Calculus Examples

$y = \arctan (\frac{x}{4}) - \frac{1}{2 (x^{2} + 16)}$

Step 1

By the Sum Rule, the derivative of $\arctan (\frac{x}{4}) - \frac{1}{2 (x^{2} + 16)}$ with respect to $x$ is $\frac{d}{d x} [\arctan (\frac{x}{4})] + \frac{d}{d x} [- \frac{1}{2 (x^{2} + 16)}]$ .

$\frac{d}{d x} [\arctan (\frac{x}{4})] + \frac{d}{d x} [- \frac{1}{2 (x^{2} + 16)}]$

Step 2

Evaluate $\frac{d}{d x} [\arctan (\frac{x}{4})]$ .

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Step 2.1

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 $g (x) = \frac{x}{4}$ .

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Step 2.1.1

To apply the Chain Rule, set $u_{1}$ as $\frac{x}{4}$ .

$\frac{d}{d u_{1}} [\arctan (u_{1})] \frac{d}{d x} [\frac{x}{4}] + \frac{d}{d x} [- \frac{1}{2 (x^{2} + 16)}]$

Step 2.1.2

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

$\frac{1}{1 + {u_{1}}^{2}} \frac{d}{d x} [\frac{x}{4}] + \frac{d}{d x} [- \frac{1}{2 (x^{2} + 16)}]$

Step 2.1.3

Replace all occurrences of $u_{1}$ with $\frac{x}{4}$ .

$\frac{1}{1 + {(\frac{x}{4})}^{2}} \frac{d}{d x} [\frac{x}{4}] + \frac{d}{d x} [- \frac{1}{2 (x^{2} + 16)}]$

Step 2.2

Since $\frac{1}{4}$ is constant with respect to $x$ , the derivative of $\frac{x}{4}$ with respect to $x$ is $\frac{1}{4} \frac{d}{d x} [x]$ .

$\frac{1}{1 + {(\frac{x}{4})}^{2}} (\frac{1}{4} \frac{d}{d x} [x]) + \frac{d}{d x} [- \frac{1}{2 (x^{2} + 16)}]$

Step 2.3

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

$\frac{1}{1 + {(\frac{x}{4})}^{2}} (\frac{1}{4} \cdot 1) + \frac{d}{d x} [- \frac{1}{2 (x^{2} + 16)}]$

Step 2.4

Multiply $\frac{1}{4}$ by $1$ .

$\frac{1}{1 + {(\frac{x}{4})}^{2}} \cdot \frac{1}{4} + \frac{d}{d x} [- \frac{1}{2 (x^{2} + 16)}]$

Step 2.5

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

$\frac{1}{(1 + {(\frac{x}{4})}^{2}) \cdot 4} + \frac{d}{d x} [- \frac{1}{2 (x^{2} + 16)}]$

Step 2.6

Move $4$ to the left of $1 + {(\frac{x}{4})}^{2}$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} + \frac{d}{d x} [- \frac{1}{2 (x^{2} + 16)}]$

Step 3

Evaluate $\frac{d}{d x} [- \frac{1}{2 (x^{2} + 16)}]$ .

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Step 3.1

Since $- \frac{1}{2}$ is constant with respect to $x$ , the derivative of $- \frac{1}{2 (x^{2} + 16)}$ with respect to $x$ is $- \frac{1}{2} \frac{d}{d x} [\frac{1}{x^{2} + 16}]$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} - \frac{1}{2} \frac{d}{d x} [\frac{1}{x^{2} + 16}]$

Step 3.2

Rewrite $\frac{1}{x^{2} + 16}$ as ${(x^{2} + 16)}^{- 1}$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} - \frac{1}{2} \frac{d}{d x} [{(x^{2} + 16)}^{- 1}]$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2} + 16$ .

