Calculus Examples

$f (x) = - | x + 1 | + 6$

Step 1

Find the first derivative of the function.

Tap for more steps...

Step 1.1

By the Sum Rule, the derivative of $- | x + 1 | + 6$ with respect to $x$ is $\frac{d}{d x} [- | x + 1 |] + \frac{d}{d x} [6]$ .

$\frac{d}{d x} [- | x + 1 |] + \frac{d}{d x} [6]$

Step 1.2

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

Tap for more steps...

Step 1.2.1

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

$- \frac{d}{d x} [| x + 1 |] + \frac{d}{d x} [6]$

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

Tap for more steps...

Step 1.2.2.1

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

$- (\frac{d}{d u} [| u |] \frac{d}{d x} [x + 1]) + \frac{d}{d x} [6]$

Step 1.2.2.2

The derivative of $| u |$ with respect to $u$ is $\frac{u}{| u |}$ .

$- (\frac{u}{| u |} \frac{d}{d x} [x + 1]) + \frac{d}{d x} [6]$

Step 1.2.2.3

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

$- (\frac{x + 1}{| x + 1 |} \frac{d}{d x} [x + 1]) + \frac{d}{d x} [6]$

Step 1.2.3

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

$- (\frac{x + 1}{| x + 1 |} (\frac{d}{d x} [x] + \frac{d}{d x} [1])) + \frac{d}{d x} [6]$

Step 1.2.4

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

$- (\frac{x + 1}{| x + 1 |} (1 + \frac{d}{d x} [1])) + \frac{d}{d x} [6]$

Step 1.2.5

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

$- (\frac{x + 1}{| x + 1 |} (1 + 0)) + \frac{d}{d x} [6]$

Step 1.2.6

Add $1$ and $0$ .

$- (\frac{x + 1}{| x + 1 |} \cdot 1) + \frac{d}{d x} [6]$

Step 1.2.7

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

$- \frac{x + 1}{| x + 1 |} + \frac{d}{d x} [6]$

Step 1.3

Differentiate using the Constant Rule.

Tap for more steps...

Step 1.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

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

Step 1.3.2

Add $- \frac{x + 1}{| x + 1 |}$ and $0$ .

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

Step 2

Find the second derivative of the function.

Tap for more steps...

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

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

Step 2.2

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 $g (x) = | x + 1 |$ .

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

Step 2.3

Differentiate.

Tap for more steps...

Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Simplify the expression.

Tap for more steps...

Step 2.3.4.1

Add $1$ and $0$ .

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

Step 2.3.4.2

Multiply $| x + 1 |$ by $1$ .

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

Step 2.4

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

Tap for more steps...

Step 2.4.1

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

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

Step 2.4.2

The derivative of $| u |$ with respect to $u$ is $\frac{u}{| u |}$ .

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

Step 2.4.3

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

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

Step 2.5

Differentiate.

Tap for more steps...

Step 2.5.1

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

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

Step 2.5.2

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

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

Step 2.5.3

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

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

Step 2.5.4

Simplify the expression.

Tap for more steps...

Step 2.5.4.1

Add $1$ and $0$ .

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

Step 2.5.4.2

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

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

Step 2.5.5

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

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

Step 2.5.6

Simplify the expression.

Tap for more steps...

Step 2.5.6.1

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

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

Step 2.5.6.2

Add $- \frac{| x + 1 | - (x + 1) \frac{x + 1}{| x + 1 |}}{{| x + 1 |}^{2}}$ and $0$ .

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

Step 2.6

Simplify.

Tap for more steps...

Step 2.6.1

Apply the distributive property.

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

Step 2.6.2

Simplify the numerator.

Tap for more steps...

Step 2.6.2.1

Simplify each term.

Tap for more steps...

Step 2.6.2.1.1

Multiply $- 1$ by $1$ .

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

Step 2.6.2.1.2

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

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

Step 2.6.2.1.3

Simplify the numerator.

Tap for more steps...

Step 2.6.2.1.3.1

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

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

Step 2.6.2.1.3.2

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 2.6.2.1.3.3

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

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

Step 2.6.2.1.3.4

Rewrite $- (x + 1)$ as $- 1 (x + 1)$ .

