Calculus Examples

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

Step 1

Write $x^{2} - x - \ln (x)$ as a function.

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

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 $x^{2} - x - \ln (x)$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [- \ln (x)]$ .

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

Step 2.1.2

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $- 1$ by $1$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.4

Reorder terms.

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

Step 3

Find the second derivative of the function.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

Multiply $2$ by $1$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

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

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

Step 3.3.5

Multiply $- 1$ by $- 1$ .

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

Step 3.3.6

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

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

Step 3.3.7

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

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

Step 3.3.8

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

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

Step 3.4

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

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

Step 3.5

Simplify.

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Step 3.5.1

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

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

Step 3.5.2

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

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

Step 3.5.3

Reorder terms.

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

Step 4

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

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

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 $x^{2} - x - \ln (x)$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [- \ln (x)]$ .

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

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 $n = 2$ .

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

Step 5.1.2

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

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Step 5.1.2.1

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

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

Step 5.1.2.2

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

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

Step 5.1.2.3

Multiply $- 1$ by $1$ .

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

Step 5.1.3

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

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Step 5.1.3.1

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

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

Step 5.1.3.2

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

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

Step 5.1.4

Reorder terms.

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

Step 5.2

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

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

Step 6

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

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Step 6.1

Set the first derivative equal to $0$ .

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

Step 6.2

Find the LCD of the terms in the equation.

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Step 6.2.1

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

$1, x, 1, 1$

Step 6.2.2

The LCM of one and any expression is the expression.

$x$

Step 6.3

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

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Step 6.3.1

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

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

Step 6.3.2

Simplify the left side.

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Step 6.3.2.1

Simplify each term.

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Step 6.3.2.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 6.3.2.1.1.1

Move $x$ .

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

Step 6.3.2.1.1.2

Multiply $x$ by $x$ .

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

Step 6.3.2.1.2

Cancel the common factor of $x$ .

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Step 6.3.2.1.2.1

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

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

Step 6.3.2.1.2.2

Cancel the common factor.

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

Step 6.3.2.1.2.3

Rewrite the expression.

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

Step 6.3.3

Simplify the right side.

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Step 6.3.3.1

Multiply $0$ by $x$ .

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

Step 6.4

Solve the equation.

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Step 6.4.1

Factor by grouping.

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Step 6.4.1.1

Reorder terms.

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

Step 6.4.1.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = - 1$ .

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Step 6.4.1.2.1

Factor $- 1$ out of $- x$ .

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

Step 6.4.1.2.2

Rewrite $- 1$ as $1$ plus $- 2$

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

Step 6.4.1.2.3

Apply the distributive property.

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

Step 6.4.1.2.4

Multiply $x$ by $1$ .

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

Step 6.4.1.3

Factor out the greatest common factor from each group.

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Step 6.4.1.3.1

Group the first two terms and the last two terms.

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

Step 6.4.1.3.2

Factor out the greatest common factor (GCF) from each group.

$x (2 x + 1) - (2 x + 1) = 0$

Step 6.4.1.4

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

$(2 x + 1) (x - 1) = 0$

Step 6.4.2

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

$2 x + 1 = 0$

$x - 1 = 0$

Step 6.4.3

Set $2 x + 1$ equal to $0$ and solve for $x$ .

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Step 6.4.3.1

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Step 6.4.3.2

Solve $2 x + 1 = 0$ for $x$ .

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Step 6.4.3.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 6.4.3.2.2

Divide each term in $2 x = - 1$ by $2$ and simplify.

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Step 6.4.3.2.2.1

Divide each term in $2 x = - 1$ by $2$ .

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

Step 6.4.3.2.2.2

Simplify the left side.

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Step 6.4.3.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.4.3.2.2.2.1.1

Cancel the common factor.

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

Step 6.4.3.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 6.4.3.2.2.3

Simplify the right side.

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Step 6.4.3.2.2.3.1

Move the negative in front of the fraction.

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

Step 6.4.4

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 6.4.4.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 6.4.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 6.4.5

The final solution is all the values that make $(2 x + 1) (x - 1) = 0$ true.

$x = - \frac{1}{2}, 1$

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.

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

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

One to any power is one.

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

Step 10.1.2

Divide $1$ by $1$ .

$1 + 2$

Step 10.2

Add $1$ and $2$ .

$3$

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.

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

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

One to any power is one.

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

Step 12.2.1.2

Multiply $- 1$ by $1$ .

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

Step 12.2.1.3

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

$f (1) = 1 - 1 - 0$

Step 12.2.1.4

Multiply $- 1$ by $0$ .

$f (1) = 1 - 1 + 0$

Step 12.2.2

Simplify by adding and subtracting.

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Step 12.2.2.1

Subtract $1$ from $1$ .

$f (1) = 0 + 0$

Step 12.2.2.2

Add $0$ and $0$ .

$f (1) = 0$

Step 12.2.3

The final answer is $0$ .

$y = 0$

Step 13

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

$(1, 0)$ is a local minima

Step 14