Calculus Examples

$g (x) = x - 2 \arctan (x)$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

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

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

Step 1.1.2

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

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

Step 1.2

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

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Step 1.2.1

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

$1 - 2 \frac{d}{d x} [\arctan (x)]$

Step 1.2.2

The derivative of $\arctan (x)$ with respect to $x$ is $\frac{1}{1 + x^{2}}$ .

$1 - 2 \frac{1}{1 + x^{2}}$

Step 1.2.3

Combine $- 2$ and $\frac{1}{1 + x^{2}}$ .

$1 + \frac{- 2}{1 + x^{2}}$

Step 1.2.4

Move the negative in front of the fraction.

$1 - \frac{2}{1 + x^{2}}$

Step 1.3

Simplify.

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Step 1.3.1

Combine terms.

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Step 1.3.1.1

Write $1$ as a fraction with a common denominator.

$\frac{1 + x^{2}}{1 + x^{2}} - \frac{2}{1 + x^{2}}$

Step 1.3.1.2

Combine the numerators over the common denominator.

$\frac{1 + x^{2} - 2}{1 + x^{2}}$

Step 1.3.1.3

Subtract $2$ from $1$ .

$\frac{x^{2} - 1}{1 + x^{2}}$

Step 1.3.2

Reorder terms.

$\frac{x^{2} - 1}{x^{2} + 1}$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x^{2} - 1$ and $g (x) = x^{2} + 1$ .

$g'' (x) = \frac{(x^{2} + 1) \frac{d}{d x} (x^{2} - 1) - (x^{2} - 1) \frac{d}{d x} (x^{2} + 1)}{{(x^{2} + 1)}^{2}}$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $x^{2} - 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 1]$ .

$g'' (x) = \frac{(x^{2} + 1) (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 1)) - (x^{2} - 1) \frac{d}{d x} (x^{2} + 1)}{{(x^{2} + 1)}^{2}}$

Step 2.2.2

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

$g'' (x) = \frac{(x^{2} + 1) (2 x + \frac{d}{d x} (- 1)) - (x^{2} - 1) \frac{d}{d x} (x^{2} + 1)}{{(x^{2} + 1)}^{2}}$

Step 2.2.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$g'' (x) = \frac{(x^{2} + 1) (2 x + 0) - (x^{2} - 1) \frac{d}{d x} (x^{2} + 1)}{{(x^{2} + 1)}^{2}}$

Step 2.2.4

Simplify the expression.

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Step 2.2.4.1

Add $2 x$ and $0$ .

$g'' (x) = \frac{(x^{2} + 1) (2 x) - (x^{2} - 1) \frac{d}{d x} (x^{2} + 1)}{{(x^{2} + 1)}^{2}}$

Step 2.2.4.2

Move $2$ to the left of $x^{2} + 1$ .

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

Step 2.2.5

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$g'' (x) = \frac{2 (x^{2} + 1) x - (x^{2} - 1) (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (1))}{{(x^{2} + 1)}^{2}}$

Step 2.2.6

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

$g'' (x) = \frac{2 (x^{2} + 1) x - (x^{2} - 1) (2 x + \frac{d}{d x} (1))}{{(x^{2} + 1)}^{2}}$

Step 2.2.7

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$g'' (x) = \frac{2 (x^{2} + 1) x - (x^{2} - 1) (2 x + 0)}{{(x^{2} + 1)}^{2}}$

Step 2.2.8

Simplify the expression.

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Step 2.2.8.1

Add $2 x$ and $0$ .

$g'' (x) = \frac{2 (x^{2} + 1) x - (x^{2} - 1) (2 x)}{{(x^{2} + 1)}^{2}}$

Step 2.2.8.2

Multiply $2$ by $- 1$ .

$g'' (x) = \frac{2 (x^{2} + 1) x - 2 (x^{2} - 1) x}{{(x^{2} + 1)}^{2}}$

Step 2.3

Simplify.