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

Step 2.6.2.1.3.5

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

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

Step 2.6.2.1.3.6

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

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

Step 2.6.2.1.3.7

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

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

Step 2.6.2.1.3.8

Add $1$ and $1$ .

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

Step 2.6.2.1.4

Move the negative in front of the fraction.

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

Step 2.6.2.2

To write $| x + 1 |$ as a fraction with a common denominator, multiply by $\frac{| x + 1 |}{| x + 1 |}$ .

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

Step 2.6.2.3

Combine the numerators over the common denominator.

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

Step 2.6.2.4

Simplify the numerator.

Tap for more steps...

Step 2.6.2.4.1

Multiply $| x + 1 | | x + 1 |$ .

Tap for more steps...

Step 2.6.2.4.1.1

To multiply absolute values, multiply the terms inside each absolute value.

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

Step 2.6.2.4.1.2

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

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

Step 2.6.2.4.1.3

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

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

Step 2.6.2.4.1.4

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

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

Step 2.6.2.4.1.5

Add $1$ and $1$ .

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

Step 2.6.2.4.2

Rewrite ${(x + 1)}^{2}$ as $(x + 1) (x + 1)$ .

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

Step 2.6.2.4.3

Expand $(x + 1) (x + 1)$ using the FOIL Method.

Tap for more steps...

Step 2.6.2.4.3.1

Apply the distributive property.

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

Step 2.6.2.4.3.2

Apply the distributive property.

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

Step 2.6.2.4.3.3

Apply the distributive property.

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

Step 2.6.2.4.4

Simplify and combine like terms.

Tap for more steps...

Step 2.6.2.4.4.1

Simplify each term.

Tap for more steps...

Step 2.6.2.4.4.1.1

Multiply $x$ by $x$ .

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

Step 2.6.2.4.4.1.2

Multiply $x$ by $1$ .

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

Step 2.6.2.4.4.1.3

Multiply $x$ by $1$ .

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

Step 2.6.2.4.4.1.4

Multiply $1$ by $1$ .

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

Step 2.6.2.4.4.2

Add $x$ and $x$ .

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

Step 2.6.2.4.5

Rewrite ${(x + 1)}^{2}$ as $(x + 1) (x + 1)$ .

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

Step 2.6.2.4.6

Expand $(x + 1) (x + 1)$ using the FOIL Method.

Tap for more steps...

Step 2.6.2.4.6.1

Apply the distributive property.

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

Step 2.6.2.4.6.2

Apply the distributive property.

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

Step 2.6.2.4.6.3

Apply the distributive property.

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

Step 2.6.2.4.7

Simplify and combine like terms.

Tap for more steps...

Step 2.6.2.4.7.1

Simplify each term.

Tap for more steps...

Step 2.6.2.4.7.1.1

Multiply $x$ by $x$ .

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

Step 2.6.2.4.7.1.2

Multiply $x$ by $1$ .

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

Step 2.6.2.4.7.1.3

Multiply $x$ by $1$ .

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

Step 2.6.2.4.7.1.4

Multiply $1$ by $1$ .

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

Step 2.6.2.4.7.2

Add $x$ and $x$ .

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

Step 2.6.2.4.8

Apply the distributive property.

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

Step 2.6.2.4.9

Simplify.

Tap for more steps...

Step 2.6.2.4.9.1

Multiply $2$ by $- 1$ .

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

Step 2.6.2.4.9.2

Multiply $- 1$ by $1$ .

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

Step 2.6.2.4.10

Reorder terms.

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

Step 2.6.2.4.11

Rewrite $- x^{2} - 2 x + | x^{2} + 2 x + 1 | - 1$ in a factored form.

Tap for more steps...

Step 2.6.2.4.11.1

Regroup terms.

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

Step 2.6.2.4.11.2

Factor $- 1$ out of $- x^{2} - 2 x + | x^{2} + 2 x + 1 |$ .

Tap for more steps...

Step 2.6.2.4.11.2.1

Factor $- 1$ out of $- x^{2}$ .

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

Step 2.6.2.4.11.2.2

Factor $- 1$ out of $- 2 x$ .

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

Step 2.6.2.4.11.2.3

Factor $- 1$ out of $| x^{2} + 2 x + 1 |$ .

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

Step 2.6.2.4.11.2.4

Factor $- 1$ out of $- (x^{2}) - (2 x)$ .

