Calculus Examples

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

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 $x + 1 - \sqrt{x^{2} + x}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1] + \frac{d}{d x} [- \sqrt{x^{2} + x}]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [1] + \frac{d}{d x} [- \sqrt{x^{2} + x}]$

Step 1.1.2

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

$1 + \frac{d}{d x} [1] + \frac{d}{d x} [- \sqrt{x^{2} + x}]$

Step 1.1.3

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

$1 + 0 + \frac{d}{d x} [- \sqrt{x^{2} + x}]$

Step 1.2

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

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Step 1.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} + x}$ as ${(x^{2} + x)}^{\frac{1}{2}}$ .

$1 + 0 + \frac{d}{d x} [- {(x^{2} + x)}^{\frac{1}{2}}]$

Step 1.2.2

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

$1 + 0 - \frac{d}{d x} [{(x^{2} + x)}^{\frac{1}{2}}]$

Step 1.2.3

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

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Step 1.2.3.1

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

$1 + 0 - (\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [x^{2} + x])$

Step 1.2.3.2

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

$1 + 0 - (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} + x])$

Step 1.2.3.3

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

$1 + 0 - (\frac{1}{2} {(x^{2} + x)}^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} + x])$

Step 1.2.4

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

$1 + 0 - (\frac{1}{2} {(x^{2} + x)}^{\frac{1}{2} - 1} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [x]))$

Step 1.2.5

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

$1 + 0 - (\frac{1}{2} {(x^{2} + x)}^{\frac{1}{2} - 1} (2 x + \frac{d}{d x} [x]))$

Step 1.2.6

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

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

Step 1.2.7

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 1.2.8

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

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

Step 1.2.9

Combine the numerators over the common denominator.

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

Step 1.2.10

Simplify the numerator.

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Step 1.2.10.1

Multiply $- 1$ by $2$ .

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

Step 1.2.10.2

Subtract $2$ from $1$ .

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

Step 1.2.11

Move the negative in front of the fraction.

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

Step 1.2.12

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

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

Step 1.2.13

Move ${(x^{2} + x)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.3

Simplify.

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Step 1.3.1

Add $1$ and $0$ .

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

Step 1.3.2

Reorder terms.

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

Step 2

Find the second derivative of the function.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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Step 2.2.5.1

To apply the Chain Rule, set $u_{1}$ as ${(x^{2} + x)}^{\frac{1}{2}}$ .

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

Step 2.2.5.2

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

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

Step 2.2.5.3

Replace all occurrences of $u_{1}$ with ${(x^{2} + x)}^{\frac{1}{2}}$ .

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

Step 2.2.6

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

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Step 2.2.6.1

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

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

Step 2.2.6.2

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

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

Step 2.2.6.3

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

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

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

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

Step 2.2.10

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

Step 2.2.11

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

Step 2.2.12

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

Step 2.2.13

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

Step 2.2.14

Multiply the exponents in ${({(x^{2} + x)}^{\frac{1}{2}})}^{- 2}$ .

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Step 2.2.14.1

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

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

Step 2.2.14.2

Cancel the common factor of $2$ .

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Step 2.2.14.2.1

Factor $2$ out of $- 2$ .

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

Step 2.2.14.2.2

Cancel the common factor.

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

Step 2.2.14.2.3

Rewrite the expression.

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

Step 2.2.15

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 2.2.16

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

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

Step 2.2.17

Combine the numerators over the common denominator.

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

Step 2.2.18

Simplify the numerator.

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Step 2.2.18.1

Multiply $- 1$ by $2$ .

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

Step 2.2.18.2

Subtract $2$ from $1$ .

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

Step 2.2.19

Move the negative in front of the fraction.

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

Step 2.2.20

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

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

Step 2.2.21

Move ${(x^{2} + x)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.2.22

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

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

Step 2.2.23

Move ${(x^{2} + x)}^{- 1}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.2.24

Multiply ${(x^{2} + x)}^{\frac{1}{2}}$ by $x^{2} + x$ by adding the exponents.

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Step 2.2.24.1

Move $x^{2} + x$ .

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

Step 2.2.24.2

Multiply $x^{2} + x$ by ${(x^{2} + x)}^{\frac{1}{2}}$ .

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Step 2.2.24.2.1

Raise $x^{2} + x$ to the power of $1$ .

Step 2.2.24.2.2

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

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

Step 2.2.24.3

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

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

Step 2.2.24.4

Combine the numerators over the common denominator.