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Step 3.3.1

To apply the Chain Rule, set $u_{2}$ as $x^{2} + 16$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} - \frac{1}{2} (\frac{d}{d u_{2}} [{u_{2}}^{- 1}] \frac{d}{d x} [x^{2} + 16])$

Step 3.3.2

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

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} - \frac{1}{2} (- {u_{2}}^{- 2} \frac{d}{d x} [x^{2} + 16])$

Step 3.3.3

Replace all occurrences of $u_{2}$ with $x^{2} + 16$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} - \frac{1}{2} (- {(x^{2} + 16)}^{- 2} \frac{d}{d x} [x^{2} + 16])$

Step 3.4

By the Sum Rule, the derivative of $x^{2} + 16$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [16]$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} - \frac{1}{2} (- {(x^{2} + 16)}^{- 2} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [16]))$

Step 3.5

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

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} - \frac{1}{2} (- {(x^{2} + 16)}^{- 2} (2 x + \frac{d}{d x} [16]))$

Step 3.6

Since $16$ is constant with respect to $x$ , the derivative of $16$ with respect to $x$ is $0$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} - \frac{1}{2} (- {(x^{2} + 16)}^{- 2} (2 x + 0))$

Step 3.7

Add $2 x$ and $0$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} - \frac{1}{2} (- {(x^{2} + 16)}^{- 2} (2 x))$

Step 3.8

Multiply $2$ by $- 1$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} - \frac{1}{2} (- 2 {(x^{2} + 16)}^{- 2} x)$

Step 3.9

Multiply $- 2$ by $- 1$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} + 2 (\frac{1}{2}) ({(x^{2} + 16)}^{- 2} x)$

Step 3.10

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

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} + \frac{2}{2} ({(x^{2} + 16)}^{- 2} x)$

Step 3.11

Combine ${(x^{2} + 16)}^{- 2}$ and $\frac{2}{2}$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} + \frac{{(x^{2} + 16)}^{- 2} \cdot 2}{2} x$

Step 3.12

Combine $\frac{{(x^{2} + 16)}^{- 2} \cdot 2}{2}$ and $x$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} + \frac{{(x^{2} + 16)}^{- 2} \cdot 2 x}{2}$

Step 3.13

Move $2$ to the left of ${(x^{2} + 16)}^{- 2}$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} + \frac{2 \cdot {(x^{2} + 16)}^{- 2} x}{2}$

Step 3.14

Move ${(x^{2} + 16)}^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} + \frac{2 x}{2 {(x^{2} + 16)}^{2}}$

Step 3.15

Cancel the common factor of $2$ .

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Step 3.15.1

Cancel the common factor.

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} + \frac{2 x}{2 {(x^{2} + 16)}^{2}}$

Step 3.15.2

Rewrite the expression.

$\frac{1}{4 (1 + {(\frac{x}{4})}^{2})} + \frac{x}{{(x^{2} + 16)}^{2}}$

Step 4

Simplify.

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Step 4.1

Apply the product rule to $\frac{x}{4}$ .

$\frac{1}{4 (1 + \frac{x^{2}}{4^{2}})} + \frac{x}{{(x^{2} + 16)}^{2}}$

Step 4.2

Apply the distributive property.

$\frac{1}{4 \cdot 1 + 4 \frac{x^{2}}{4^{2}}} + \frac{x}{{(x^{2} + 16)}^{2}}$

Step 4.3

Combine terms.

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Step 4.3.1

Multiply $4$ by $1$ .

$\frac{1}{4 + 4 \frac{x^{2}}{4^{2}}} + \frac{x}{{(x^{2} + 16)}^{2}}$

Step 4.3.2

Raise $4$ to the power of $2$ .

$\frac{1}{4 + 4 \frac{x^{2}}{16}} + \frac{x}{{(x^{2} + 16)}^{2}}$

Step 4.3.3

Combine $4$ and $\frac{x^{2}}{16}$ .

$\frac{1}{4 + \frac{4 x^{2}}{16}} + \frac{x}{{(x^{2} + 16)}^{2}}$

Step 4.3.4

Cancel the common factor of $4$ and $16$ .