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

Step 2.6.2.4.11.2.5

Factor $- 1$ out of $- (x^{2} + 2 x) - 1 (- | x^{2} + 2 x + 1 |)$ .

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

Step 2.6.2.4.11.3

Factor $- 1$ out of $- 1 - (x^{2} + 2 x - | x^{2} + 2 x + 1 |)$ .

Tap for more steps...

Step 2.6.2.4.11.3.1

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 2.6.2.4.11.3.2

Factor $- 1$ out of $- 1 (1) - (x^{2} + 2 x - | x^{2} + 2 x + 1 |)$ .

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

Step 2.6.2.4.11.3.3

Rewrite $- 1 (1 + x^{2} + 2 x - | x^{2} + 2 x + 1 |)$ as $- (1 + x^{2} + 2 x - | x^{2} + 2 x + 1 |)$ .

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

Step 2.6.2.4.11.4

Reorder terms.

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

Step 2.6.2.5

Move the negative in front of the fraction.

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

Step 2.6.3

Combine terms.

Tap for more steps...

Step 2.6.3.1

Rewrite $\frac{- \frac{x^{2} + 2 x + 1 - | x^{2} + 2 x + 1 |}{| x + 1 |}}{{| x + 1 |}^{2}}$ as a product.

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

Step 2.6.3.2

Multiply $\frac{1}{{| x + 1 |}^{2}}$ by $\frac{x^{2} + 2 x + 1 - | x^{2} + 2 x + 1 |}{| x + 1 |}$ .

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

Step 2.6.3.3

Multiply ${| x + 1 |}^{2}$ by $| x + 1 |$ by adding the exponents.

Tap for more steps...

Step 2.6.3.3.1

Multiply ${| x + 1 |}^{2}$ by $| x + 1 |$ .

Tap for more steps...

Step 2.6.3.3.1.1

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

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

Step 2.6.3.3.1.2

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

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

Step 2.6.3.3.2

Add $2$ and $1$ .

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

Step 2.6.3.4

Multiply $- 1$ by $- 1$ .

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

Step 2.6.3.5

Multiply $\frac{x^{2} + 2 x + 1 - | x^{2} + 2 x + 1 |}{{| x + 1 |}^{3}}$ by $1$ .

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

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.

Tap for more steps...

Step 4.1

Find the first derivative.

Tap for more steps...

Step 4.1.1

By the Sum Rule, the derivative of $- | x + 1 | + 6$ with respect to $x$ is $\frac{d}{d x} [- | x + 1 |] + \frac{d}{d x} [6]$ .

$\frac{d}{d x} [- | x + 1 |] + \frac{d}{d x} [6]$

Step 4.1.2

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

Tap for more steps...

Step 4.1.2.1

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

$- \frac{d}{d x} [| x + 1 |] + \frac{d}{d x} [6]$

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

Tap for more steps...

Step 4.1.2.2.1

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

$- (\frac{d}{d u} [| u |] \frac{d}{d x} [x + 1]) + \frac{d}{d x} [6]$

Step 4.1.2.2.2

The derivative of $| u |$ with respect to $u$ is $\frac{u}{| u |}$ .

$- (\frac{u}{| u |} \frac{d}{d x} [x + 1]) + \frac{d}{d x} [6]$

Step 4.1.2.2.3

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

$- (\frac{x + 1}{| x + 1 |} \frac{d}{d x} [x + 1]) + \frac{d}{d x} [6]$

Step 4.1.2.3

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

$- (\frac{x + 1}{| x + 1 |} (\frac{d}{d x} [x] + \frac{d}{d x} [1])) + \frac{d}{d x} [6]$

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 $n = 1$ .

$- (\frac{x + 1}{| x + 1 |} (1 + \frac{d}{d x} [1])) + \frac{d}{d x} [6]$

Step 4.1.2.5

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

$- (\frac{x + 1}{| x + 1 |} (1 + 0)) + \frac{d}{d x} [6]$

Step 4.1.2.6

Add $1$ and $0$ .

$- (\frac{x + 1}{| x + 1 |} \cdot 1) + \frac{d}{d x} [6]$

Step 4.1.2.7

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

$- \frac{x + 1}{| x + 1 |} + \frac{d}{d x} [6]$

Step 4.1.3

Differentiate using the Constant Rule.