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

Step 2.2.24.5

Add $2$ and $1$ .

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

Step 2.2.25

Multiply $\frac{1}{2}$ by $\frac{1}{2 {(x^{2} + x)}^{\frac{3}{2}}}$ .

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

Step 2.2.26

Multiply $2$ by $2$ .

$f'' (x) = - ((2 x + 1) (- \frac{1}{4 {(x^{2} + x)}^{\frac{3}{2}}} \cdot (2 x + 1)) + \frac{1}{2 {(x^{2} + x)}^{\frac{1}{2}}} \cdot (2 \cdot 1 + 0)) + \frac{d}{d x} (1)$

Step 2.2.27

Raise $2 x + 1$ to the power of $1$ .

$f'' (x) = - (- \frac{1}{4 {(x^{2} + x)}^{\frac{3}{2}}} \cdot ((2 x + 1) (2 x + 1)) + \frac{1}{2 {(x^{2} + x)}^{\frac{1}{2}}} \cdot (2 \cdot 1 + 0)) + \frac{d}{d x} (1)$

Step 2.2.28

Raise $2 x + 1$ to the power of $1$ .

$f'' (x) = - (- \frac{1}{4 {(x^{2} + x)}^{\frac{3}{2}}} \cdot ((2 x + 1) (2 x + 1)) + \frac{1}{2 {(x^{2} + x)}^{\frac{1}{2}}} \cdot (2 \cdot 1 + 0)) + \frac{d}{d x} (1)$

Step 2.2.29

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

$f'' (x) = - (- \frac{1}{4 {(x^{2} + x)}^{\frac{3}{2}}} \cdot {(2 x + 1)}^{1 + 1} + \frac{1}{2 {(x^{2} + x)}^{\frac{1}{2}}} \cdot (2 \cdot 1 + 0)) + \frac{d}{d x} (1)$

Step 2.2.30

Add $1$ and $1$ .

$f'' (x) = - (- \frac{1}{4 {(x^{2} + x)}^{\frac{3}{2}}} \cdot {(2 x + 1)}^{2} + \frac{1}{2 {(x^{2} + x)}^{\frac{1}{2}}} \cdot (2 \cdot 1 + 0)) + \frac{d}{d x} (1)$

Step 2.2.31

Combine ${(2 x + 1)}^{2}$ and $\frac{1}{4 {(x^{2} + x)}^{\frac{3}{2}}}$ .

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{1}{2 {(x^{2} + x)}^{\frac{1}{2}}} \cdot (2 \cdot 1 + 0)) + \frac{d}{d x} (1)$

Step 2.2.32

Multiply $2$ by $1$ .

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{1}{2 {(x^{2} + x)}^{\frac{1}{2}}} \cdot (2 + 0)) + \frac{d}{d x} (1)$

Step 2.2.33

Add $2$ and $0$ .

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{1}{2 {(x^{2} + x)}^{\frac{1}{2}}} \cdot 2) + \frac{d}{d x} (1)$

Step 2.2.34

Combine $\frac{1}{2 {(x^{2} + x)}^{\frac{1}{2}}}$ and $2$ .

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{2}{2 {(x^{2} + x)}^{\frac{1}{2}}}) + \frac{d}{d x} (1)$

Step 2.2.35

Cancel the common factor.

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{2}{2 {(x^{2} + x)}^{\frac{1}{2}}}) + \frac{d}{d x} (1)$

Step 2.2.36

Rewrite the expression.

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{1}{{(x^{2} + x)}^{\frac{1}{2}}}) + \frac{d}{d x} (1)$

Step 2.2.37

To write $\frac{1}{{(x^{2} + x)}^{\frac{1}{2}}}$ as a fraction with a common denominator, multiply by $\frac{4 {(x^{2} + x)}^{\frac{2}{2}}}{4 {(x^{2} + x)}^{\frac{2}{2}}}$ .

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{1}{{(x^{2} + x)}^{\frac{1}{2}}} \cdot \frac{4 {(x^{2} + x)}^{\frac{2}{2}}}{4 {(x^{2} + x)}^{\frac{2}{2}}}) + \frac{d}{d x} (1)$

Step 2.2.38

Write each expression with a common denominator of $4 {(x^{2} + x)}^{\frac{3}{2}}$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.2.38.1

Multiply $\frac{1}{{(x^{2} + x)}^{\frac{1}{2}}}$ by $\frac{4 {(x^{2} + x)}^{\frac{2}{2}}}{4 {(x^{2} + x)}^{\frac{2}{2}}}$ .

