Calculus Examples

$y = \arctan (\sqrt{\frac{1 + x}{1 - x}})$

Step 1

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

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

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 3

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

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Step 3.1

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

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

Step 3.2

Cancel the common factor of $2$ .

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Step 3.2.1

Cancel the common factor.

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

Step 3.2.2

Rewrite the expression.

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

Step 4

Simplify.

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

Step 5

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

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

Step 6

Combine the numerators over the common denominator.

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

Step 7

Add $1$ and $1$ .

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

Step 8

Add $- x$ and $x$ .

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

Step 9

Add $0$ and $2$ .

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

Step 10

Multiply by the reciprocal of the fraction to divide by $\frac{2}{1 - x}$ .

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

Step 11

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

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

Step 12

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

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Step 12.1

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

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

Step 12.2

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

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

Step 12.3

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

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

Step 13

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

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

Step 14

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

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

Step 15

Combine the numerators over the common denominator.

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

Step 16

Simplify the numerator.

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Step 16.1

Multiply $- 1$ by $2$ .

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

Step 16.2

Subtract $2$ from $1$ .

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

Step 17

Combine fractions.

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Step 17.1

Move the negative in front of the fraction.

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

Step 17.2

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

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

Step 17.3

Multiply $2$ by $2$ .

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

Step 18

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

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

Step 19

Differentiate.

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Step 19.1

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

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

Step 19.2

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

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

Step 19.3

Add $0$ and $\frac{d}{d x} [x]$ .

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

Step 19.4

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

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

Step 19.5

Multiply $1 - x$ by $1$ .

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

Step 19.6

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

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

Step 19.7

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

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

Step 19.8

Add $0$ and $\frac{d}{d x} [- x]$ .

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

Step 19.9

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

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

Step 19.10

Multiply.

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Step 19.10.1

Multiply $- 1$ by $- 1$ .

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

Step 19.10.2

Multiply $1 + x$ by $1$ .

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

Step 19.11

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

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

Step 19.12

Simplify terms.

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Step 19.12.1

Multiply $1 + x$ by $1$ .

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

Step 19.12.2

Add $1$ and $1$ .

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

Step 19.12.3

Add $- x$ and $x$ .

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

Step 19.12.4

Add $0$ and $2$ .

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

Step 19.12.5

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

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

Step 19.12.6

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

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

Step 20

Cancel the common factors.

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Step 20.1

Factor $2$ out of $4 {(1 - x)}^{2}$ .

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

Step 20.2

Cancel the common factor.

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

Step 20.3

Rewrite the expression.

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

Step 21

Cancel the common factor of $1 - x$ and ${(1 - x)}^{2}$ .

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Step 21.1

Multiply by $1$ .

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

Step 21.2

Cancel the common factors.

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Step 21.2.1

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

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

Step 21.2.2

Cancel the common factor.

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

Step 21.2.3

Rewrite the expression.

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

Step 22

Simplify.

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Step 22.1

Change the sign of the exponent by rewriting the base as its reciprocal.

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

Step 22.2

Apply the product rule to $\frac{1 - x}{1 + x}$ .

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

Step 22.3

Apply the distributive property.

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

Step 22.4

Combine terms.

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Step 22.4.1

Multiply $2$ by $1$ .

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

Step 22.4.2

Multiply $- 1$ by $2$ .

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

Step 22.4.3

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

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

Step 22.5

Reorder terms.

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

Step 22.6

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

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Step 22.6.1

Factor $2$ out of $2$ .

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

Step 22.6.2

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

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

Step 22.6.3

Factor $2$ out of $2 (1) + 2 (- x)$ .

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

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

Step 22.7

Reorder terms.

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

Step 22.8

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

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

Step 22.9

Cancel the common factors.

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Step 22.9.1

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

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

Step 22.9.2

Cancel the common factor.

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

Step 22.9.3

Rewrite the expression.

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

Step 22.10

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

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

Step 22.11

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

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