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Step 4.3.4.1

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

$\frac{1}{4 + \frac{4 (x^{2})}{16}} + \frac{x}{{(x^{2} + 16)}^{2}}$

Step 4.3.4.2

Cancel the common factors.

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Step 4.3.4.2.1

Factor $4$ out of $16$ .

$\frac{1}{4 + \frac{4 x^{2}}{4 \cdot 4}} + \frac{x}{{(x^{2} + 16)}^{2}}$

Step 4.3.4.2.2

Cancel the common factor.

$\frac{1}{4 + \frac{4 x^{2}}{4 \cdot 4}} + \frac{x}{{(x^{2} + 16)}^{2}}$

Step 4.3.4.2.3

Rewrite the expression.

$\frac{1}{4 + \frac{x^{2}}{4}} + \frac{x}{{(x^{2} + 16)}^{2}}$

Step 4.3.5

To write $\frac{1}{4 + \frac{x^{2}}{4}}$ as a fraction with a common denominator, multiply by $\frac{{(x^{2} + 16)}^{2}}{{(x^{2} + 16)}^{2}}$ .

$\frac{1}{4 + \frac{x^{2}}{4}} \cdot \frac{{(x^{2} + 16)}^{2}}{{(x^{2} + 16)}^{2}} + \frac{x}{{(x^{2} + 16)}^{2}}$

Step 4.3.6

To write $\frac{x}{{(x^{2} + 16)}^{2}}$ as a fraction with a common denominator, multiply by $\frac{4 + \frac{x^{2}}{4}}{4 + \frac{x^{2}}{4}}$ .

$\frac{1}{4 + \frac{x^{2}}{4}} \cdot \frac{{(x^{2} + 16)}^{2}}{{(x^{2} + 16)}^{2}} + \frac{x}{{(x^{2} + 16)}^{2}} \cdot \frac{4 + \frac{x^{2}}{4}}{4 + \frac{x^{2}}{4}}$

Step 4.3.7

Write each expression with a common denominator of $(4 + \frac{x^{2}}{4}) {(x^{2} + 16)}^{2}$ , by multiplying each by an appropriate factor of $1$ .

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Step 4.3.7.1

Multiply $\frac{1}{4 + \frac{x^{2}}{4}}$ by $\frac{{(x^{2} + 16)}^{2}}{{(x^{2} + 16)}^{2}}$ .

$\frac{{(x^{2} + 16)}^{2}}{(4 + \frac{x^{2}}{4}) {(x^{2} + 16)}^{2}} + \frac{x}{{(x^{2} + 16)}^{2}} \cdot \frac{4 + \frac{x^{2}}{4}}{4 + \frac{x^{2}}{4}}$

Step 4.3.7.2

Multiply $\frac{x}{{(x^{2} + 16)}^{2}}$ by $\frac{4 + \frac{x^{2}}{4}}{4 + \frac{x^{2}}{4}}$ .

$\frac{{(x^{2} + 16)}^{2}}{(4 + \frac{x^{2}}{4}) {(x^{2} + 16)}^{2}} + \frac{x (4 + \frac{x^{2}}{4})}{{(x^{2} + 16)}^{2} (4 + \frac{x^{2}}{4})}$

Step 4.3.7.3

Reorder the factors of $(4 + \frac{x^{2}}{4}) {(x^{2} + 16)}^{2}$ .

$\frac{{(x^{2} + 16)}^{2}}{{(x^{2} + 16)}^{2} (4 + \frac{x^{2}}{4})} + \frac{x (4 + \frac{x^{2}}{4})}{{(x^{2} + 16)}^{2} (4 + \frac{x^{2}}{4})}$

Step 4.3.8

Combine the numerators over the common denominator.

$\frac{{(x^{2} + 16)}^{2} + x (4 + \frac{x^{2}}{4})}{{(x^{2} + 16)}^{2} (4 + \frac{x^{2}}{4})}$