Calculus Examples

$f (x) = x \cdot | x - 1 | + x + 1$

Step 1

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

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

Step 2

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

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

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

Step 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 $g (x) = x - 1$ .

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

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

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

Step 2.7

Add $1$ and $0$ .

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

Step 2.8

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

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

Step 2.9

Combine $x$ and $\frac{x - 1}{| x - 1 |}$ .

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

Step 2.10

Multiply $| x - 1 |$ by $1$ .

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

Step 2.11

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

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

Step 2.12

Combine the numerators over the common denominator.

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

Step 2.13

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

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

Step 2.14

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

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

Step 2.15

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

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

Step 2.16

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

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

Step 2.17

Add $1$ and $1$ .

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

Step 3

Differentiate.

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Step 3.1

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

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

Step 3.2

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

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

Step 4

Simplify.

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Step 4.1

Apply the distributive property.

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

Step 4.2

Combine terms.

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Step 4.2.1

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

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

Step 4.2.2

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

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

Step 4.2.3

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

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

Step 4.2.4

Add $1$ and $1$ .

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

Step 4.2.5

Move $- 1$ to the left of $x$ .

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

Step 4.2.6

Rewrite $- 1 x$ as $- x$ .

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

Step 4.2.7

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

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

Step 4.2.8

Combine the numerators over the common denominator.

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

Step 4.2.9

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

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