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Step 2.3.1

Apply the distributive property.

$g'' (x) = \frac{(2 x^{2} + 2 \cdot 1) x - 2 (x^{2} - 1) x}{{(x^{2} + 1)}^{2}}$

Step 2.3.2

Apply the distributive property.

$g'' (x) = \frac{2 x^{2} x + 2 \cdot (1 x) - 2 (x^{2} - 1) x}{{(x^{2} + 1)}^{2}}$

Step 2.3.3

Apply the distributive property.

$g'' (x) = \frac{2 x^{2} x + 2 \cdot (1 x) + (- 2 x^{2} - 2 \cdot - 1) x}{{(x^{2} + 1)}^{2}}$

Step 2.3.4

Apply the distributive property.

$g'' (x) = \frac{2 x^{2} x + 2 \cdot (1 x) - 2 x^{2} x - 2 \cdot (- 1 x)}{{(x^{2} + 1)}^{2}}$

Step 2.3.5

Simplify the numerator.

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Step 2.3.5.1

Combine the opposite terms in $2 x^{2} x + 2 \cdot 1 x - 2 x^{2} x - 2 \cdot - 1 x$ .

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Step 2.3.5.1.1

Subtract $2 x^{2} x$ from $2 x^{2} x$ .

$g'' (x) = \frac{2 \cdot (1 x) + 0 - 2 \cdot (- 1 x)}{{(x^{2} + 1)}^{2}}$

Step 2.3.5.1.2

Add $2 \cdot 1 x$ and $0$ .

$g'' (x) = \frac{2 \cdot (1 x) - 2 \cdot (- 1 x)}{{(x^{2} + 1)}^{2}}$

Step 2.3.5.2

Simplify each term.

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Step 2.3.5.2.1

Multiply $2$ by $1$ .

$g'' (x) = \frac{2 x - 2 \cdot (- 1 x)}{{(x^{2} + 1)}^{2}}$

Step 2.3.5.2.2

Multiply $- 2$ by $- 1$ .

$g'' (x) = \frac{2 x + 2 x}{{(x^{2} + 1)}^{2}}$

Step 2.3.5.3

Add $2 x$ and $2 x$ .

$g'' (x) = \frac{4 x}{{(x^{2} + 1)}^{2}}$

Step 3

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

$\frac{x^{2} - 1}{x^{2} + 1} = 0$

Step 4

Set the numerator equal to zero.

$x^{2} - 1 = 0$

Step 5

Solve the equation for $x$ .

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Step 5.1

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 5.2

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

$x = \pm \sqrt{1}$

Step 5.3

Any root of $1$ is $1$ .

$x = \pm 1$

Step 5.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Step 5.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

Step 5.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Step 6

Evaluate the second derivative at $x = 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{4 (1)}{{({(1)}^{2} + 1)}^{2}}$

Step 7

Evaluate the second derivative.

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Step 7.1

Multiply $4$ by $1$ .

$\frac{4}{{(1^{2} + 1)}^{2}}$

Step 7.2

Simplify the denominator.

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Step 7.2.1

One to any power is one.

$\frac{4}{{(1 + 1)}^{2}}$

Step 7.2.2

Add $1$ and $1$ .

$\frac{4}{2^{2}}$

Step 7.2.3

Raise $2$ to the power of $2$ .

$\frac{4}{4}$

Step 7.3

Divide $4$ by $4$ .

$1$

Step 8

$x = 1$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 1$ is a local minimum

Step 9

Find the y-value when $x = 1$ .

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Step 9.1

Replace the variable $x$ with $1$ in the expression.

$g (1) = (1) - 2 \arctan (1)$

Step 9.2

Simplify the result.

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Step 9.2.1

Simplify each term.

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Step 9.2.1.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$g (1) = 1 - 2 \frac{π}{4}$

Step 9.2.1.2

Cancel the common factor of $2$ .

