Find the Function f(x) = square root of x+6

$f (x) = \sqrt{x + 6}$

The function $F (x)$ can be found by evaluating the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Let $u = x + 6$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

Let $u = x + 6$ . Find $\frac{d u}{d x}$ .

Differentiate $x + 6$ .

$\frac{d}{d x} [x + 6]$

By the Sum Rule, the derivative of $x + 6$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [6]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [6]$

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

$1 + \frac{d}{d x} [6]$

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$1 + 0$

Add $1$ and $0$ .

$1$

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

$\int \sqrt{u} d u$

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

$\int u^{\frac{1}{2}} d u$

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

$\frac{2}{3} u^{\frac{3}{2}} + C$

Replace all occurrences of $u$ with $x + 6$ .

$\frac{2}{3} {(x + 6)}^{\frac{3}{2}} + C$

The function $F$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

$F (x) = \frac{2}{3} \cdot {(x + 6)}^{\frac{3}{2}} + C$