Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Divide each term in $\sqrt{1 - 4 x^{2}} \frac{d y}{d x} = x$ by $\sqrt{1 - 4 x^{2}}$ and simplify.

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Step 1.1.1

Divide each term in $\sqrt{1 - 4 x^{2}} \frac{d y}{d x} = x$ by $\sqrt{1 - 4 x^{2}}$ .

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

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $\sqrt{1 - 4 x^{2}}$ .

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Step 1.1.2.1.1

Cancel the common factor.

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

Step 1.1.2.1.2

Divide $\frac{d y}{d x}$ by $1$ .

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

Step 1.1.3

Simplify the right side.

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Step 1.1.3.1

Simplify the denominator.

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Step 1.1.3.1.1

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

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

Step 1.1.3.1.2

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

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

Step 1.1.3.1.3

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 $b = 2 x$ .

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

Step 1.1.3.1.4

Multiply $2$ by $- 1$ .

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

Step 1.1.3.2

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

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

Step 1.1.3.3

Combine and simplify the denominator.

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Step 1.1.3.3.1

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

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

Step 1.1.3.3.2

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

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

Step 1.1.3.3.3

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

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

Step 1.1.3.3.4

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

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

Step 1.1.3.3.5

Add $1$ and $1$ .

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

Step 1.1.3.3.6

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

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Step 1.1.3.3.6.1

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

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

Step 1.1.3.3.6.2

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

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

Step 1.1.3.3.6.3

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

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

Step 1.1.3.3.6.4

Cancel the common factor of $2$ .

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Step 1.1.3.3.6.4.1

Cancel the common factor.

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

Step 1.1.3.3.6.4.2

Rewrite the expression.

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

Step 1.1.3.3.6.5

Simplify.

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

Step 1.2

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u = (1 + 2 x) (1 - 2 x)$ . Then $d u = - 8 x d x$ , so $- \frac{1}{8} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.1.1

Let $u = (1 + 2 x) (1 - 2 x)$ . Find $\frac{d u}{d x}$ .

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Step 2.3.1.1.1

Differentiate $(1 + 2 x) (1 - 2 x)$ .

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

Step 2.3.1.1.2

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 $f (x) = 1 + 2 x$ and $g (x) = 1 - 2 x$ .

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

Step 2.3.1.1.3

Differentiate.

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Step 2.3.1.1.3.1

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

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

Step 2.3.1.1.3.2

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

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

Step 2.3.1.1.3.3

Add $0$ and $\frac{d}{d x} [- 2 x]$ .

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

Step 2.3.1.1.3.4

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

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

Step 2.3.1.1.3.5

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

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

Step 2.3.1.1.3.6

Simplify the expression.

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Step 2.3.1.1.3.6.1

Multiply $- 2$ by $1$ .

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

Step 2.3.1.1.3.6.2

Move $- 2$ to the left of $1 + 2 x$ .

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

Step 2.3.1.1.3.7

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

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

Step 2.3.1.1.3.8

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

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

Step 2.3.1.1.3.9

Add $0$ and $\frac{d}{d x} [2 x]$ .

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

Step 2.3.1.1.3.10

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

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

Step 2.3.1.1.3.11

Move $2$ to the left of $1 - 2 x$ .

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

Step 2.3.1.1.3.12

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

$- 2 (1 + 2 x) + 2 (1 - 2 x) \cdot 1$

Step 2.3.1.1.3.13

Multiply $2$ by $1$ .

$- 2 (1 + 2 x) + 2 (1 - 2 x)$

Step 2.3.1.1.4

Simplify.

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Step 2.3.1.1.4.1

Apply the distributive property.

$- 2 \cdot 1 - 2 (2 x) + 2 (1 - 2 x)$

Step 2.3.1.1.4.2

Apply the distributive property.

$- 2 \cdot 1 - 2 (2 x) + 2 \cdot 1 + 2 (- 2 x)$

Step 2.3.1.1.4.3

Combine terms.

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Step 2.3.1.1.4.3.1

Multiply $- 2$ by $1$ .

$- 2 - 2 (2 x) + 2 \cdot 1 + 2 (- 2 x)$

Step 2.3.1.1.4.3.2

Multiply $2$ by $- 2$ .

$- 2 - 4 x + 2 \cdot 1 + 2 (- 2 x)$

Step 2.3.1.1.4.3.3

Multiply $2$ by $1$ .

$- 2 - 4 x + 2 + 2 (- 2 x)$

Step 2.3.1.1.4.3.4

Multiply $- 2$ by $2$ .

$- 2 - 4 x + 2 - 4 x$

Step 2.3.1.1.4.3.5

Add $- 2$ and $2$ .

$- 4 x + 0 - 4 x$

Step 2.3.1.1.4.3.6

Add $- 4 x$ and $0$ .

$- 4 x - 4 x$

Step 2.3.1.1.4.3.7

Subtract $4 x$ from $- 4 x$ .

$- 8 x$

Step 2.3.1.2

Rewrite the problem using $u$ and $d u$ .

$y + C_{1} = \int \frac{\sqrt{u}}{u} \cdot \frac{1}{- 8} d u$

Step 2.3.2

Simplify.

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Step 2.3.2.1

Move the negative in front of the fraction.

$y + C_{1} = \int \frac{\sqrt{u}}{u} (- \frac{1}{8}) d u$

Step 2.3.2.2

Multiply $\frac{\sqrt{u}}{u}$ by $\frac{1}{8}$ .

