Calculus Examples

$\frac{d y}{d x} = 2 x \sqrt{4 - y^{2}}$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{\sqrt{4 - y^{2}}}$ .

$\frac{1}{\sqrt{4 - y^{2}}} \frac{d y}{d x} = \frac{1}{\sqrt{4 - y^{2}}} (2 x \sqrt{4 - y^{2}})$

Step 1.2

Simplify.

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Step 1.2.1

Rewrite using the commutative property of multiplication.

$\frac{1}{\sqrt{4 - y^{2}}} \frac{d y}{d x} = 2 \frac{1}{\sqrt{4 - y^{2}}} (x \sqrt{4 - y^{2}})$

Step 1.2.2

Simplify the denominator.

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Step 1.2.2.1

Rewrite $4$ as $2^{2}$ .

$\frac{1}{\sqrt{4 - y^{2}}} \frac{d y}{d x} = 2 \frac{1}{\sqrt{2^{2} - y^{2}}} (x \sqrt{4 - y^{2}})$

Step 1.2.2.2

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

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

Step 1.2.3

Multiply $\frac{1}{\sqrt{(2 + y) (2 - y)}}$ by $\frac{\sqrt{(2 + y) (2 - y)}}{\sqrt{(2 + y) (2 - y)}}$ .

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

Step 1.2.4

Combine and simplify the denominator.

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Step 1.2.4.1

Multiply $\frac{1}{\sqrt{(2 + y) (2 - y)}}$ by $\frac{\sqrt{(2 + y) (2 - y)}}{\sqrt{(2 + y) (2 - y)}}$ .

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

Step 1.2.4.2

Raise $\sqrt{(2 + y) (2 - y)}$ to the power of $1$ .

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

Step 1.2.4.3

Raise $\sqrt{(2 + y) (2 - y)}$ to the power of $1$ .

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

Step 1.2.4.4

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

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

Step 1.2.4.5

Add $1$ and $1$ .

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

Step 1.2.4.6

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

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Step 1.2.4.6.1

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

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

Step 1.2.4.6.2

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

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

Step 1.2.4.6.3

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

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

Step 1.2.4.6.4

Cancel the common factor of $2$ .

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Step 1.2.4.6.4.1

Cancel the common factor.

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

Step 1.2.4.6.4.2

Rewrite the expression.

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

Step 1.2.4.6.5

Simplify.

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

Step 1.2.5

Combine $2$ and $\frac{\sqrt{(2 + y) (2 - y)}}{(2 + y) (2 - y)}$ .

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

Step 1.2.6

Rewrite $4$ as $2^{2}$ .

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

Step 1.2.7

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

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

Step 1.2.8

Multiply $\frac{2 \sqrt{(2 + y) (2 - y)}}{(2 + y) (2 - y)} (x \sqrt{(2 + y) (2 - y)})$ .

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Step 1.2.8.1

Combine $x$ and $\frac{2 \sqrt{(2 + y) (2 - y)}}{(2 + y) (2 - y)}$ .

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

Step 1.2.8.2

Combine $\frac{x (2 \sqrt{(2 + y) (2 - y)})}{(2 + y) (2 - y)}$ and $\sqrt{(2 + y) (2 - y)}$ .

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

Step 1.2.8.3

Raise $\sqrt{(2 + y) (2 - y)}$ to the power of $1$ .

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

Step 1.2.8.4

Raise $\sqrt{(2 + y) (2 - y)}$ to the power of $1$ .

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

Step 1.2.8.5

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

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

Step 1.2.8.6

Add $1$ and $1$ .

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

Step 1.2.9

Simplify the numerator.

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Step 1.2.9.1

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

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Step 1.2.9.1.1

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

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

Step 1.2.9.1.2

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

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

Step 1.2.9.1.3

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

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

Step 1.2.9.1.4

Cancel the common factor of $2$ .

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Step 1.2.9.1.4.1

Cancel the common factor.

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

Step 1.2.9.1.4.2

Rewrite the expression.

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

Step 1.2.9.1.5

Simplify.

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

Step 1.2.9.2

Expand $(2 + y) (2 - y)$ using the FOIL Method.

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Step 1.2.9.2.1

Apply the distributive property.

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

Step 1.2.9.2.2

Apply the distributive property.

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

Step 1.2.9.2.3

Apply the distributive property.

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

Step 1.2.9.3

Simplify and combine like terms.

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Step 1.2.9.3.1

Simplify each term.

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Step 1.2.9.3.1.1

Multiply $2$ by $2$ .

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

Step 1.2.9.3.1.2

Multiply $- 1$ by $2$ .

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

Step 1.2.9.3.1.3

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

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

Step 1.2.9.3.1.4

Rewrite using the commutative property of multiplication.

