Calculus Examples

$f (x) = \sec (2 x)$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sec (x)$ and $g (x) = 2 x$ .

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Step 1.1.1

To apply the Chain Rule, set $u$ as $2 x$ .

$\frac{d}{d u} [\sec (u)] \frac{d}{d x} [2 x]$

Step 1.1.2

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

$\sec (u) \tan (u) \frac{d}{d x} [2 x]$

Step 1.1.3

Replace all occurrences of $u$ with $2 x$ .

$\sec (2 x) \tan (2 x) \frac{d}{d x} [2 x]$

Step 1.2

Differentiate.

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Step 1.2.1

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

$\sec (2 x) \tan (2 x) (2 \frac{d}{d x} [x])$

Step 1.2.2

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

$\sec (2 x) \tan (2 x) (2 \cdot 1)$

Step 1.2.3

Simplify the expression.

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Step 1.2.3.1

Multiply $2$ by $1$ .

$\sec (2 x) \tan (2 x) \cdot 2$

Step 1.2.3.2

Move $2$ to the left of $\sec (2 x) \tan (2 x)$ .

$2 \sec (2 x) \tan (2 x)$

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 $2 \sec (2 x) \tan (2 x)$ with respect to $x$ is $2 \frac{d}{d x} [\sec (2 x) \tan (2 x)]$ .

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

Step 2.2

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 $f (x) = \sec (2 x)$ and $g (x) = \tan (2 x)$ .

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

Step 2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = 2 x$ .

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Step 2.3.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

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

Step 2.3.2

The derivative of $\tan (u_{1})$ with respect to $u_{1}$ is $\sec^{2} (u_{1})$ .

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

Step 2.3.3

Replace all occurrences of $u_{1}$ with $2 x$ .

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

Step 2.4

Raise $\sec (2 x)$ to the power of $1$ .

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

Step 2.5

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

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

Step 2.6

Differentiate.

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Step 2.6.1

Add $2$ and $1$ .

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

Step 2.6.2

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

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

Step 2.6.3

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

$f'' (x) = 2 (\sec^{3} (2 x) (2 \cdot 1) + \tan (2 x) \frac{d}{d x} (\sec (2 x)))$

Step 2.6.4

Simplify the expression.

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Step 2.6.4.1

Multiply $2$ by $1$ .

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

Step 2.6.4.2

Move $2$ to the left of $\sec^{3} (2 x)$ .

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

Step 2.7

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sec (x)$ and $g (x) = 2 x$ .

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Step 2.7.1

To apply the Chain Rule, set $u_{2}$ as $2 x$ .

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

Step 2.7.2

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

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

Step 2.7.3

Replace all occurrences of $u_{2}$ with $2 x$ .

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

Step 2.8

Raise $\tan (2 x)$ to the power of $1$ .

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

Step 2.9

Raise $\tan (2 x)$ to the power of $1$ .

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

Step 2.10

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

$f'' (x) = 2 (2 \sec^{3} (2 x) + {\tan (2 x)}^{1 + 1} (\sec (2 x) \frac{d}{d x} (2 x)))$

Step 2.11

Add $1$ and $1$ .

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

Step 2.12

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

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

Step 2.13

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

$f'' (x) = 2 (2 \sec^{3} (2 x) + \tan^{2} (2 x) \sec (2 x) (2 \cdot 1))$

Step 2.14

Simplify the expression.

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Step 2.14.1

Multiply $2$ by $1$ .

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

Step 2.14.2

Move $2$ to the left of $\tan^{2} (2 x) \sec (2 x)$ .

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

Step 2.15

Simplify.

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Step 2.15.1

Apply the distributive property.

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

Step 2.15.2

Combine terms.

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Step 2.15.2.1

Multiply $2$ by $2$ .

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

Step 2.15.2.2

Multiply $2$ by $2$ .

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

Step 2.15.3

Reorder terms.

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

Step 3

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

$2 \sec (2 x) \tan (2 x) = 0$

Step 4

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

$\sec (2 x) = 0$

$\tan (2 x) = 0$

Step 5

Set $\sec (2 x)$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $\sec (2 x)$ equal to $0$ .

$\sec (2 x) = 0$

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 $\tan (2 x)$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $\tan (2 x)$ equal to $0$ .

$\tan (2 x) = 0$

Step 6.2

Solve $\tan (2 x) = 0$ 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.

$2 x = \arctan (0)$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

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

$2 x = 0$

Step 6.2.3

Divide each term in $2 x = 0$ by $2$ and simplify.

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Step 6.2.3.1

Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Step 6.2.3.2

Simplify the left side.

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Step 6.2.3.2.1

Cancel the common factor of $2$ .

