Calculus Examples

$f (x) = \frac{1}{x} + \ln (x)$

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

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

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

Step 1.4

Simplify.

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Step 1.4.1

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

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

Step 1.4.2

Reorder terms.

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

Step 2

Find the second derivative of the function.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

Evaluate $\frac{d}{d x} [- \frac{1}{x^{2}}]$ .

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Step 2.3.1

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

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

Step 2.3.2

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

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

Step 2.3.3

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

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Step 2.3.3.1

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

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

Step 2.3.3.2

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

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

Step 2.3.3.3

Replace all occurrences of $u$ with $x^{2}$ .

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

Multiply the exponents in ${(x^{2})}^{- 2}$ .

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Step 2.3.6.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 2.3.6.2

Multiply $2$ by $- 2$ .

$f'' (x) = - x^{- 2} - (- x^{- 4} (2 x)) + \frac{1}{x^{2}} \cdot 0$

Step 2.3.7

Multiply $2$ by $- 1$ .

$f'' (x) = - x^{- 2} - (- 2 x^{- 4} x) + \frac{1}{x^{2}} \cdot 0$

Step 2.3.8

Raise $x$ to the power of $1$ .

$f'' (x) = - x^{- 2} - (- 2 (x \cdot x^{- 4})) + \frac{1}{x^{2}} \cdot 0$

Step 2.3.9

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

$f'' (x) = - x^{- 2} - (- 2 x^{1 - 4}) + \frac{1}{x^{2}} \cdot 0$

Step 2.3.10

Subtract $4$ from $1$ .

$f'' (x) = - x^{- 2} - (- 2 x^{- 3}) + \frac{1}{x^{2}} \cdot 0$

Step 2.3.11

Multiply $- 2$ by $- 1$ .

$f'' (x) = - x^{- 2} + 2 x^{- 3} + \frac{1}{x^{2}} \cdot 0$

Step 2.3.12

Multiply $\frac{1}{x^{2}}$ by $0$ .

$f'' (x) = - x^{- 2} + 2 x^{- 3} + 0$

Step 2.3.13

Add $2 x^{- 3}$ and $0$ .

$f'' (x) = - x^{- 2} + 2 x^{- 3}$

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

Combine $2$ and $\frac{1}{x^{3}}$ .

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

Step 3

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

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

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 $\frac{1}{x} + \ln (x)$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{x}] + \frac{d}{d x} [\ln (x)]$ .

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

Step 4.1.2

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

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Step 4.1.2.1

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

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

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 $n = - 1$ .

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

Step 4.1.3

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

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

Step 4.1.4

Simplify.

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Step 4.1.4.1

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

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

Step 4.1.4.2

Reorder terms.

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

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{1}{x} - \frac{1}{x^{2}}$ .

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

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{1}{x} - \frac{1}{x^{2}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

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

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.

$x, x^{2}, 1$

Step 5.2.2

Since $x, x^{2}, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1, 1$ then find LCM for the variable part $x^{1}, x^{2}$ .

Step 5.2.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 5.2.4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 5.2.5

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 5.2.6

The factor for $x^{1}$ is $x$ itself.

$x^{1} = x$

$x$ occurs $1$ time.

Step 5.2.7

The factors for $x^{2}$ are $x \cdot x$ , which is $x$ multiplied by each other $2$ times.

$x^{2} = x \cdot x$

$x$ occurs $2$ times.

Step 5.2.8

The LCM of $x^{1}, x^{2}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x \cdot x$

Step 5.2.9

Multiply $x$ by $x$ .

$x^{2}$

Step 5.3

Multiply each term in $\frac{1}{x} - \frac{1}{x^{2}} = 0$ by $x^{2}$ to eliminate the fractions.

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Step 5.3.1

Multiply each term in $\frac{1}{x} - \frac{1}{x^{2}} = 0$ by $x^{2}$ .

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

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

Cancel the common factor of $x$ .

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Step 5.3.2.1.1.1

Factor $x$ out of $x^{2}$ .

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

Step 5.3.2.1.1.2

Cancel the common factor.

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

Step 5.3.2.1.1.3

Rewrite the expression.

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

Step 5.3.2.1.2

Cancel the common factor of $x^{2}$ .

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Step 5.3.2.1.2.1

Move the leading negative in $- \frac{1}{x^{2}}$ into the numerator.

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

Step 5.3.2.1.2.2

Cancel the common factor.

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

Step 5.3.2.1.2.3

Rewrite the expression.

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Multiply $0$ by $x^{2}$ .

$x - 1 = 0$

Step 5.4

Add $1$ to both sides of the equation.

$x = 1$

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 6.2

Set the denominator in $\frac{1}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Step 6.3

Solve for $x$ .

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Step 6.3.1

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

$x = \pm \sqrt{0}$

Step 6.3.2

Simplify $\pm \sqrt{0}$ .

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Step 6.3.2.1

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

$x = \pm \sqrt{0^{2}}$

Step 6.3.2.2

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

$x = \pm 0$

Step 6.3.2.3

Plus or minus $0$ is $0$ .

$x = 0$

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.

$- \frac{1}{{(1)}^{2}} + \frac{2}{{(1)}^{3}}$

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.

$- \frac{1}{1} + \frac{2}{{(1)}^{3}}$

Step 9.1.2

Cancel the common factor of $1$ .

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Step 9.1.2.1

Cancel the common factor.

$- \frac{1}{1} + \frac{2}{{(1)}^{3}}$

Step 9.1.2.2

Rewrite the expression.

$- 1 \cdot 1 + \frac{2}{{(1)}^{3}}$

Step 9.1.3

Multiply $- 1$ by $1$ .

$- 1 + \frac{2}{{(1)}^{3}}$

Step 9.1.4

One to any power is one.

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

Step 9.1.5

Divide $2$ by $1$ .

$- 1 + 2$

Step 9.2

Add $- 1$ and $2$ .

$1$

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.

$f (1) = \frac{1}{1} + \ln (1)$

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

Divide $1$ by $1$ .

$f (1) = 1 + \ln (1)$

Step 11.2.1.2

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

$f (1) = 1 + 0$

Step 11.2.2

Add $1$ and $0$ .

$f (1) = 1$

Step 11.2.3

The final answer is $1$ .

$y = 1$

Step 12

These are the local extrema for $f (x) = \frac{1}{x} + \ln (x)$ .

$(1, 1)$ is a local minima

Step 13