Question

7．函數(shù)f（x）=$\sqrt{x\sqrt{x\sqrt{x}}}$（x＞0）的導(dǎo)數(shù)為f′（x）=$\frac{7}{8}$•${x}^{-\frac{1}{8}}$．

Answer 1

分析 化簡去根號f（x）=$（x\sqrt{x\sqrt{x}}）^{\frac{1}{2}}$=${x}^{\frac{1}{2}}$•${x}^{\frac{1}{4}}$•${x}^{\frac{1}{8}}$=${x}^{\frac{7}{8}}$，從而求導(dǎo)．

解答 解：f（x）=$\sqrt{x\sqrt{x\sqrt{x}}}$
=$（x\sqrt{x\sqrt{x}}）^{\frac{1}{2}}$
=${x}^{\frac{1}{2}}$$（x\sqrt{x}）^{\frac{1}{4}}$
=${x}^{\frac{1}{2}}$•${x}^{\frac{1}{4}}$•${x}^{\frac{1}{8}}$=${x}^{\frac{7}{8}}$，
故f′（x）=$\frac{7}{8}$•${x}^{-\frac{1}{8}}$，
故答案為：$\frac{7}{8}$•${x}^{-\frac{1}{8}}$．

點(diǎn)評 本題考查了學(xué)生的化簡運(yùn)算能力及導(dǎo)數(shù)的計(jì)算．