Calculus Examples

$\arcsin (x y) = \frac{2}{3} \cdot \arctan (4 x)$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (\arcsin (x y)) = \frac{d}{d x} (\frac{2}{3} \cdot \arctan (4 x))$

Step 2

Differentiate the left side of the equation.

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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 $f (x) = \arcsin (x)$ and $g (x) = x y$ .

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Step 2.1.1

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

$\frac{d}{d u} [\arcsin (u)] \frac{d}{d x} [x y]$

Step 2.1.2

The derivative of $\arcsin (u)$ with respect to $u$ is $\frac{1}{\sqrt{1 - u^{2}}}$ .

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

Step 2.1.3

Replace all occurrences of $u$ with $x y$ .

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

Step 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 $f (x) = x$ and $g (x) = y$ .

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

Step 2.3

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

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

Step 2.4

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

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

Step 2.5

Multiply $y$ by $1$ .

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

Step 2.6

Simplify.

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Step 2.6.1

Apply the product rule to $x y$ .

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

Step 2.6.2

Reorder the factors of $\frac{1}{\sqrt{1 - x^{2} y^{2}}} (x y' + y)$ .

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $\frac{2}{3}$ is constant with respect to $x$ , the derivative of $\frac{2}{3} \cdot \arctan (4 x)$ with respect to $x$ is $\frac{2}{3} \frac{d}{d x} [\arctan (4 x)]$ .

$\frac{2}{3} \frac{d}{d x} [\arctan (4 x)]$

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 $f (x) = \arctan (x)$ and $g (x) = 4 x$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $4 x$ .

$\frac{2}{3} (\frac{d}{d u} [\arctan (u)] \frac{d}{d x} [4 x])$

Step 3.2.2

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

$\frac{2}{3} (\frac{1}{1 + u^{2}} \frac{d}{d x} [4 x])$

Step 3.2.3

Replace all occurrences of $u$ with $4 x$ .

$\frac{2}{3} (\frac{1}{1 + {(4 x)}^{2}} \frac{d}{d x} [4 x])$

Step 3.3

Differentiate.

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Step 3.3.1

Factor $4$ out of $4 x$ .

$\frac{2}{3} (\frac{1}{1 + {(4 (x))}^{2}} \frac{d}{d x} [4 x])$

Step 3.3.2

Combine fractions.

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Step 3.3.2.1

Simplify the expression.

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Step 3.3.2.1.1

Apply the product rule to $4 (x)$ .

$\frac{2}{3} (\frac{1}{1 + 4^{2} x^{2}} \frac{d}{d x} [4 x])$

Step 3.3.2.1.2

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

$\frac{2}{3} (\frac{1}{1 + 16 x^{2}} \frac{d}{d x} [4 x])$

Step 3.3.2.2

Multiply $\frac{1}{1 + 16 x^{2}}$ by $\frac{2}{3}$ .

$\frac{2}{(1 + 16 x^{2}) \cdot 3} \frac{d}{d x} [4 x]$

Step 3.3.2.3

Move $3$ to the left of $1 + 16 x^{2}$ .

$\frac{2}{3 \cdot (1 + 16 x^{2})} \frac{d}{d x} [4 x]$

Step 3.3.3

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

$\frac{2}{3 (1 + 16 x^{2})} (4 \frac{d}{d x} [x])$

Step 3.3.4

Combine fractions.

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Step 3.3.4.1

Combine $4$ and $\frac{2}{3 (1 + 16 x^{2})}$ .

$\frac{4 \cdot 2}{3 (1 + 16 x^{2})} \frac{d}{d x} [x]$

Step 3.3.4.2

Multiply $4$ by $2$ .

$\frac{8}{3 (1 + 16 x^{2})} \frac{d}{d x} [x]$

Step 3.3.5

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

$\frac{8}{3 (1 + 16 x^{2})} \cdot 1$

Step 3.3.6

Multiply $\frac{8}{3 (1 + 16 x^{2})}$ by $1$ .

$\frac{8}{3 (1 + 16 x^{2})}$

Step 3.4

Simplify.

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Step 3.4.1

Apply the distributive property.

$\frac{8}{3 \cdot 1 + 3 (16 x^{2})}$

Step 3.4.2

Combine terms.

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Step 3.4.2.1

Multiply $3$ by $1$ .

$\frac{8}{3 + 3 (16 x^{2})}$

Step 3.4.2.2

Multiply $16$ by $3$ .

$\frac{8}{3 + 48 x^{2}}$

Step 3.4.3

Reorder terms.