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

Step 1.2.9.3.1.5

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

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Step 1.2.9.3.1.5.1

Move $y$ .

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

Step 1.2.9.3.1.5.2

Multiply $y$ by $y$ .

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

Step 1.2.9.3.2

Add $- 2 y$ and $2 y$ .

$\frac{1}{\sqrt{4 - y^{2}}} \frac{d y}{d x} = \frac{x \cdot 2 (4 + 0 - y^{2})}{(2 + y) (2 - y)}$

Step 1.2.9.3.3

Add $4$ and $0$ .

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

Step 1.2.9.4

Rewrite $4$ as $2^{2}$ .

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

Step 1.2.9.5

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

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

Step 1.2.10

Cancel the common factor of $2 + y$ .

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Step 1.2.10.1

Cancel the common factor.

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

Step 1.2.10.2

Rewrite the expression.

$\frac{1}{\sqrt{4 - y^{2}}} \frac{d y}{d x} = \frac{x \cdot 2 (2 - y)}{2 - y}$

Step 1.2.11

Cancel the common factor of $2 - y$ .

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Step 1.2.11.1

Cancel the common factor.

$\frac{1}{\sqrt{4 - y^{2}}} \frac{d y}{d x} = \frac{x \cdot 2 (2 - y)}{2 - y}$

Step 1.2.11.2

Divide $x \cdot 2$ by $1$ .

$\frac{1}{\sqrt{4 - y^{2}}} \frac{d y}{d x} = x \cdot 2$

Step 1.2.12

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

$\frac{1}{\sqrt{4 - y^{2}}} \frac{d y}{d x} = 2 x$

Step 1.3

Rewrite the equation.

$\frac{1}{\sqrt{4 - y^{2}}} d y = 2 x d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{\sqrt{4 - y^{2}}} d y = \int 2 x d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Rewrite $4$ as $2^{2}$ .

$\int \frac{1}{\sqrt{2^{2} - y^{2}}} d y = \int 2 x d x$

Step 2.2.2

The integral of $\frac{1}{\sqrt{2^{2} - y^{2}}}$ with respect to $y$ is $\arcsin (\frac{y}{2})$

$\arcsin (\frac{y}{2}) + C_{1} = \int 2 x d x$

Step 2.2.3

Rewrite $\arcsin (\frac{y}{2}) + C_{1}$ as $\arcsin (\frac{1}{2} y) + C_{1}$ .

$\arcsin (\frac{1}{2} y) + C_{1} = \int 2 x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$\arcsin (\frac{1}{2} y) + C_{1} = 2 \int x d x$

Step 2.3.2

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$\arcsin (\frac{1}{2} y) + C_{1} = 2 (\frac{1}{2} x^{2} + C_{2})$

Step 2.3.3

Simplify the answer.

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Step 2.3.3.1

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

$\arcsin (\frac{1}{2} y) + C_{1} = 2 (\frac{1}{2}) x^{2} + C_{2}$

Step 2.3.3.2

Simplify.

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Step 2.3.3.2.1

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

$\arcsin (\frac{1}{2} y) + C_{1} = \frac{2}{2} x^{2} + C_{2}$

Step 2.3.3.2.2

Cancel the common factor of $2$ .

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Step 2.3.3.2.2.1

Cancel the common factor.

$\arcsin (\frac{1}{2} y) + C_{1} = \frac{2}{2} x^{2} + C_{2}$

Step 2.3.3.2.2.2

Rewrite the expression.

$\arcsin (\frac{1}{2} y) + C_{1} = 1 x^{2} + C_{2}$

Step 2.3.3.2.3

Multiply $x^{2}$ by $1$ .

$\arcsin (\frac{1}{2} y) + C_{1} = x^{2} + C_{2}$

Step 2.4

Group the constant of integration on the right side as $K$ .

$\arcsin (\frac{1}{2} y) = x^{2} + K$

Step 3

Solve for $y$ .

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Step 3.1

Take the inverse arcsine of both sides of the equation to extract $y$ from inside the arcsine.

$\frac{1}{2} y = \sin (x^{2} + K)$

Step 3.2

Simplify the left side.

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Step 3.2.1

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

$\frac{y}{2} = \sin (x^{2} + K)$

Step 3.3

Multiply both sides of the equation by $2$ .

$2 \frac{y}{2} = 2 \sin (x^{2} + K)$

Step 3.4

Simplify the left side.

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Step 3.4.1

Cancel the common factor of $2$ .

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Step 3.4.1.1

Cancel the common factor.

$2 \frac{y}{2} = 2 \sin (x^{2} + K)$

Step 3.4.1.2

Rewrite the expression.

$y = 2 \sin (x^{2} + K)$