Calculus Examples

$y = 2 x - \tan (x)$

Step 1

Write $y = 2 x - \tan (x)$ as a function.

$f (x) = 2 x - \tan (x)$

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

By the Sum Rule, the derivative of $2 x - \tan (x)$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- \tan (x)]$ .

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

Step 2.1.2

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 2.1.2.1

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

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

Step 2.1.2.2

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

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

Step 2.1.2.3

Multiply $2$ by $1$ .

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

Step 2.1.3

Evaluate $\frac{d}{d x} [- \tan (x)]$ .

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Step 2.1.3.1

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

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

Step 2.1.3.2

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

$f' (x) = 2 - \sec^{2} (x)$

Step 2.2

Find the second derivative.

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Step 2.2.1

Differentiate.

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Step 2.2.1.1

By the Sum Rule, the derivative of $2 - \sec^{2} (x)$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [- \sec^{2} (x)]$ .

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

Step 2.2.1.2

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$f'' (x) = 0 + \frac{d}{d x} (- \sec^{2} (x))$

Step 2.2.2

Evaluate $\frac{d}{d x} [- \sec^{2} (x)]$ .

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Step 2.2.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \sec^{2} (x)$ with respect to $x$ is $- \frac{d}{d x} [\sec^{2} (x)]$ .

$f'' (x) = 0 - \frac{d}{d x} \sec^{2} (x)$

Step 2.2.2.2

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

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Step 2.2.2.2.1

To apply the Chain Rule, set $u$ as $\sec (x)$ .

$f'' (x) = 0 - (\frac{d}{d u} (u^{2}) \frac{d}{d x} (\sec (x)))$

Step 2.2.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

$f'' (x) = 0 - (2 u \frac{d}{d x} (\sec (x)))$

Step 2.2.2.2.3

Replace all occurrences of $u$ with $\sec (x)$ .

$f'' (x) = 0 - (2 \sec (x) \frac{d}{d x} (\sec (x)))$

Step 2.2.2.3

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

$f'' (x) = 0 - (2 \sec (x) (\sec (x) \tan (x)))$

Step 2.2.2.4

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

$f'' (x) = 0 - (2 (\sec (x) \sec (x)) \tan (x))$

Step 2.2.2.5

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

$f'' (x) = 0 - (2 (\sec (x) \sec (x)) \tan (x))$

Step 2.2.2.6

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

$f'' (x) = 0 - (2 {\sec (x)}^{1 + 1} \tan (x))$

Step 2.2.2.7

Add $1$ and $1$ .

$f'' (x) = 0 - (2 \sec^{2} (x) \tan (x))$

Step 2.2.2.8

Multiply $2$ by $- 1$ .

$f'' (x) = 0 - 2 \sec^{2} (x) \tan (x)$

Step 2.2.3

Subtract $2 \sec^{2} (x) \tan (x)$ from $0$ .

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

Step 2.3

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

$- 2 \sec^{2} (x) \tan (x)$

Step 3

Set the second derivative equal to $0$ then solve the equation $- 2 \sec^{2} (x) \tan (x) = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$- 2 \sec^{2} (x) \tan (x) = 0$

Step 3.2

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

$\sec^{2} (x) = 0$

$\tan (x) = 0$

Step 3.3

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

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Step 3.3.1

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

$\sec^{2} (x) = 0$

Step 3.3.2

Solve $\sec^{2} (x) = 0$ for $x$ .

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Step 3.3.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\sec (x) = \pm \sqrt{0}$

Step 3.3.2.2

Simplify $\pm \sqrt{0}$ .

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Step 3.3.2.2.1

Rewrite $0$ as $0^{2}$ .

$\sec (x) = \pm \sqrt{0^{2}}$

Step 3.3.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$\sec (x) = \pm 0$

Step 3.3.2.2.3

Plus or minus $0$ is $0$ .

$\sec (x) = 0$

Step 3.3.2.3

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 3.4

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

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Step 3.4.1

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

$\tan (x) = 0$

Step 3.4.2

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

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Step 3.4.2.1

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

$x = \arctan (0)$

Step 3.4.2.2

Simplify the right side.

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Step 3.4.2.2.1

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

$x = 0$

Step 3.4.2.3

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 3.4.2.4

Add $π$ and $0$ .

$x = π$

Step 3.4.2.5

Find the period of $\tan (x)$ .

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Step 3.4.2.5.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 3.4.2.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 3.4.2.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 3.4.2.5.4

Divide $π$ by $1$ .

$π$

Step 3.4.2.6

The period of the $\tan (x)$ function is $π$ so values will repeat every $π$ radians in both directions.

$x = π n, π + π n$ , for any integer $n$

Step 3.5

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

$x = π n, π + π n$ , for any integer $n$

Step 3.6

Consolidate the answers.

$x = π n$ , for any integer $n$

Step 4

The point found by substituting in $f (x) = 2 x - \tan (x)$ is $(,)$ . This point can be an inflection point.

$(,)$

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty,) \cup (, \infty)$

Step 6

Substitute a value from the interval $(- \infty,)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 0.1$ in the expression.

$f'' (- 0.1) = - 2 \sec^{2} (- 0.1) \tan (- 0.1)$

Step 6.2

The final answer is $- 2 \sec^{2} (- 0.1) \tan (- 0.1)$ .

$- 2 \sec^{2} (- 0.1) \tan (- 0.1)$

Step 6.3

At $- 0.1$ , the second derivative is $0.20268949$ . Since this is positive, the second derivative is increasing on the interval $(- \infty,)$ .

Increasing on $(- \infty,)$ since $f'' (x) > 0$

Step 7

Substitute a value from the interval $(, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0.1$ in the expression.

$f'' (0.1) = - 2 \sec^{2} (0.1) \tan (0.1)$

Step 7.2

The final answer is $- 2 \sec^{2} (0.1) \tan (0.1)$ .

$- 2 \sec^{2} (0.1) \tan (0.1)$

Step 7.3

At $0.1$ , the second derivative is $- 0.20268949$ . Since this is negative, the second derivative is decreasing on the interval $(, \infty)$

Decreasing on $(, \infty)$ since $f'' (x) < 0$

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(,)$ .

$(,)$

Step 9