Calculus Examples

$\frac{x^{2}}{2} - \ln (x)$

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

Combine $2$ and $\frac{1}{2}$ .

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

Step 1.2.4

Combine $\frac{2}{2}$ and $x$ .

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

Step 1.2.5

Cancel the common factor of $2$ .

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Step 1.2.5.1

Cancel the common factor.

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

Step 1.2.5.2

Divide $x$ by $1$ .

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

Step 1.3

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

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Step 1.3.1

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

$x - \frac{d}{d x} [\ln (x)]$

Step 1.3.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$x - \frac{1}{x}$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.2

Evaluate $\frac{d}{d x} [- \frac{1}{x}]$ .

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Step 2.2.1

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 $f (x) = - 1$ and $g (x) = \frac{1}{x}$ .

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

Step 2.2.2

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $- 1$ by $- 1$ .

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

Step 2.2.6

Multiply $x^{- 2}$ by $1$ .

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

Step 2.2.7

Multiply $\frac{1}{x}$ by $0$ .

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

Step 2.2.8

Add $x^{- 2}$ and $0$ .

$f'' (x) = 1 + x^{- 2}$

Step 2.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.4

Reorder terms.

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

Step 3

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

$x - \frac{1}{x} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

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

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

Step 4.1.2

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

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Step 4.1.2.1

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

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

Step 4.1.2.2

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

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

Step 4.1.2.3

Combine $2$ and $\frac{1}{2}$ .

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

Step 4.1.2.4

Combine $\frac{2}{2}$ and $x$ .

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

Step 4.1.2.5

Cancel the common factor of $2$ .

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Step 4.1.2.5.1

Cancel the common factor.

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

Step 4.1.2.5.2

Divide $x$ by $1$ .

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

Step 4.1.3

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

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Step 4.1.3.1

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

$x - \frac{d}{d x} [\ln (x)]$

Step 4.1.3.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$f' (x) = x - \frac{1}{x}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $x - \frac{1}{x}$ .

$x - \frac{1}{x}$

Step 5

Set the first derivative equal to $0$ then solve the equation $x - \frac{1}{x} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$x - \frac{1}{x} = 0$

Step 5.2

Find the LCD of the terms in the equation.

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Step 5.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x, 1$

Step 5.2.2

The LCM of one and any expression is the expression.

$x$

Step 5.3

Multiply each term in $x - \frac{1}{x} = 0$ by $x$ to eliminate the fractions.

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Step 5.3.1

Multiply each term in $x - \frac{1}{x} = 0$ by $x$ .

$x \cdot x - \frac{1}{x} x = 0 x$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Simplify each term.

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Step 5.3.2.1.1

Multiply $x$ by $x$ .

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

Step 5.3.2.1.2

Cancel the common factor of $x$ .

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Step 5.3.2.1.2.1

Move the leading negative in $- \frac{1}{x}$ into the numerator.

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

Step 5.3.2.1.2.2

Cancel the common factor.

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

Step 5.3.2.1.2.3

Rewrite the expression.

$x^{2} - 1 = 0 x$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Multiply $0$ by $x$ .

$x^{2} - 1 = 0$

Step 5.4

Solve the equation.

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Step 5.4.1

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 5.4.2

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

$x = \pm \sqrt{1}$

Step 5.4.3

Any root of $1$ is $1$ .

$x = \pm 1$

Step 5.4.4

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

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Step 5.4.4.1

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

$x = 1$

Step 5.4.4.2

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

$x = - 1$

Step 5.4.4.3

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

$x = 1, - 1$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{1}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 7

Critical points to evaluate.

$x = 1$

Step 8

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.

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

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

One to any power is one.

$\frac{1}{1} + 1$

Step 9.1.2

Divide $1$ by $1$ .

$1 + 1$

Step 9.2

Add $1$ and $1$ .

$2$

Step 10

$x = 1$ 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 11

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

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Step 11.1

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

$f (1) = \frac{{(1)}^{2}}{2} - \ln (1)$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

One to any power is one.

$f (1) = \frac{1}{2} - \ln (1)$

Step 11.2.1.2

The natural logarithm of $1$ is $0$ .

$f (1) = \frac{1}{2} - 0$

Step 11.2.1.3

Multiply $- 1$ by $0$ .

$f (1) = \frac{1}{2} + 0$

Step 11.2.2

Add $\frac{1}{2}$ and $0$ .

$f (1) = \frac{1}{2}$

Step 11.2.3

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

$y = \frac{1}{2}$

Step 12

These are the local extrema for $f (x) = \frac{x^{2}}{2} - \ln (x)$ .

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

Step 13