdot 1) + 4 \sec^{3} (π)$

Step 14.1.8

Multiply $0 (- 1 \cdot 1)$ .

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Step 14.1.8.1

Multiply $- 1$ by $1$ .

$0 \cdot - 1 + 4 \sec^{3} (π)$

Step 14.1.8.2

Multiply $0$ by $- 1$ .

$0 + 4 \sec^{3} (π)$

Step 14.1.9

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the second quadrant.

$0 + 4 {(- \sec (0))}^{3}$

Step 14.1.10

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

$0 + 4 {(- 1 \cdot 1)}^{3}$

Step 14.1.11

Multiply $- 1$ by $1$ .

$0 + 4 {(- 1)}^{3}$

Step 14.1.12

Raise $- 1$ to the power of $3$ .

$0 + 4 \cdot - 1$

Step 14.1.13

Multiply $4$ by $- 1$ .

$0 - 4$

Step 14.2

Subtract $4$ from $0$ .

$- 4$

Step 15

$x = π$ 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 16

Find the y-value when $x = π$ .

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Step 16.1

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

$f (π) = 4 \sec (π)$

Step 16.2

Simplify the result.

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Step 16.2.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the second quadrant.

$f (π) = 4 (- \sec (0))$

Step 16.2.2

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

$f (π) = 4 (- 1 \cdot 1)$

Step 16.2.3

Multiply $4 (- 1 \cdot 1)$ .

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Step 16.2.3.1

Multiply $- 1$ by $1$ .

$f (π) = 4 \cdot - 1$

Step 16.2.3.2

Multiply $4$ by $- 1$ .

$f (π) = - 4$

Step 16.2.4

The final answer is $- 4$ .

$y = - 4$

Step 17

These are the local extrema for $f (x) = 4 \sec (x)$ .

$(0, 4)$ is a local minima

$(π, - 4)$ is a local maxima

Step 18