Calculus Examples

$y = x \arctan (2 x) - \frac{1}{4} \cdot \ln (1 + 4 x^{2})$

Step 1

By the Sum Rule, the derivative of $x \arctan (2 x) - \frac{1}{4} \cdot \ln (1 + 4 x^{2})$ with respect to $x$ is $\frac{d}{d x} [x \arctan (2 x)] + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$ .

$\frac{d}{d x} [x \arctan (2 x)] + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2

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

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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 $g (x) = \arctan (2 x)$ .

$x \frac{d}{d x} [\arctan (2 x)] + \arctan (2 x) \frac{d}{d x} [x] + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

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 $f (x) = \arctan (x)$ and $g (x) = 2 x$ .

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Step 2.2.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

$x (\frac{d}{d u_{1}} [\arctan (u_{1})] \frac{d}{d x} [2 x]) + \arctan (2 x) \frac{d}{d x} [x] + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.2.2

The derivative of $\arctan (u_{1})$ with respect to $u_{1}$ is $\frac{1}{1 + {u_{1}}^{2}}$ .

$x (\frac{1}{1 + {u_{1}}^{2}} \frac{d}{d x} [2 x]) + \arctan (2 x) \frac{d}{d x} [x] + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.2.3

Replace all occurrences of $u_{1}$ with $2 x$ .

$x (\frac{1}{1 + {(2 x)}^{2}} \frac{d}{d x} [2 x]) + \arctan (2 x) \frac{d}{d x} [x] + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.3

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

$x (\frac{1}{1 + {(2 x)}^{2}} (2 \frac{d}{d x} [x])) + \arctan (2 x) \frac{d}{d x} [x] + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.4

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

$x (\frac{1}{1 + {(2 x)}^{2}} (2 \cdot 1)) + \arctan (2 x) \frac{d}{d x} [x] + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.5

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

$x (\frac{1}{1 + {(2 x)}^{2}} (2 \cdot 1)) + \arctan (2 x) \cdot 1 + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.6

Factor $2$ out of $2 x$ .

$x (\frac{1}{1 + {(2 (x))}^{2}} (2 \cdot 1)) + \arctan (2 x) \cdot 1 + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.7

Apply the product rule to $2 (x)$ .

$x (\frac{1}{1 + 2^{2} x^{2}} (2 \cdot 1)) + \arctan (2 x) \cdot 1 + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.8

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

$x (\frac{1}{1 + 4 x^{2}} (2 \cdot 1)) + \arctan (2 x) \cdot 1 + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.9

Multiply $2$ by $1$ .

$x (\frac{1}{1 + 4 x^{2}} \cdot 2) + \arctan (2 x) \cdot 1 + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.10

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

$x \frac{2}{1 + 4 x^{2}} + \arctan (2 x) \cdot 1 + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.11

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

$\frac{x \cdot 2}{1 + 4 x^{2}} + \arctan (2 x) \cdot 1 + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.12

Move $2$ to the left of $x$ .

$\frac{2 \cdot x}{1 + 4 x^{2}} + \arctan (2 x) \cdot 1 + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.13

Multiply $\arctan (2 x)$ by $1$ .

$\frac{2 x}{1 + 4 x^{2}} + \arctan (2 x) + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.14

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

$\frac{2 x}{1 + 4 x^{2}} + \frac{\arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 2.15

Combine the numerators over the common denominator.

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} + \frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$

Step 3

Evaluate $\frac{d}{d x} [- \frac{1}{4} \cdot \ln (1 + 4 x^{2})]$ .

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Step 3.1

Since $- \frac{1}{4}$ is constant with respect to $x$ , the derivative of $- \frac{1}{4} \cdot \ln (1 + 4 x^{2})$ with respect to $x$ is $- \frac{1}{4} \frac{d}{d x} [\ln (1 + 4 x^{2})]$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{1}{4} \frac{d}{d x} [\ln (1 + 4 x^{2})]$

Step 3.2

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

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Step 3.2.1

To apply the Chain Rule, set $u_{2}$ as $1 + 4 x^{2}$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{1}{4} (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [1 + 4 x^{2}])$

Step 3.2.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{1}{4} (\frac{1}{u_{2}} \frac{d}{d x} [1 + 4 x^{2}])$