$\frac{8}{48 x^{2} + 3}$

Step 3.4.4

Factor $3$ out of $48 x^{2} + 3$ .

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Step 3.4.4.1

Factor $3$ out of $48 x^{2}$ .

$\frac{8}{3 (16 x^{2}) + 3}$

Step 3.4.4.2

Factor $3$ out of $3$ .

$\frac{8}{3 (16 x^{2}) + 3 (1)}$

Step 3.4.4.3

Factor $3$ out of $3 (16 x^{2}) + 3 (1)$ .

$\frac{8}{3 (16 x^{2} + 1)}$

Step 4

Reform the equation by setting the left side equal to the right side.

$(x y' + y) (\frac{1}{\sqrt{1 - x^{2} y^{2}}}) = \frac{8}{3 (16 x^{2} + 1)}$

Step 5

Solve for $y'$ .

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Step 5.1

Multiply each term in $(x y' + y) \frac{1}{\sqrt{1 - x^{2} y^{2}}} = \frac{8}{3 (16 x^{2} + 1)}$ by $\sqrt{1 - x^{2} y^{2}}$ to eliminate the fractions.

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Step 5.1.1

Multiply each term in $(x y' + y) \frac{1}{\sqrt{1 - x^{2} y^{2}}} = \frac{8}{3 (16 x^{2} + 1)}$ by $\sqrt{1 - x^{2} y^{2}}$ .

$(x y' + y) \frac{1}{\sqrt{1 - x^{2} y^{2}}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2

Simplify the left side.

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Step 5.1.2.1

Simplify the denominator.

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Step 5.1.2.1.1

Rewrite $1$ as $1^{2}$ .

$(x y' + y) \frac{1}{\sqrt{1^{2} - x^{2} y^{2}}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.1.2

Rewrite $x^{2} y^{2}$ as ${(x y)}^{2}$ .

$(x y' + y) \frac{1}{\sqrt{1^{2} - {(x y)}^{2}}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.1.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x y$ .

$(x y' + y) \frac{1}{\sqrt{(1 + x y) (1 - x y)}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.2

Multiply $\frac{1}{\sqrt{(1 + x y) (1 - x y)}}$ by $\frac{\sqrt{(1 + x y) (1 - x y)}}{\sqrt{(1 + x y) (1 - x y)}}$ .

$(x y' + y) (\frac{1}{\sqrt{(1 + x y) (1 - x y)}} \cdot \frac{\sqrt{(1 + x y) (1 - x y)}}{\sqrt{(1 + x y) (1 - x y)}}) \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3

Combine fractions.

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Step 5.1.2.3.1

Combine and simplify the denominator.

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Step 5.1.2.3.1.1

Multiply $\frac{1}{\sqrt{(1 + x y) (1 - x y)}}$ by $\frac{\sqrt{(1 + x y) (1 - x y)}}{\sqrt{(1 + x y) (1 - x y)}}$ .

$(x y' + y) \frac{\sqrt{(1 + x y) (1 - x y)}}{\sqrt{(1 + x y) (1 - x y)} \sqrt{(1 + x y) (1 - x y)}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.1.2

Raise $\sqrt{(1 + x y) (1 - x y)}$ to the power of $1$ .

$(x y' + y) \frac{\sqrt{(1 + x y) (1 - x y)}}{{\sqrt{(1 + x y) (1 - x y)}}^{1} \sqrt{(1 + x y) (1 - x y)}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.1.3

Raise $\sqrt{(1 + x y) (1 - x y)}$ to the power of $1$ .

$(x y' + y) \frac{\sqrt{(1 + x y) (1 - x y)}}{{\sqrt{(1 + x y) (1 - x y)}}^{1} {\sqrt{(1 + x y) (1 - x y)}}^{1}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.1.4

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

$(x y' + y) \frac{\sqrt{(1 + x y) (1 - x y)}}{{\sqrt{(1 + x y) (1 - x y)}}^{1 + 1}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.1.5

Add $1$ and $1$ .

$(x y' + y) \frac{\sqrt{(1 + x y) (1 - x y)}}{{\sqrt{(1 + x y) (1 - x y)}}^{2}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.1.6

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

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Step 5.1.2.3.1.6.1

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

$(x y' + y) \frac{\sqrt{(1 + x y) (1 - x y)}}{{({((1 + x y) (1 - x y))}^{\frac{1}{2}})}^{2}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.1.6.2

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

$(x y' + y) \frac{\sqrt{(1 + x y) (1 - x y)}}{{((1 + x y) (1 - x y))}^{\frac{1}{2} \cdot 2}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.1.6.3

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

$(x y' + y) \frac{\sqrt{(1 + x y) (1 - x y)}}{{((1 + x y) (1 - x y))}^{\frac{2}{2}}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.1.6.4

Cancel the common factor of $2$ .

