Calculus Examples

$y = x - \ln (x)$

Step 1

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

$f (x) = x - \ln (x)$

Step 2

Find the first 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 $x - \ln (x)$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- \ln (x)]$ .

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

Step 2.1.2

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

Reorder terms.

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

Step 3

Find the second derivative of the function.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.2.4

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

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

Step 3.2.5

Multiply $- 1$ by $- 1$ .

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

Step 3.2.6

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

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

Step 3.2.7

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

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

Step 3.2.8

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

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

Step 3.3

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

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

Step 3.4

Simplify.

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Step 3.4.1

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

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

Step 3.4.2

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

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

Step 4

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

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

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

Differentiate.

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Step 5.1.1.1

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

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

Step 5.1.1.2

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

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

Step 5.1.2

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

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Step 5.1.2.1

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

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

Step 5.1.2.2

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

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

Step 5.1.3

Reorder terms.

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

Step 5.2

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

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

Step 6

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

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Step 6.1

Set the first derivative equal to $0$ .

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

Step 6.2

Subtract $1$ from both sides of the equation.

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

Step 6.3

Find the LCD of the terms in the equation.

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Step 6.3.1

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

$x, 1$

Step 6.3.2

The LCM of one and any expression is the expression.

$x$

Step 6.4

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

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Step 6.4.1

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

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

Step 6.4.2

Simplify the left side.

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Step 6.4.2.1

Cancel the common factor of $x$ .

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Step 6.4.2.1.1

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

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

Step 6.4.2.1.2

Cancel the common factor.

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

Step 6.4.2.1.3

Rewrite the expression.

$- 1 = - x$

Step 6.5

Solve the equation.

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Step 6.5.1

Rewrite the equation as $- x = - 1$ .

$- x = - 1$

Step 6.5.2

Divide each term in $- x = - 1$ by $- 1$ and simplify.

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Step 6.5.2.1

Divide each term in $- x = - 1$ by $- 1$ .

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

Step 6.5.2.2

Simplify the left side.

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Step 6.5.2.2.1

Dividing two negative values results in a positive value.

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

Step 6.5.2.2.2

Divide $x$ by $1$ .

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

Step 6.5.2.3

Simplify the right side.

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Step 6.5.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

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

$x = 0$

Step 8

Critical points to evaluate.

$x = 1$

Step 9

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.

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

Step 10

Evaluate the second derivative.

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Step 10.1

One to any power is one.

$\frac{1}{1}$

Step 10.2

Divide $1$ by $1$ .

$1$

Step 11

$x = 1$ 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 12

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

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Step 12.1

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

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

Step 12.2

Simplify the result.

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Step 12.2.1

Simplify each term.

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Step 12.2.1.1

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

$f (1) = 1 - 0$

Step 12.2.1.2

Multiply $- 1$ by $0$ .

$f (1) = 1 + 0$

Step 12.2.2

Add $1$ and $0$ .

$f (1) = 1$

Step 12.2.3

The final answer is $1$ .

$y = 1$

Step 13

These are the local extrema for $f (x) = x - \ln (x)$ .

$(1, 1)$ is a local minima

Step 14