Calculus Examples

$y = \sec (x)$

Step 1

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

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

Step 2

Find the second derivative of the function.

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Step 2.1

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 $g (x) = \tan (x)$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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Step 2.3.1.1

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

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

Step 2.3.1.2

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

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

Step 2.3.2

Add $1$ and $2$ .

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

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

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

Step 2.7

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

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

Step 2.8

Add $1$ and $1$ .

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

Step 2.9

Reorder terms.

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

Step 3

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

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

Step 4

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 5

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

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Step 5.1

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

$\sec (x) = 0$

Step 5.2

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 6

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

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Step 6.1

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

$\tan (x) = 0$

Step 6.2

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

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Step 6.2.1

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

$x = \arctan (0)$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

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

$x = 0$

Step 6.2.3

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 6.2.4

Add $π$ and $0$ .

$x = π$

Step 6.2.5

The solution to the equation $x = 0$ .

$x = 0, π$

Step 7

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

$x = 0, π$

Step 8

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.

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

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

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

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

Step 9.1.2

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

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

Step 9.1.3

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

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

Step 9.1.4

Multiply $0$ by $1$ .

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

Step 9.1.5

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

$0 + 1^{3}$

Step 9.1.6

One to any power is one.

$0 + 1$

Step 9.2

Add $0$ and $1$ .

$1$

Step 10

$x = 0$ 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 11

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

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Step 11.1

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

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

Step 11.2

Simplify the result.

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Step 11.2.1

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

$f (0) = 1$

Step 11.2.2

The final answer is $1$ .

$y = 1$

Step 12

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.

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

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify each term.

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Step 13.1.1

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.

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

Step 13.1.2

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

${(- 0)}^{2} \sec (π) + \sec^{3} (π)$

Step 13.1.3

Multiply $- 1$ by $0$ .

$0^{2} \sec (π) + \sec^{3} (π)$

Step 13.1.4

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

$0 \sec (π) + \sec^{3} (π)$

Step 13.1.5

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.

$0 (- \sec (0)) + \sec^{3} (π)$

Step 13.1.6

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

$0 (- 1 \cdot 1) + \sec^{3} (π)$

Step 13.1.7

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

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Step 13.1.7.1

Multiply $- 1$ by $1$ .

$0 \cdot - 1 + \sec^{3} (π)$

Step 13.1.7.2

Multiply $0$ by $- 1$ .

$0 + \sec^{3} (π)$

Step 13.1.8

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.

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

Step 13.1.9

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

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

Step 13.1.10

Multiply $- 1$ by $1$ .

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

Step 13.1.11

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

$0 - 1$

Step 13.2

Subtract $1$ from $0$ .

$- 1$

Step 14

$x = π$ 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 15

Find the y-value when $x = π$ .

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Step 15.1

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

$f (π) = \sec (π)$

Step 15.2

Simplify the result.

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Step 15.2.1

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.

$f (π) = - \sec (0)$

Step 15.2.2

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

$f (π) = - 1 \cdot 1$

Step 15.2.3

Multiply $- 1$ by $1$ .

$f (π) = - 1$

Step 15.2.4

The final answer is $- 1$ .

$y = - 1$

Step 16

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

$(0, 1)$ is a local minima

$(π, - 1)$ is a local maxima

Step 17