Calculus Examples

$\frac{d y}{d x} = \sqrt{x + 49}$

Step 1

Rewrite the equation.

$d y = \sqrt{x + 49} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int \sqrt{x + 49} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \sqrt{x + 49} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u = x + 49$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.1.1

Let $u = x + 49$ . Find $\frac{d u}{d x}$ .

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Step 2.3.1.1.1

Differentiate $x + 49$ .

$\frac{d}{d x} [x + 49]$

Step 2.3.1.1.2

By the Sum Rule, the derivative of $x + 49$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [49]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [49]$

Step 2.3.1.1.3

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

$1 + \frac{d}{d x} [49]$

Step 2.3.1.1.4

Since $49$ is constant with respect to $x$ , the derivative of $49$ with respect to $x$ is $0$ .

$1 + 0$

Step 2.3.1.1.5

Add $1$ and $0$ .

$1$

Step 2.3.1.2

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Replace all occurrences of $u$ with $x + 49$ .

$y + C_{1} = \frac{2}{3} {(x + 49)}^{\frac{3}{2}} + C_{2}$

Step 2.4

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

$y = \frac{2}{3} {(x + 49)}^{\frac{3}{2}} + K$