# How do you find the second derivative of sqrtx?

Jul 27, 2016

The Second Derivative$= - \frac{1}{4 {\left(\sqrt{x}\right)}^{3}}$.

#### Explanation:

Let $y = f \left(x\right) = \sqrt{x}$. Then, $y = f \left(x\right) = {x}^{\frac{1}{2}}$.

Now, we can use the Standard Form $\left(1\right) : \frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n \cdot {x}^{n - 1}$.

In our case, as n=1/2, we have,

The First Derivative of $y = f \left(x\right) = {x}^{\frac{1}{2}}$ denoted by $\frac{\mathrm{dy}}{\mathrm{dx}}$, or, $f ' \left(x\right)$, is given by,

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

Now, the Second Derivative, denoted by, $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 \mathmr{and} , f ' ' \left(x\right)$ is defined by,

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left\{\frac{\mathrm{dy}}{\mathrm{dx}}\right\} , \mathmr{and} , \left(f '\right) ' \left(x\right)$.

It simply means that to find the second derivative of a given fun. $f$, we have to differentiate $f '$, i.e., the (first) derivative of $f$ again.

So, for $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$, we will find $\frac{d}{\mathrm{dx}} \left\{\frac{1}{2} \cdot {x}^{- \frac{1}{2}}\right\}$, by using the

Std. Form $\left(1\right)$ and, the Working Rule$\left(2\right) : \frac{d}{\mathrm{dx}} \left\{k \cdot u\right\} = k \cdot \frac{\mathrm{du}}{\mathrm{dx}}$, where, $k$ is a const., &, $u$, a fun. of $x$.

Hence, $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left\{\frac{1}{2} \cdot {x}^{- \frac{1}{2}}\right\} = \frac{1}{2} \left\{- \frac{1}{2} \cdot {x}^{- \frac{1}{2} - 1}\right\} = - \frac{1}{4} \cdot {x}^{- \frac{3}{2}}$.

Using radicals, $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{4 {\left(\sqrt{x}\right)}^{3}}$.