Tap for more steps...

Step 4.1.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

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

Step 4.1.3.2

Add $- \frac{x + 1}{| x + 1 |}$ and $0$ .

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

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

Tap for more steps...

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

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5.4

Exclude the solutions that do not make $- \frac{x + 1}{| x + 1 |} = 0$ true.

$No solution$

Step 6

Find the values where the derivative is undefined.

Tap for more steps...

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

Solve for $x$ .

Tap for more steps...

Step 6.2.1

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x + 1 = \pm 0$

Step 6.2.2

Plus or minus $0$ is $0$ .

$x + 1 = 0$

Step 6.2.3

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.

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

Step 9

Evaluate the second derivative.

Tap for more steps...

Step 9.1

Add $- 1$ and $1$ .

$\frac{{(- 1)}^{2} + 2 (- 1) + 1 - | {(- 1)}^{2} + 2 (- 1) + 1 |}{{| 0 |}^{3}}$

Step 9.2

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$\frac{{(- 1)}^{2} + 2 (- 1) + 1 - | {(- 1)}^{2} + 2 (- 1) + 1 |}{0^{3}}$

Step 9.3

Raising $0$ to any positive power yields $0$ .

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

Step 9.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

Step 10

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

Tap for more steps...

Step 10.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, - 1) \cup (- 1, \infty)$

Step 10.2

Substitute any number, such as $- 4$ , from the interval $(- \infty, - 1)$ in the first derivative $- \frac{x + 1}{| x + 1 |}$ to check if the result is negative or positive.

Tap for more steps...

Step 10.2.1

Replace the variable $x$ with $- 4$ in the expression.

$f' (- 4) = - \frac{(- 4) + 1}{| (- 4) + 1 |}$

Step 10.2.2

Simplify the result.

Tap for more steps...

Step 10.2.2.1

Add $- 4$ and $1$ .

$f' (- 4) = - \frac{- 3}{| - 4 + 1 |}$

Step 10.2.2.2

Simplify the denominator.

Tap for more steps...

Step 10.2.2.2.1

Add $- 4$ and $1$ .

$f' (- 4) = - \frac{- 3}{| - 3 |}$

Step 10.2.2.2.2

The absolute value is the distance between a number and zero. The distance between $- 3$ and $0$ is $3$ .

$f' (- 4) = - \frac{- 3}{3}$

Step 10.2.2.3

Simplify the expression.

Tap for more steps...

Step 10.2.2.3.1

Divide $- 3$ by $3$ .

$f' (- 4) = 1$

Step 10.2.2.3.2

Multiply $- 1$ by $- 1$ .

$f' (- 4) = 1$

Step 10.2.2.4

The final answer is $1$ .

$1$

Step 10.3

Substitute any number, such as $0$ , from the interval $(- 1, \infty)$ in the first derivative $- \frac{x + 1}{| x + 1 |}$ to check if the result is negative or positive.

Tap for more steps...

Step 10.3.1

Replace the variable $x$ with $0$ in the expression.

$f' (0) = - \frac{(0) + 1}{| (0) + 1 |}$

Step 10.3.2

Simplify the result.

Tap for more steps...

Step 10.3.2.1

Add $0$ and $1$ .

$f' (0) = - \frac{1}{| 0 + 1 |}$

Step 10.3.2.2

Simplify the denominator.

Tap for more steps...

Step 10.3.2.2.1

Add $0$ and $1$ .

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

Step 10.3.2.2.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

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

Step 10.3.2.3

Reduce the expression by cancelling the common factors.

Tap for more steps...

Step 10.3.2.3.1

Cancel the common factor of $1$ .

Tap for more steps...

Step 10.3.2.3.1.1

Cancel the common factor.

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

Step 10.3.2.3.1.2

Rewrite the expression.

$f' (0) = - 1 \cdot 1$

Step 10.3.2.3.2

Multiply $- 1$ by $1$ .

$f' (0) = - 1$

Step 10.3.2.4

The final answer is $- 1$ .

$- 1$

Step 10.4

Since the first derivative changed signs from positive to negative around $x = - 1$ , then $x = - 1$ is a local maximum.

$x = - 1$ is a local maximum

Step 11