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Step 5.1.2.3.1.6.4.1

Cancel the common factor.

$(x y' + y) \frac{\sqrt{(1 + x y) (1 - x y)}}{{((1 + x y) (1 - x y))}^{\frac{2}{2}}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.1.6.4.2

Rewrite the expression.

$(x y' + y) \frac{\sqrt{(1 + x y) (1 - x y)}}{{((1 + x y) (1 - x y))}^{1}} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.1.6.5

Simplify.

$(x y' + y) \frac{\sqrt{(1 + x y) (1 - x y)}}{(1 + x y) (1 - x y)} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.2

Multiply $x y' + y$ by $\frac{\sqrt{(1 + x y) (1 - x y)}}{(1 + x y) (1 - x y)}$ .

$\frac{(x y' + y) \sqrt{(1 + x y) (1 - x y)}}{(1 + x y) (1 - x y)} \sqrt{1 - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.3

Write the expression using exponents.

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Step 5.1.2.3.3.1

Rewrite $1$ as $1^{2}$ .

$\frac{(x y' + y) \sqrt{(1 + x y) (1 - x y)}}{(1 + x y) (1 - x y)} \sqrt{1^{2} - x^{2} y^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.3.3.2

Rewrite $x^{2} y^{2}$ as ${(x y)}^{2}$ .

$\frac{(x y' + y) \sqrt{(1 + x y) (1 - x y)}}{(1 + x y) (1 - x y)} \sqrt{1^{2} - {(x y)}^{2}} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x y$ .

$\frac{(x y' + y) \sqrt{(1 + x y) (1 - x y)}}{(1 + x y) (1 - x y)} \sqrt{(1 + x y) (1 - (x y))} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.5

Multiply $\frac{(x y' + y) \sqrt{(1 + x y) (1 - x y)}}{(1 + x y) (1 - x y)} \sqrt{(1 + x y) (1 - x y)}$ .

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Step 5.1.2.5.1

Combine $\frac{(x y' + y) \sqrt{(1 + x y) (1 - x y)}}{(1 + x y) (1 - x y)}$ and $\sqrt{(1 + x y) (1 - x y)}$ .

$\frac{(x y' + y) \sqrt{(1 + x y) (1 - x y)} \sqrt{(1 + x y) (1 - x y)}}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.5.2

Raise $\sqrt{(1 + x y) (1 - x y)}$ to the power of $1$ .

$\frac{(x y' + y) ({\sqrt{(1 + x y) (1 - x y)}}^{1} \sqrt{(1 + x y) (1 - x y)})}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.5.3

Raise $\sqrt{(1 + x y) (1 - x y)}$ to the power of $1$ .

$\frac{(x y' + y) ({\sqrt{(1 + x y) (1 - x y)}}^{1} {\sqrt{(1 + x y) (1 - x y)}}^{1})}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.5.4

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

$\frac{(x y' + y) {\sqrt{(1 + x y) (1 - x y)}}^{1 + 1}}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.5.5

Add $1$ and $1$ .

$\frac{(x y' + y) {\sqrt{(1 + x y) (1 - x y)}}^{2}}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6

Simplify the numerator.

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Step 5.1.2.6.1

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

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Step 5.1.2.6.1.1

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

$\frac{(x y' + y) {({((1 + x y) (1 - x y))}^{\frac{1}{2}})}^{2}}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.1.2

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

$\frac{(x y' + y) {((1 + x y) (1 - x y))}^{\frac{1}{2} \cdot 2}}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.1.3

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

$\frac{(x y' + y) {((1 + x y) (1 - x y))}^{\frac{2}{2}}}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.1.4

Cancel the common factor of $2$ .

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Step 5.1.2.6.1.4.1

Cancel the common factor.

$\frac{(x y' + y) {((1 + x y) (1 - x y))}^{\frac{2}{2}}}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.1.4.2

Rewrite the expression.

$\frac{(x y' + y) {((1 + x y) (1 - x y))}^{1}}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.1.5

Simplify.

$\frac{(x y' + y) ((1 + x y) (1 - x y))}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.2

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

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Step 5.1.2.6.2.1

Apply the distributive property.

$\frac{(x y' + y) (1 (1 - x y) + x y (1 - x y))}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.2.2

Apply the distributive property.

$\frac{(x y' + y) (1 \cdot 1 + 1 (- x y) + x y (1 - x y))}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.2.3

Apply the distributive property.