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{4 {(x^{2} + x)}^{\frac{2}{2}}}{{(x^{2} + x)}^{\frac{1}{2}} (4 {(x^{2} + x)}^{\frac{2}{2}})}) + \frac{d}{d x} (1)$

Step 2.2.38.2

Multiply ${(x^{2} + x)}^{\frac{1}{2}}$ by ${(x^{2} + x)}^{\frac{2}{2}}$ by adding the exponents.

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Step 2.2.38.2.1

Move ${(x^{2} + x)}^{\frac{2}{2}}$ .

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{4 {(x^{2} + x)}^{\frac{2}{2}}}{{(x^{2} + x)}^{\frac{2}{2}} {(x^{2} + x)}^{\frac{1}{2}} \cdot 4}) + \frac{d}{d x} (1)$

Step 2.2.38.2.2

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

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{4 {(x^{2} + x)}^{\frac{2}{2}}}{{(x^{2} + x)}^{\frac{2}{2} + \frac{1}{2}} \cdot 4}) + \frac{d}{d x} (1)$

Step 2.2.38.2.3

Combine the numerators over the common denominator.

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{4 {(x^{2} + x)}^{\frac{2}{2}}}{{(x^{2} + x)}^{\frac{2 + 1}{2}} \cdot 4}) + \frac{d}{d x} (1)$

Step 2.2.38.2.4

Add $2$ and $1$ .

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{4 {(x^{2} + x)}^{\frac{2}{2}}}{{(x^{2} + x)}^{\frac{3}{2}} \cdot 4}) + \frac{d}{d x} (1)$

Step 2.2.38.3

Reorder the factors of ${(x^{2} + x)}^{\frac{3}{2}} \cdot 4$ .

$f'' (x) = - (- \frac{{(2 x + 1)}^{2}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{4 {(x^{2} + x)}^{\frac{2}{2}}}{4 {(x^{2} + x)}^{\frac{3}{2}}}) + \frac{d}{d x} (1)$

Step 2.2.39

Combine the numerators over the common denominator.

$f'' (x) = - \frac{- {(2 x + 1)}^{2} + 4 {(x^{2} + x)}^{\frac{2}{2}}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{d}{d x} (1)$

Step 2.2.40

Cancel the common factor of $2$ .

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Step 2.2.40.1

Cancel the common factor.

$f'' (x) = - \frac{- {(2 x + 1)}^{2} + 4 {(x^{2} + x)}^{\frac{2}{2}}}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{d}{d x} (1)$

Step 2.2.40.2

Rewrite the expression.

$f'' (x) = - \frac{- {(2 x + 1)}^{2} + 4 (x^{2} + x)}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{d}{d x} (1)$

Step 2.2.41

Simplify.

$f'' (x) = - \frac{- {(2 x + 1)}^{2} + 4 (x^{2} + x)}{4 {(x^{2} + x)}^{\frac{3}{2}}} + \frac{d}{d x} (1)$

Step 2.3

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

$f'' (x) = - \frac{- {(2 x + 1)}^{2} + 4 (x^{2} + x)}{4 {(x^{2} + x)}^{\frac{3}{2}}} + 0$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$f'' (x) = - \frac{- {(2 x + 1)}^{2} + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}} + 0$

Step 2.4.2

Add $- \frac{- {(2 x + 1)}^{2} + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$ and $0$ .

$f'' (x) = - \frac{- {(2 x + 1)}^{2} + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3

Simplify the numerator.

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Step 2.4.3.1

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

$f'' (x) = - \frac{- ((2 x + 1) (2 x + 1)) + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.2

Expand $(2 x + 1) (2 x + 1)$ using the FOIL Method.

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Step 2.4.3.2.1

Apply the distributive property.

$f'' (x) = - \frac{- (2 x (2 x + 1) + 1 (2 x + 1)) + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.2.2

Apply the distributive property.

$f'' (x) = - \frac{- (2 x (2 x) + 2 x \cdot 1 + 1 (2 x + 1)) + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.2.3

Apply the distributive property.

$f'' (x) = - \frac{- (2 x (2 x) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1) + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.3

Simplify and combine like terms.

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Step 2.4.3.3.1

Simplify each term.

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Step 2.4.3.3.1.1

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{- (2 \cdot (2 x \cdot x) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1) + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.3.1.2

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

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Step 2.4.3.3.1.2.1

Move $x$ .