$y + C_{1} = \int - \frac{\sqrt{u}}{u \cdot 8} d u$

Step 2.3.2.3

Move $8$ to the left of $u$ .

$y + C_{1} = \int - \frac{\sqrt{u}}{8 u} d u$

Step 2.3.3

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

$y + C_{1} = - \int \frac{\sqrt{u}}{8 u} d u$

Step 2.3.4

Since $\frac{1}{8}$ is constant with respect to $u$ , move $\frac{1}{8}$ out of the integral.

$y + C_{1} = - (\frac{1}{8} \int \frac{\sqrt{u}}{u} d u)$

Step 2.3.5

Simplify the expression.

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Step 2.3.5.1

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

$y + C_{1} = - \frac{1}{8} \int \frac{u^{\frac{1}{2}}}{u} d u$

Step 2.3.5.2

Simplify.

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Step 2.3.5.2.1

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

$y + C_{1} = - \frac{1}{8} \int \frac{1}{u \cdot u^{- \frac{1}{2}}} d u$

Step 2.3.5.2.2

Multiply $u$ by $u^{- \frac{1}{2}}$ by adding the exponents.

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Step 2.3.5.2.2.1

Multiply $u$ by $u^{- \frac{1}{2}}$ .

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Step 2.3.5.2.2.1.1

Raise $u$ to the power of $1$ .

$y + C_{1} = - \frac{1}{8} \int \frac{1}{u^{1} u^{- \frac{1}{2}}} d u$

Step 2.3.5.2.2.1.2

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

$y + C_{1} = - \frac{1}{8} \int \frac{1}{u^{1 - \frac{1}{2}}} d u$

Step 2.3.5.2.2.2

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

$y + C_{1} = - \frac{1}{8} \int \frac{1}{u^{\frac{2}{2} - \frac{1}{2}}} d u$

Step 2.3.5.2.2.3

Combine the numerators over the common denominator.

$y + C_{1} = - \frac{1}{8} \int \frac{1}{u^{\frac{2 - 1}{2}}} d u$

Step 2.3.5.2.2.4

Subtract $1$ from $2$ .

$y + C_{1} = - \frac{1}{8} \int \frac{1}{u^{\frac{1}{2}}} d u$

Step 2.3.5.3

Apply basic rules of exponents.

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Step 2.3.5.3.1

Move $u^{\frac{1}{2}}$ out of the denominator by raising it to the $- 1$ power.

$y + C_{1} = - \frac{1}{8} \int {(u^{\frac{1}{2}})}^{- 1} d u$

Step 2.3.5.3.2

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

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Step 2.3.5.3.2.1

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

$y + C_{1} = - \frac{1}{8} \int u^{\frac{1}{2} \cdot - 1} d u$

Step 2.3.5.3.2.2

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

$y + C_{1} = - \frac{1}{8} \int u^{\frac{- 1}{2}} d u$

Step 2.3.5.3.2.3

Move the negative in front of the fraction.

$y + C_{1} = - \frac{1}{8} \int u^{- \frac{1}{2}} d u$

Step 2.3.6

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

$y + C_{1} = - \frac{1}{8} (2 u^{\frac{1}{2}} + C_{2})$

Step 2.3.7

Simplify.

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Step 2.3.7.1

Rewrite $- \frac{1}{8} (2 u^{\frac{1}{2}} + C_{2})$ as $- \frac{1}{8} \cdot 2 u^{\frac{1}{2}} + C_{2}$ .

$y + C_{1} = - \frac{1}{8} \cdot 2 u^{\frac{1}{2}} + C_{2}$

Step 2.3.7.2

Simplify.

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Step 2.3.7.2.1

Multiply $2$ by $- 1$ .

$y + C_{1} = - 2 (\frac{1}{8}) u^{\frac{1}{2}} + C_{2}$

Step 2.3.7.2.2

Combine $- 2$ and $\frac{1}{8}$ .

$y + C_{1} = \frac{- 2}{8} u^{\frac{1}{2}} + C_{2}$

Step 2.3.7.2.3

Cancel the common factor of $- 2$ and $8$ .

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Step 2.3.7.2.3.1

Factor $2$ out of $- 2$ .

$y + C_{1} = \frac{2 (- 1)}{8} u^{\frac{1}{2}} + C_{2}$

Step 2.3.7.2.3.2

Cancel the common factors.

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Step 2.3.7.2.3.2.1

Factor $2$ out of $8$ .

$y + C_{1} = \frac{2 \cdot - 1}{2 \cdot 4} u^{\frac{1}{2}} + C_{2}$

Step 2.3.7.2.3.2.2

Cancel the common factor.

$y + C_{1} = \frac{2 \cdot - 1}{2 \cdot 4} u^{\frac{1}{2}} + C_{2}$

Step 2.3.7.2.3.2.3

Rewrite the expression.

$y + C_{1} = \frac{- 1}{4} u^{\frac{1}{2}} + C_{2}$

Step 2.3.7.2.4

Move the negative in front of the fraction.

$y + C_{1} = - \frac{1}{4} u^{\frac{1}{2}} + C_{2}$

Step 2.3.8

Replace all occurrences of $u$ with $(1 + 2 x) (1 - 2 x)$ .

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

Step 2.4

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

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