Answer
$\dfrac{(-1)^n}{2^n} \times \dfrac{1 \cdot 3 \cdot 5 ... (2n-1)}{x^{\frac{2n+1}{2}}}$
Work Step by Step
Since $ f(x)=x^{-1/2}$, then we have
$ f'(x)=(-\dfrac{1}{2})x^{-3/2} \quad f''(x)=(-\dfrac{1}{2})(-\dfrac{3}{2})x^{-5/2},\quad f'''(x)=(-\dfrac{1}{2})(-\dfrac{3}{2})(-\dfrac{5}{2})x^{-7/2}$
Hence, we get the general formula for $ f^{(n)}(x)$ as follows
$ f^{(n)}(x)=(-\dfrac{1}{2})(-\dfrac{3}{2})(-\dfrac{5}{2})(-\dfrac{(2n-1)}{2})x^{-1/2-n}\\=\dfrac{(-1)^n}{2^n} \times \dfrac{1 \cdot 3 \cdot 5... (2n-1)}{x^{\frac{2n+1}{2}}}$