$\frac{(x y' + y) (1 \cdot 1 + 1 (- x y) + x y \cdot 1 + x y (- x y))}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.3

Simplify and combine like terms.

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Step 5.1.2.6.3.1

Simplify each term.

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Step 5.1.2.6.3.1.1

Multiply $1$ by $1$ .

$\frac{(x y' + y) (1 + 1 (- x y) + x y \cdot 1 + x y (- x y))}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.3.1.2

Multiply $- x y$ by $1$ .

$\frac{(x y' + y) (1 - x y + x y \cdot 1 + x y (- x y))}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.3.1.3

Multiply $x$ by $1$ .

$\frac{(x y' + y) (1 - x y + x y + x y (- x y))}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.3.1.4

Rewrite using the commutative property of multiplication.

$\frac{(x y' + y) (1 - x y + x y - x y (x y))}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.3.1.5

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

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Step 5.1.2.6.3.1.5.1

Move $x$ .

$\frac{(x y' + y) (1 - x y + x y - (x \cdot x) y \cdot y)}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.3.1.5.2

Multiply $x$ by $x$ .

$\frac{(x y' + y) (1 - x y + x y - x^{2} y \cdot y)}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.3.1.6

Multiply $y$ by $y$ by adding the exponents.

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Step 5.1.2.6.3.1.6.1

Move $y$ .

$\frac{(x y' + y) (1 - x y + x y - x^{2} (y \cdot y))}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.3.1.6.2

Multiply $y$ by $y$ .

$\frac{(x y' + y) (1 - x y + x y - x^{2} y^{2})}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.3.2

Add $- x y$ and $x y$ .

$\frac{(x y' + y) (1 + 0 - x^{2} y^{2})}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.3.3

Add $1$ and $0$ .

$\frac{(x y' + y) (1 - x^{2} y^{2})}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.4

Rewrite $1$ as $1^{2}$ .

$\frac{(x y' + y) (1^{2} - x^{2} y^{2})}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.5

Rewrite $x^{2} y^{2}$ as ${(x y)}^{2}$ .

$\frac{(x y' + y) (1^{2} - {(x y)}^{2})}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.6

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x y$ .

$\frac{(x y' + y) ((1 + x y) (1 - (x y)))}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.6.7

Remove unnecessary parentheses.

$\frac{(x y' + y) (1 + x y) (1 - x y)}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.7

Reduce the expression by cancelling the common factors.

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Step 5.1.2.7.1

Cancel the common factor of $1 + x y$ .

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Step 5.1.2.7.1.1

Cancel the common factor.

$\frac{(x y' + y) (1 + x y) (1 - x y)}{(1 + x y) (1 - x y)} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.7.1.2

Rewrite the expression.

$\frac{(x y' + y) (1 - x y)}{1 - x y} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.7.2

Cancel the common factor of $1 - x y$ .

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Step 5.1.2.7.2.1

Cancel the common factor.

$\frac{(x y' + y) (1 - x y)}{1 - x y} = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.2.7.2.2

Divide $x y' + y$ by $1$ .

$x y' + y = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1 - x^{2} y^{2}}$

Step 5.1.3

Simplify the right side.

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Step 5.1.3.1

Write the expression using exponents.

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Step 5.1.3.1.1

Rewrite $1$ as $1^{2}$ .

$x y' + y = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1^{2} - x^{2} y^{2}}$

Step 5.1.3.1.2

Rewrite $x^{2} y^{2}$ as ${(x y)}^{2}$ .

$x y' + y = \frac{8}{3 (16 x^{2} + 1)} \sqrt{1^{2} - {(x y)}^{2}}$

Step 5.1.3.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x y$ .

$x y' + y = \frac{8}{3 (16 x^{2} + 1)} \sqrt{(1 + x y) (1 - (x y))}$

Step 5.1.3.3

Combine $\frac{8}{3 (16 x^{2} + 1)}$ and $\sqrt{(1 + x y) (1 - x y)}$ .

$x y' + y = \frac{8 \sqrt{(1 + x y) (1 - x y)}}{3 (16 x^{2} + 1)}$

Step 5.2

Subtract $y$ from both sides of the equation.

$x y' = \frac{8 \sqrt{(1 + x y) (1 - x y)}}{3 (16 x^{2} + 1)} - y$

Step 5.3

Divide each term in $x y' = \frac{8 \sqrt{(1 + x y) (1 - x y)}}{3 (16 x^{2} + 1)} - y$ by $x$ and simplify.

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Step 5.3.1

Divide each term in $x y' = \frac{8 \sqrt{(1 + x y) (1 - x y)}}{3 (16 x^{2} + 1)} - y$ by $x$ .

