## Task Condition

Find the $n$-th order derivative.

$y=\sqrt{x}$
## Solution

$y^{\prime}=(\sqrt{x})^{\prime}=\left(x^{\frac{1}{2}}\right)^{\prime}=\frac{1}{2} \cdot x^{-\frac{1}{2}}$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1}{2} \cdot x^{-\frac{1}{2}}\right)^{\prime}=\frac{1}{2} \cdot\left(-\frac{1}{2}\right) \cdot x^{-\frac{3}{2}}=-\frac{1}{2^{2}} \cdot x^{-\frac{3}{2}}$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-\frac{1}{2^{2}} \cdot x^{-\frac{3}{2}}\right)^{\prime}=-\frac{1}{2^{2}} \cdot\left(-\frac{3}{2}\right) \cdot x^{-\frac{5}{2}}=\frac{1 \cdot 3}{2^{4}} \cdot x^{-\frac{5}{2}}$

$y^{(I V)}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(\frac{3}{2^{4}} \cdot x^{-\frac{5}{2}}\right)^{\prime}=\frac{3}{2^{4}} \cdot\left(-\frac{5}{2}\right) \cdot x^{-\frac{7}{2}}=\frac{1 \cdot 3 \cdot 5}{2^{5}} \cdot x^{-\frac{7}{2}}$

$\cdots$

Obviously,

$y^{(n)}=(-1)^{n-1} \cdot \frac{\prod_{k=1}^{n-1}(2 k-1)}{2^{(n+1)}} \cdot x^{-\frac{2 n-1}{2}}=\frac{(-1)^{n-1} \cdot \prod_{k=1}^{n-1}(2 k-1)}{2^{(n+1)} \cdot x^{n-1} \cdot \sqrt{x}}$

## Problem Kuznetsov Differentiation 18-9