Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $y^{2}$ .

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

Step 1.2

Cancel the common factor of $y^{2}$ .

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

By the Power Rule, the integral of $y^{2}$ with respect to $y$ is $\frac{1}{3} y^{3}$ .

$\frac{1}{3} y^{3} + C_{1} = \int \sqrt{4 x - 1} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u = 4 x - 1$ . Then $d u = 4 d x$ , so $\frac{1}{4} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.1.1

Let $u = 4 x - 1$ . Find $\frac{d u}{d x}$ .

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Step 2.3.1.1.1

Differentiate $4 x - 1$ .

$\frac{d}{d x} [4 x - 1]$

Step 2.3.1.1.2

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

$\frac{d}{d x} [4 x] + \frac{d}{d x} [- 1]$

Step 2.3.1.1.3

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

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Step 2.3.1.1.3.1

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

$4 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 2.3.1.1.3.2

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

$4 \cdot 1 + \frac{d}{d x} [- 1]$

Step 2.3.1.1.3.3

Multiply $4$ by $1$ .

$4 + \frac{d}{d x} [- 1]$

Step 2.3.1.1.4

Differentiate using the Constant Rule.

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Step 2.3.1.1.4.1

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

$4 + 0$

Step 2.3.1.1.4.2

Add $4$ and $0$ .

$4$

Step 2.3.1.2

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

$\frac{1}{3} y^{3} + C_{1} = \int \sqrt{u} \frac{1}{4} d u$

Step 2.3.2

Combine $\sqrt{u}$ and $\frac{1}{4}$ .

$\frac{1}{3} y^{3} + C_{1} = \int \frac{\sqrt{u}}{4} d u$

Step 2.3.3

Since $\frac{1}{4}$ is constant with respect to $u$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{3} y^{3} + C_{1} = \frac{1}{4} \int \sqrt{u} d u$

Step 2.3.4

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

$\frac{1}{3} y^{3} + C_{1} = \frac{1}{4} \int u^{\frac{1}{2}} d u$

Step 2.3.5

By the Power Rule, the integral of $u^{\frac{1}{2}}$ with respect to $u$ is $\frac{2}{3} u^{\frac{3}{2}}$ .

$\frac{1}{3} y^{3} + C_{1} = \frac{1}{4} (\frac{2}{3} u^{\frac{3}{2}} + C_{2})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Rewrite $\frac{1}{4} (\frac{2}{3} u^{\frac{3}{2}} + C_{2})$ as $\frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}} + C_{2}$ .

$\frac{1}{3} y^{3} + C_{1} = \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}} + C_{2}$

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

Multiply $\frac{1}{4}$ by $\frac{2}{3}$ .

$\frac{1}{3} y^{3} + C_{1} = \frac{2}{4 \cdot 3} u^{\frac{3}{2}} + C_{2}$

Step 2.3.6.2.2

Multiply $4$ by $3$ .

$\frac{1}{3} y^{3} + C_{1} = \frac{2}{12} u^{\frac{3}{2}} + C_{2}$

Step 2.3.6.2.3

Cancel the common factor of $2$ and $12$ .

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Step 2.3.6.2.3.1

Factor $2$ out of $2$ .

$\frac{1}{3} y^{3} + C_{1} = \frac{2 (1)}{12} u^{\frac{3}{2}} + C_{2}$

Step 2.3.6.2.3.2

Cancel the common factors.

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Step 2.3.6.2.3.2.1

Factor $2$ out of $12$ .

$\frac{1}{3} y^{3} + C_{1} = \frac{2 \cdot 1}{2 \cdot 6} u^{\frac{3}{2}} + C_{2}$

Step 2.3.6.2.3.2.2

Cancel the common factor.

$\frac{1}{3} y^{3} + C_{1} = \frac{2 \cdot 1}{2 \cdot 6} u^{\frac{3}{2}} + C_{2}$

Step 2.3.6.2.3.2.3

Rewrite the expression.

$\frac{1}{3} y^{3} + C_{1} = \frac{1}{6} u^{\frac{3}{2}} + C_{2}$

Step 2.3.7

Replace all occurrences of $u$ with $4 x - 1$ .

$\frac{1}{3} y^{3} + C_{1} = \frac{1}{6} {(4 x - 1)}^{\frac{3}{2}} + C_{2}$

Step 2.4

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

$\frac{1}{3} y^{3} = \frac{1}{6} {(4 x - 1)}^{\frac{3}{2}} + K$