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Step 6.2.3.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Step 6.2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Step 6.2.3.3

Simplify the right side.

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Step 6.2.3.3.1

Divide $0$ by $2$ .

$x = 0$

Step 6.2.4

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.

$2 x = π + 0$

Step 6.2.5

Solve for $x$ .

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Step 6.2.5.1

Add $π$ and $0$ .

$2 x = π$

Step 6.2.5.2

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

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Step 6.2.5.2.1

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

$\frac{2 x}{2} = \frac{π}{2}$

Step 6.2.5.2.2

Simplify the left side.

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Step 6.2.5.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.5.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Step 6.2.5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{2}$

Step 6.2.6

The solution to the equation $2 x = 0$ .

$x = 0, \frac{π}{2}$

Step 7

The final solution is all the values that make $2 \sec (2 x) \tan (2 x) = 0$ true.

$x = 0, \frac{π}{2}$

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.

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

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

Multiply $2$ by $0$ .

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

Step 9.1.2

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

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

Step 9.1.3

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

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

Step 9.1.4

Multiply $4$ by $0$ .

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

Step 9.1.5

Multiply $2$ by $0$ .

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

Step 9.1.6

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

$0 \cdot 1 + 4 \sec^{3} (2 (0))$

Step 9.1.7

Multiply $0$ by $1$ .

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

Step 9.1.8

Multiply $2$ by $0$ .

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

Step 9.1.9

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

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

Step 9.1.10

One to any power is one.

$0 + 4 \cdot 1$

Step 9.1.11

Multiply $4$ by $1$ .

$0 + 4$

Step 9.2

Add $0$ and $4$ .

$4$

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.

$f (0) = \sec (2 (0))$

Step 11.2

Simplify the result.

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Step 11.2.1

Multiply $2$ by $0$ .

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

Step 11.2.2

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

$f (0) = 1$

Step 11.2.3

The final answer is $1$ .

$y = 1$

Step 12

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.

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

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify each term.

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Step 13.1.1

Cancel the common factor of $2$ .

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Step 13.1.1.1

Cancel the common factor.

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

Step 13.1.1.2

Rewrite the expression.

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

Step 13.1.2

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.

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

Step 13.1.3

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

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

Step 13.1.4

Multiply $- 1$ by $0$ .

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

Step 13.1.5

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

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

Step 13.1.6

Multiply $4$ by $0$ .

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

Step 13.1.7

Cancel the common factor of $2$ .

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Step 13.1.7.1

Cancel the common factor.

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

Step 13.1.7.2

Rewrite the expression.

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

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.

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

Step 13.1.9

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

$0 (- 1 \cdot 1) + 4 \sec^{3} (2 (\frac{π}{2}))$

Step 13.1.10

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

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Step 13.1.10.1

Multiply $- 1$ by $1$ .

$0 \cdot - 1 + 4 \sec^{3} (2 (\frac{π}{2}))$

Step 13.1.10.2

Multiply $0$ by $- 1$ .

$0 + 4 \sec^{3} (2 (\frac{π}{2}))$

Step 13.1.11

Cancel the common factor of $2$ .

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Step 13.1.11.1

Cancel the common factor.

$0 + 4 \sec^{3} (2 \frac{π}{2})$

Step 13.1.11.2

Rewrite the expression.

$0 + 4 \sec^{3} (π)$

Step 13.1.12

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 13.1.13

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

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

Step 13.1.14

Multiply $- 1$ by $1$ .

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

Step 13.1.15

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

$0 + 4 \cdot - 1$

Step 13.1.16

Multiply $4$ by $- 1$ .

$0 - 4$

Step 13.2

Subtract $4$ from $0$ .

$- 4$

Step 14

$x = \frac{π}{2}$ 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 15

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

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Step 15.1

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

$f (\frac{π}{2}) = \sec (2 (\frac{π}{2}))$

Step 15.2

Simplify the result.

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Step 15.2.1

Cancel the common factor of $2$ .

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Step 15.2.1.1

Cancel the common factor.

$f (\frac{π}{2}) = \sec (2 (\frac{π}{2}))$

Step 15.2.1.2

Rewrite the expression.

$f (\frac{π}{2}) = \sec (π)$

Step 15.2.2

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.

$f (\frac{π}{2}) = - \sec (0)$

Step 15.2.3

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

$f (\frac{π}{2}) = - 1 \cdot 1$

Step 15.2.4

Multiply $- 1$ by $1$ .

$f (\frac{π}{2}) = - 1$

Step 15.2.5

The final answer is $- 1$ .

$y = - 1$

Step 16

These are the local extrema for $f (x) = \sec (2 x)$ .

$(0, 1)$ is a local minima

$(\frac{π}{2}, - 1)$ is a local maxima

Step 17