Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.5 Higher Derivatives - Exercises - Page 136: 33


$\dfrac{(-1)^n}{2^n} \times \dfrac{1 \cdot 3 \cdot 5 ... (2n-1)}{x^{\frac{2n+1}{2}}}$

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}}}$
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