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Step 9.2.1.2.1

Factor $2$ out of $- 2$ .

$g (1) = 1 + 2 (- 1) (\frac{π}{4})$

Step 9.2.1.2.2

Factor $2$ out of $4$ .

$g (1) = 1 + 2 \cdot (- 1 \frac{π}{2 \cdot 2})$

Step 9.2.1.2.3

Cancel the common factor.

$g (1) = 1 + 2 \cdot (- 1 \frac{π}{2 \cdot 2})$

Step 9.2.1.2.4

Rewrite the expression.

$g (1) = 1 - 1 \frac{π}{2}$

Step 9.2.1.3

Rewrite $- 1 \frac{π}{2}$ as $- \frac{π}{2}$ .

$g (1) = 1 - \frac{π}{2}$

Step 9.2.2

The final answer is $1 - \frac{π}{2}$ .

$y = 1 - \frac{π}{2}$

Step 10

Evaluate the second derivative at $x = - 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{4 (- 1)}{{({(- 1)}^{2} + 1)}^{2}}$

Step 11

Evaluate the second derivative.

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Step 11.1

Multiply $4$ by $- 1$ .

$\frac{- 4}{{({(- 1)}^{2} + 1)}^{2}}$

Step 11.2

Simplify the denominator.

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Step 11.2.1

Raise $- 1$ to the power of $2$ .

$\frac{- 4}{{(1 + 1)}^{2}}$

Step 11.2.2

Add $1$ and $1$ .

$\frac{- 4}{2^{2}}$

Step 11.2.3

Raise $2$ to the power of $2$ .

$\frac{- 4}{4}$

Step 11.3

Divide $- 4$ by $4$ .

$- 1$

Step 12

$x = - 1$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 1$ is a local maximum

Step 13

Find the y-value when $x = - 1$ .

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Step 13.1

Replace the variable $x$ with $- 1$ in the expression.

$g (- 1) = (- 1) - 2 \arctan (- 1)$

Step 13.2

Simplify the result.

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Step 13.2.1

Simplify each term.

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Step 13.2.1.1

The exact value of $\arctan (- 1)$ is $- \frac{π}{4}$ .

$g (- 1) = - 1 - 2 (- \frac{π}{4})$

Step 13.2.1.2

Cancel the common factor of $2$ .

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Step 13.2.1.2.1

Move the leading negative in $- \frac{π}{4}$ into the numerator.

$g (- 1) = - 1 - 2 \frac{- π}{4}$

Step 13.2.1.2.2

Factor $2$ out of $- 2$ .

$g (- 1) = - 1 + 2 (- 1) (\frac{- π}{4})$

Step 13.2.1.2.3

Factor $2$ out of $4$ .

$g (- 1) = - 1 + 2 \cdot (- 1 \frac{- π}{2 \cdot 2})$

Step 13.2.1.2.4

Cancel the common factor.

$g (- 1) = - 1 + 2 \cdot (- 1 \frac{- π}{2 \cdot 2})$

Step 13.2.1.2.5

Rewrite the expression.

$g (- 1) = - 1 - 1 \frac{- π}{2}$

Step 13.2.1.3

Move the negative in front of the fraction.

$g (- 1) = - 1 - 1 (- \frac{π}{2})$

Step 13.2.1.4

Multiply $- 1 (- \frac{π}{2})$ .

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Step 13.2.1.4.1

Multiply $- 1$ by $- 1$ .

$g (- 1) = - 1 + 1 (\frac{π}{2})$

Step 13.2.1.4.2

Multiply $\frac{π}{2}$ by $1$ .

$g (- 1) = - 1 + \frac{π}{2}$

Step 13.2.2

The final answer is $- 1 + \frac{π}{2}$ .

$y = - 1 + \frac{π}{2}$

Step 14

These are the local extrema for $g (x) = x - 2 \arctan (x)$ .

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

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

Step 15