# What is the antiderivative of sqrtx?

Jun 6, 2015

One law of exponentials states that ${a}^{\frac{m}{n}} = \sqrt[n]{{a}^{m}}$

Thus, we can rewrite $\sqrt{x}$ as ${x}^{\frac{1}{2}}$

Derivating it using the product rule, which states $y = {a}^{n}$, thus $y ' = n \cdot {a}^{n - 1}$, we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\frac{1}{2} - 1} = {x}^{-} \left(\frac{1}{2}\right)$

However, as another law of exponentials states, ${a}^{-} n = \frac{1}{a} ^ n$. Thus,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x} ^ \left(\frac{1}{2}\right) = \frac{1}{\sqrt{x}}$

Jun 6, 2015

You can write the square root as:
$\sqrt{x} = {x}^{\frac{1}{2}}$
Now integrate this using the fact that:
$\int {x}^{n} \mathrm{dx} = \frac{{x}^{n + 1}}{n + 1} + c$
Getting:
$\int {x}^{\frac{1}{2}} \mathrm{dx} = \frac{{x}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + c =$
$= \frac{{x}^{\frac{3}{2}}}{\frac{3}{2}} + c = \frac{2}{3} \left(x \sqrt{x}\right) + c$