$f'' (x) = - \frac{- (2 \cdot (2 (x \cdot x)) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1) + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.3.1.2.2

Multiply $x$ by $x$ .

$f'' (x) = - \frac{- (2 \cdot (2 x^{2}) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1) + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.3.1.3

Multiply $2$ by $2$ .

$f'' (x) = - \frac{- (4 x^{2} + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1) + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.3.1.4

Multiply $2$ by $1$ .

$f'' (x) = - \frac{- (4 x^{2} + 2 x + 1 (2 x) + 1 \cdot 1) + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.3.1.5

Multiply $2 x$ by $1$ .

$f'' (x) = - \frac{- (4 x^{2} + 2 x + 2 x + 1 \cdot 1) + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.3.1.6

Multiply $1$ by $1$ .

$f'' (x) = - \frac{- (4 x^{2} + 2 x + 2 x + 1) + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.3.2

Add $2 x$ and $2 x$ .

$f'' (x) = - \frac{- (4 x^{2} + 4 x + 1) + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.4

Apply the distributive property.

$f'' (x) = - \frac{- (4 x^{2}) - (4 x) - 1 \cdot 1 + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.5

Simplify.

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Step 2.4.3.5.1

Multiply $4$ by $- 1$ .

$f'' (x) = - \frac{- 4 x^{2} - (4 x) - 1 \cdot 1 + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.5.2

Multiply $4$ by $- 1$ .

$f'' (x) = - \frac{- 4 x^{2} - 4 x - 1 \cdot 1 + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.5.3

Multiply $- 1$ by $1$ .

$f'' (x) = - \frac{- 4 x^{2} - 4 x - 1 + 4 x^{2} + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.6

Add $- 4 x^{2}$ and $4 x^{2}$ .

$f'' (x) = - \frac{- 4 x - 1 + 0 + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.7

Add $- 4 x$ and $0$ .

$f'' (x) = - \frac{- 4 x - 1 + 4 x}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.8

Add $- 4 x$ and $4 x$ .

$f'' (x) = - \frac{0 - 1}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.3.9

Subtract $1$ from $0$ .

$f'' (x) = - \frac{- 1}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.4

Move the negative in front of the fraction.

$f'' (x) = \frac{1}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 2.4.5

Multiply $- - \frac{1}{4 {(x^{2} + x)}^{\frac{3}{2}}}$ .

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Step 2.4.5.1

Multiply $- 1$ by $- 1$ .

$f'' (x) = 1 (\frac{1}{4 {(x^{2} + x)}^{\frac{3}{2}}})$

Step 2.4.5.2

Multiply $\frac{1}{4 {(x^{2} + x)}^{\frac{3}{2}}}$ by $1$ .

$f'' (x) = \frac{1}{4 {(x^{2} + x)}^{\frac{3}{2}}}$

Step 3

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

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

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 $x + 1 - \sqrt{x^{2} + x}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1] + \frac{d}{d x} [- \sqrt{x^{2} + x}]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [1] + \frac{d}{d x} [- \sqrt{x^{2} + x}]$

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 $n = 1$ .

$1 + \frac{d}{d x} [1] + \frac{d}{d x} [- \sqrt{x^{2} + x}]$

Step 4.1.1.3

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

$1 + 0 + \frac{d}{d x} [- \sqrt{x^{2} + x}]$

Step 4.1.2

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

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Step 4.1.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} + x}$ as ${(x^{2} + x)}^{\frac{1}{2}}$ .

$1 + 0 + \frac{d}{d x} [- {(x^{2} + x)}^{\frac{1}{2}}]$

Step 4.1.2.2

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

$1 + 0 - \frac{d}{d x} [{(x^{2} + x)}^{\frac{1}{2}}]$

Step 4.1.2.3

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

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Step 4.1.2.3.1

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

$1 + 0 - (\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [x^{2} + x])$

Step 4.1.2.3.2

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

$1 + 0 - (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} + x])$

Step 4.1.2.3.3

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

$1 + 0 - (\frac{1}{2} {(x^{2} + x)}^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} + x])$

Step 4.1.2.4

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

$1 + 0 - (\frac{1}{2} {(x^{2} + x)}^{\frac{1}{2} - 1} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [x]))$

Step 4.1.2.5

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

$1 + 0 - (\frac{1}{2} {(x^{2} + x)}^{\frac{1}{2} - 1} (2 x + \frac{d}{d x} [x]))$

Step 4.1.2.6

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

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

Step 4.1.2.7

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 4.1.2.8

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

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

Step 4.1.2.9

Combine the numerators over the common denominator.

