Calculus Examples

$f (x) = x - 2 \ln (x + 1)$

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 $x - 2 \ln (x + 1)$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 2 \ln (x + 1)]$ .

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

Step 1.1.2

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

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

Step 1.2

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

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Step 1.2.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 \ln (x + 1)$ with respect to $x$ is $- 2 \frac{d}{d x} [\ln (x + 1)]$ .

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

Step 1.2.2

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

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Step 1.2.2.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 1.2.2.2

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

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

Step 1.2.2.3

Replace all occurrences of $u$ with $x + 1$ .

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

Step 1.2.3

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

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

Step 1.2.4

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

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

Step 1.2.5

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

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

Step 1.2.6

Add $1$ and $0$ .

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

Step 1.2.7

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

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

Step 1.2.8

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

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

Step 1.2.9

Move the negative in front of the fraction.

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

Step 1.3

Combine terms.

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Step 1.3.1

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

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

Step 1.3.2

Combine the numerators over the common denominator.

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

Step 1.3.3

Subtract $2$ from $1$ .

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

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 $f (x) = x - 1$ and $g (x) = x + 1$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

Simplify the expression.

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Step 2.2.4.1

Add $1$ and $0$ .

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

Step 2.2.4.2

Multiply $x + 1$ by $1$ .

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.2.8

Simplify the expression.

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Step 2.2.8.1

Add $1$ and $0$ .

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

Step 2.2.8.2

Multiply $- 1$ by $1$ .

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

Step 2.3

Simplify.

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Step 2.3.1

Apply the distributive property.

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

Step 2.3.2

Simplify the numerator.

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Step 2.3.2.1

Combine the opposite terms in $x + 1 - x - - 1$ .

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Step 2.3.2.1.1

Subtract $x$ from $x$ .

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

Step 2.3.2.1.2

Add $0$ and $1$ .

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

Step 2.3.2.2

Multiply $- 1$ by $- 1$ .

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

Step 2.3.2.3

Add $1$ and $1$ .

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

Step 3

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

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

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

Differentiate.

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Step 4.1.1.1

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

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

Step 4.1.1.2

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

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

Step 4.1.2

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

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Step 4.1.2.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 \ln (x + 1)$ with respect to $x$ is $- 2 \frac{d}{d x} [\ln (x + 1)]$ .

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

Step 4.1.2.2

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

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Step 4.1.2.2.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 4.1.2.2.2

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

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

Step 4.1.2.2.3

Replace all occurrences of $u$ with $x + 1$ .

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

Step 4.1.2.3

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

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

Step 4.1.2.4

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

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

Step 4.1.2.5

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

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

Step 4.1.2.6

Add $1$ and $0$ .

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

Step 4.1.2.7

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

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

Step 4.1.2.8

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

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

Step 4.1.2.9

Move the negative in front of the fraction.

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

Step 4.1.3

Combine terms.

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Step 4.1.3.1

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

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

Step 4.1.3.2

Combine the numerators over the common denominator.

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

Step 4.1.3.3

Subtract $2$ from $1$ .

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

Step 4.2

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

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

Step 5

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

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Step 5.1

Set the first derivative equal to $0$ .

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

Step 5.2

Set the numerator equal to zero.

$x - 1 = 0$

Step 5.3

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 $\frac{x - 1}{x + 1}$ equal to $0$ to find where the expression is undefined.

$x + 1 = 0$

Step 6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

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.

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

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the denominator.

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Step 9.1.1

Add $1$ and $1$ .

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

Step 9.1.2

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

$\frac{2}{4}$

Step 9.2

Cancel the common factor of $2$ and $4$ .

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Step 9.2.1

Factor $2$ out of $2$ .

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

Step 9.2.2

Cancel the common factors.

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Step 9.2.2.1

Factor $2$ out of $4$ .

$\frac{2 \cdot 1}{2 \cdot 2}$

Step 9.2.2.2

Cancel the common factor.

$\frac{2 \cdot 1}{2 \cdot 2}$

Step 9.2.2.3

Rewrite the expression.

$\frac{1}{2}$

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.

$f (1) = (1) - 2 \ln ((1) + 1)$

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

Add $1$ and $1$ .

$f (1) = 1 - 2 \ln (2)$

Step 11.2.1.2

Simplify $- 2 \ln (2)$ by moving $2$ inside the logarithm.

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

Step 11.2.1.3

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

$f (1) = 1 - \ln (4)$

Step 11.2.2

The final answer is $1 - \ln (4)$ .

$y = 1 - \ln (4)$

Step 12

These are the local extrema for $f (x) = x - 2 \ln (x + 1)$ .

$(1, 1 - \ln (4))$ is a local minima

Step 13