# How do you find the antiderivative of 4/sqrtx?

Nov 6, 2016

Rewrite as $4 {x}^{- \frac{1}{2}}$.

#### Explanation:

The first step is to recognize that $\frac{1}{\sqrt{x}}$ is equivalent to ${x}^{- \frac{1}{2}}$ by the laws of exponents, where $\frac{1}{x} ^ a$ is equivalent to ${x}^{-} a$. Once we realize this, it becomes much easier to manage.

When we take the derivative of an exponential term, we bring down the power (i.e. multiply the term you are deriving by the value of its exponent) and reduce the power by one. When we take an antiderivative of an exponential term, we want to do the opposite. In essence, we are "undoing" the derivative. Thus, we need to increase the power by one, and multiply the term by the reciprocal of that final exponent.

Because this function we want to "anti-derive" has a negative exponent, we need to make sure we pay a little extra attention. The exponent is $- \frac{1}{2}$, and adding one to this would result in an exponent of $\frac{1}{2}$. If we were to derive ${x}^{\frac{1}{2}}$ we would bring down the $\frac{1}{2}$ power and reduce it by one. This would result in $4 \left(\frac{1}{2} {x}^{- \frac{1}{2}}\right)$.

However, we can see that our original function does not have that $\frac{1}{2}$ constant, so we'll need to do something to counteract that. We can multiply the term by the reciprocal of it's power, which we determined to be $\frac{1}{2}$. The reciprocal of $\frac{1}{2}$ is $2$.

Because this would be an indefinite integral (i.e. having no specific values for which its variables should be defined), we need to account for any constants that could have been present when the derivative was taken, since when we take the derivative of a constant, it reduces to 0. How could we know that the antiderivative is $4 \cdot 2 {x}^{\frac{1}{2}}$ for example, and not $\left(4 \cdot 2 {x}^{\frac{1}{2}}\right) + 42$? After all, if we took the derivative of that, we would end up with the original function we wanted to "anti-derive" also: ${x}^{- \frac{1}{2}}$. This is the case for any constant you might pick.

We can account for this by adding $+ C$ at the end of our answer.
$8 {x}^{\frac{1}{2}} + C$