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

Step 4.1.2.10

Simplify the numerator.

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Step 4.1.2.10.1

Multiply $- 1$ by $2$ .

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

Step 4.1.2.10.2

Subtract $2$ from $1$ .

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

Step 4.1.2.11

Move the negative in front of the fraction.

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

Step 4.1.2.12

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

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

Step 4.1.2.13

Move ${(x^{2} + x)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 4.1.3

Simplify.

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Step 4.1.3.1

Add $1$ and $0$ .

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

Step 4.1.3.2

Reorder terms.

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

Step 4.2

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

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

Step 5

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

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Step 5.1

Set the first derivative equal to $0$ .

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

Step 5.2

Simplify $- (2 x + 1) \frac{1}{2 {(x^{2} + x)}^{\frac{1}{2}}} + 1$ .

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Apply the distributive property.

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

Step 5.2.1.2

Multiply $2$ by $- 1$ .

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

Step 5.2.1.3

Multiply $- 1$ by $1$ .

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

Step 5.2.1.4

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

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

Step 5.2.2

Simplify terms.

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Step 5.2.2.1

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

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

Step 5.2.2.2

Combine the numerators over the common denominator.

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

Step 5.2.2.3

Reorder terms.

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

Step 5.2.2.4

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

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

Step 5.2.2.5

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

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

Step 5.2.2.6

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

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

Step 5.2.2.7

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

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

Step 5.2.2.8

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

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

Step 5.2.2.9

Simplify the expression.

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Step 5.2.2.9.1

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

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

Step 5.2.2.9.2

Move the negative in front of the fraction.

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

Step 5.3

Graph each side of the equation. The solution is the x-value of the point of intersection.

No solution

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Convert expressions with fractional exponents to radicals.

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Step 6.1.1

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

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

Step 6.1.2

Anything raised to $1$ is the base itself.

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

Step 6.2

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

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

Step 6.3

Solve for $x$ .

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Step 6.3.1

To remove the radical on the left side of the equation, square both sides of the equation.

${(2 \sqrt{x^{2} + x})}^{2} = 0^{2}$

Step 6.3.2

Simplify each side of the equation.

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Step 6.3.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} + x}$ as ${(x^{2} + x)}^{\frac{1}{2}}$ .

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

Step 6.3.2.2

Simplify the left side.

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Step 6.3.2.2.1

Simplify ${(2 {(x^{2} + x)}^{\frac{1}{2}})}^{2}$ .

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Step 6.3.2.2.1.1

Apply the product rule to $2 {(x^{2} + x)}^{\frac{1}{2}}$ .

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

Step 6.3.2.2.1.2

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

$4 {({(x^{2} + x)}^{\frac{1}{2}})}^{2} = 0^{2}$

Step 6.3.2.2.1.3

Multiply the exponents in ${({(x^{2} + x)}^{\frac{1}{2}})}^{2}$ .

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Step 6.3.2.2.1.3.1

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

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

Step 6.3.2.2.1.3.2

Cancel the common factor of $2$ .

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Step 6.3.2.2.1.3.2.1

Cancel the common factor.

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

Step 6.3.2.2.1.3.2.2

Rewrite the expression.

$4 {(x^{2} + x)}^{1} = 0^{2}$

Step 6.3.2.2.1.4

Simplify.

$4 (x^{2} + x) = 0^{2}$

Step 6.3.2.2.1.5

Apply the distributive property.

$4 x^{2} + 4 x = 0^{2}$

Step 6.3.2.3

Simplify the right side.

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Step 6.3.2.3.1

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

$4 x^{2} + 4 x = 0$

Step 6.3.3

Solve for $x$ .

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Step 6.3.3.1

Factor $4 x$ out of $4 x^{2} + 4 x$ .

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Step 6.3.3.1.1

Factor $4 x$ out of $4 x^{2}$ .

$4 x (x) + 4 x = 0$

Step 6.3.3.1.2

Factor $4 x$ out of $4 x$ .

$4 x (x) + 4 x (1) = 0$

Step 6.3.3.1.3

Factor $4 x$ out of $4 x (x) + 4 x (1)$ .

$4 x (x + 1) = 0$

Step 6.3.3.2

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

$x = 0$

$x + 1 = 0$

Step 6.3.3.3

Set $x$ equal to $0$ .