Step 3.2.3

Replace all occurrences of $u_{2}$ with $1 + 4 x^{2}$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{1}{4} (\frac{1}{1 + 4 x^{2}} \frac{d}{d x} [1 + 4 x^{2}])$

Step 3.3

By the Sum Rule, the derivative of $1 + 4 x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [4 x^{2}]$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{1}{4} (\frac{1}{1 + 4 x^{2}} (\frac{d}{d x} [1] + \frac{d}{d x} [4 x^{2}]))$

Step 3.4

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

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{1}{4} (\frac{1}{1 + 4 x^{2}} (0 + \frac{d}{d x} [4 x^{2}]))$

Step 3.5

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{1}{4} (\frac{1}{1 + 4 x^{2}} (0 + 4 \frac{d}{d x} [x^{2}]))$

Step 3.6

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

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

Step 3.7

Multiply $2$ by $4$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{1}{4} (\frac{1}{1 + 4 x^{2}} (0 + 8 x))$

Step 3.8

Add $0$ and $8 x$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{1}{4} (\frac{1}{1 + 4 x^{2}} (8 x))$

Step 3.9

Combine $8$ and $\frac{1}{1 + 4 x^{2}}$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{1}{4} (\frac{8}{1 + 4 x^{2}} x)$

Step 3.10

Combine $\frac{8}{1 + 4 x^{2}}$ and $x$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{1}{4} \cdot \frac{8 x}{1 + 4 x^{2}}$

Step 3.11

Multiply $\frac{8 x}{1 + 4 x^{2}}$ by $\frac{1}{4}$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{8 x}{(1 + 4 x^{2}) \cdot 4}$

Step 3.12

Move $4$ to the left of $1 + 4 x^{2}$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{8 x}{4 \cdot (1 + 4 x^{2})}$

Step 3.13

Cancel the common factor of $8$ and $4$ .

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Step 3.13.1

Factor $4$ out of $8 x$ .

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{4 (2 x)}{4 (1 + 4 x^{2})}$

Step 3.13.2

Cancel the common factors.

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Step 3.13.2.1

Cancel the common factor.

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{4 (2 x)}{4 (1 + 4 x^{2})}$

Step 3.13.2.2

Rewrite the expression.

$\frac{2 x + \arctan (2 x) (1 + 4 x^{2})}{1 + 4 x^{2}} - \frac{2 x}{1 + 4 x^{2}}$

Step 4

Simplify.

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Step 4.1

Apply the distributive property.

$\frac{2 x + \arctan (2 x) \cdot 1 + \arctan (2 x) (4 x^{2})}{1 + 4 x^{2}} - \frac{2 x}{1 + 4 x^{2}}$

Step 4.2

Combine terms.

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Step 4.2.1

Multiply $\arctan (2 x)$ by $1$ .

$\frac{2 x + \arctan (2 x) + \arctan (2 x) (4 x^{2})}{1 + 4 x^{2}} - \frac{2 x}{1 + 4 x^{2}}$

Step 4.2.2

Combine the numerators over the common denominator.

$\frac{2 x + \arctan (2 x) + \arctan (2 x) (4 x^{2}) - 2 x}{1 + 4 x^{2}}$

Step 4.2.3

Subtract $2 x$ from $2 x$ .

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

Step 4.2.4

Add $\arctan (2 x) + \arctan (2 x) (4 x^{2})$ and $0$ .

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

Step 4.3

Reorder terms.

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

Step 4.4

Factor $\arctan (2 x)$ out of $4 x^{2} \arctan (2 x) + \arctan (2 x)$ .

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Step 4.4.1

Factor $\arctan (2 x)$ out of $4 x^{2} \arctan (2 x)$ .

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

Step 4.4.2

Multiply by $1$ .

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

Step 4.4.3

Factor $\arctan (2 x)$ out of $\arctan (2 x) (4 x^{2}) + \arctan (2 x) \cdot 1$ .

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

Step 4.5

Cancel the common factor of $4 x^{2} + 1$ .

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Step 4.5.1

Cancel the common factor.

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

Step 4.5.2

Divide $\arctan (2 x)$ by $1$ .

$\arctan (2 x)$