$\frac{x y'}{x} = \frac{\frac{8 \sqrt{(1 + x y) (1 - x y)}}{3 (16 x^{2} + 1)}}{x} + \frac{- y}{x}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $x$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{x y'}{x} = \frac{\frac{8 \sqrt{(1 + x y) (1 - x y)}}{3 (16 x^{2} + 1)}}{x} + \frac{- y}{x}$

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{\frac{8 \sqrt{(1 + x y) (1 - x y)}}{3 (16 x^{2} + 1)}}{x} + \frac{- y}{x}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

$y' = \frac{\frac{8 \sqrt{(1 + x y) (1 - x y)}}{3 (16 x^{2} + 1)} - y}{x}$

Step 5.3.3.2

To write $- y$ as a fraction with a common denominator, multiply by $\frac{3 (16 x^{2} + 1)}{3 (16 x^{2} + 1)}$ .

$y' = \frac{\frac{8 \sqrt{(1 + x y) (1 - x y)}}{3 (16 x^{2} + 1)} - y \cdot \frac{3 (16 x^{2} + 1)}{3 (16 x^{2} + 1)}}{x}$

Step 5.3.3.3

Simplify terms.

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Step 5.3.3.3.1

Combine $- y$ and $\frac{3 (16 x^{2} + 1)}{3 (16 x^{2} + 1)}$ .

$y' = \frac{\frac{8 \sqrt{(1 + x y) (1 - x y)}}{3 (16 x^{2} + 1)} + \frac{- y (3 (16 x^{2} + 1))}{3 (16 x^{2} + 1)}}{x}$

Step 5.3.3.3.2

Combine the numerators over the common denominator.

$y' = \frac{\frac{8 \sqrt{(1 + x y) (1 - x y)} - y (3 (16 x^{2} + 1))}{3 (16 x^{2} + 1)}}{x}$

Step 5.3.3.4

Simplify the numerator.

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Step 5.3.3.4.1

Multiply $3$ by $- 1$ .

$y' = \frac{\frac{8 \sqrt{(1 + x y) (1 - x y)} - 3 y (16 x^{2} + 1)}{3 (16 x^{2} + 1)}}{x}$

Step 5.3.3.4.2

Apply the distributive property.

$y' = \frac{\frac{8 \sqrt{(1 + x y) (1 - x y)} - 3 y (16 x^{2}) - 3 y \cdot 1}{3 (16 x^{2} + 1)}}{x}$

Step 5.3.3.4.3

Rewrite using the commutative property of multiplication.

$y' = \frac{\frac{8 \sqrt{(1 + x y) (1 - x y)} - 3 \cdot 16 y x^{2} - 3 y \cdot 1}{3 (16 x^{2} + 1)}}{x}$

Step 5.3.3.4.4

Multiply $- 3$ by $1$ .

$y' = \frac{\frac{8 \sqrt{(1 + x y) (1 - x y)} - 3 \cdot 16 y x^{2} - 3 y}{3 (16 x^{2} + 1)}}{x}$

Step 5.3.3.4.5

Multiply $- 3$ by $16$ .

$y' = \frac{\frac{8 \sqrt{(1 + x y) (1 - x y)} - 48 y x^{2} - 3 y}{3 (16 x^{2} + 1)}}{x}$

Step 5.3.3.5

Multiply the numerator by the reciprocal of the denominator.

$y' = \frac{8 \sqrt{(1 + x y) (1 - x y)} - 48 y x^{2} - 3 y}{3 (16 x^{2} + 1)} \cdot \frac{1}{x}$

Step 5.3.3.6

Multiply $\frac{8 \sqrt{(1 + x y) (1 - x y)} - 48 y x^{2} - 3 y}{3 (16 x^{2} + 1)}$ by $\frac{1}{x}$ .

$y' = \frac{8 \sqrt{(1 + x y) (1 - x y)} - 48 y x^{2} - 3 y}{3 (16 x^{2} + 1) x}$

Step 5.3.3.7

Reorder factors in $\frac{8 \sqrt{(1 + x y) (1 - x y)} - 48 y x^{2} - 3 y}{3 (16 x^{2} + 1) x}$ .

$y' = \frac{8 \sqrt{(1 + x y) (1 - x y)} - 48 y x^{2} - 3 y}{3 x (16 x^{2} + 1)}$

Step 6

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = \frac{8 \sqrt{(1 + x y) (1 - x y)} - 48 y x^{2} - 3 y}{3 x (16 x^{2} + 1)}$