$x = 0$

Step 6.3.3.4

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

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Step 6.3.3.4.1

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

$x + 1 = 0$

Step 6.3.3.4.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6.3.3.5

The final solution is all the values that make $4 x (x + 1) = 0$ true.

$x = 0, - 1$

Step 6.4

Set the radicand in $\sqrt{x^{2} + x}$ less than $0$ to find where the expression is undefined.

$x^{2} + x < 0$

Step 6.5

Solve for $x$ .

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Step 6.5.1

Convert the inequality to an equation.

$x^{2} + x = 0$

Step 6.5.2

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

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Step 6.5.2.1

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

$x \cdot x + x = 0$

Step 6.5.2.2

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

$x \cdot x + x = 0$

Step 6.5.2.3

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

$x \cdot x + x \cdot 1 = 0$

Step 6.5.2.4

Factor $x$ out of $x \cdot x + x \cdot 1$ .

$x (x + 1) = 0$

Step 6.5.3

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

$x = 0$

$x + 1 = 0$

Step 6.5.4

Set $x$ equal to $0$ .

$x = 0$

Step 6.5.5

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

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Step 6.5.5.1

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

$x + 1 = 0$

Step 6.5.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6.5.6

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

$x = 0, - 1$

Step 6.5.7

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 0$

$x > 0$

Step 6.5.8

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 6.5.8.1

Test a value on the interval $x < - 1$ to see if it makes the inequality true.

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Step 6.5.8.1.1

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 6.5.8.1.2

Replace $x$ with $- 4$ in the original inequality.

${(- 4)}^{2} - 4 < 0$

Step 6.5.8.1.3

The left side $12$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 6.5.8.2

Test a value on the interval $- 1 < x < 0$ to see if it makes the inequality true.

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Step 6.5.8.2.1

Choose a value on the interval $- 1 < x < 0$ and see if this value makes the original inequality true.

$x = - 0.5$

Step 6.5.8.2.2

Replace $x$ with $- 0.5$ in the original inequality.

${(- 0.5)}^{2} - 0.5 < 0$

Step 6.5.8.2.3

The left side $- 0.25$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 6.5.8.3

Test a value on the interval $x > 0$ to see if it makes the inequality true.

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Step 6.5.8.3.1

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$

Step 6.5.8.3.2

Replace $x$ with $2$ in the original inequality.

${(2)}^{2} + 2 < 0$

Step 6.5.8.3.3

The left side $6$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 6.5.8.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 1$ False

$- 1 < x < 0$ True

$x > 0$ False

$x < - 1$ False

$- 1 < x < 0$ True

$x > 0$ False

Step 6.5.9

The solution consists of all of the true intervals.

$- 1 < x < 0$

Step 6.6

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$- 1 \leq x \leq 0$

$[- 1, 0]$

$- 1 \leq x \leq 0$

$[- 1, 0]$

Step 7

Critical points to evaluate.

$x = - 1, 0$

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.

$\frac{1}{4 {({(- 1)}^{2} - 1)}^{\frac{3}{2}}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the expression.

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Step 9.1.1

Remove parentheses.

$\frac{1}{4 {({(- 1)}^{2} - 1)}^{\frac{3}{2}}}$

Step 9.1.2

Raise $- 1$ to the power of $2$ .

$\frac{1}{4 {(1 - 1)}^{\frac{3}{2}}}$

Step 9.1.3

Subtract $1$ from $1$ .

$\frac{1}{4 \cdot 0^{\frac{3}{2}}}$

Step 9.1.4

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

$\frac{1}{4 \cdot {(0^{2})}^{\frac{3}{2}}}$

Step 9.1.5

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

$\frac{1}{4 \cdot 0^{2 (\frac{3}{2})}}$

Step 9.2

Cancel the common factor of $2$ .

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Step 9.2.1

Cancel the common factor.

$\frac{1}{4 \cdot 0^{2 (\frac{3}{2})}}$

Step 9.2.2

Rewrite the expression.

$\frac{1}{4 \cdot 0^{3}}$

Step 9.3

Simplify the expression.

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Step 9.3.1

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

$\frac{1}{4 \cdot 0}$

Step 9.3.2

Multiply $4$ by $0$ .

$\frac{1}{0}$

Step 9.3.3

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

Undefined

$\frac{1}{0}$

Step 9.4

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

Undefined

Step 10

Since the first derivative test failed, there are no local extrema.

No Local Extrema

Step 11