Hướng dẫn giải Bài 1 (Trang 176 SGK Toán Đại số & Giải tích 11)

Tìm các đạo hàm sau:

$a, y = \frac{x^{3}}{3} - \frac{x^{2}}{2} + x - 5 b, y = \frac{2}{x} - \frac{4}{x^{2}} + \frac{5}{x^{3}} - \frac{6}{7 x^{4}} c, y = \frac{3 x^{2} - 6 x + 7}{4 x} d, y = (\frac{2}{x} + 3 x) (\sqrt{x} - 1) e, y = \frac{1 + \sqrt{x}}{1 - \sqrt{x}} f, y = \frac{- x^{2} + 7 x + 5}{x^{2} - 3 x}$

Giải

$a, y' = (\frac{x^{3}}{3})' - (\frac{x^{2}}{2})' + (x)' - (5)' = \frac{3 x^{2}}{3} - \frac{2 x}{2} + 1 = x^{2} - x + 1 b, y' = (\frac{2}{x})' - (\frac{4}{x^{2}})' + (\frac{5}{x^{3}})' - (\frac{6}{7 x^{4}})' = - \frac{2 . (x)'}{x^{2}} + \frac{4 (x^{2})'}{x^{4}} - \frac{5 . (x^{3})'}{x^{6}} + \frac{6}{7} . \frac{(x^{4})'}{x^{8}} = \frac{- 2}{x^{2}} + \frac{4.2 x}{x^{4}} - \frac{5.3 x^{2}}{x^{6}} + \frac{6}{7} . \frac{4 x^{3}}{x^{8}} = \frac{- 2}{x^{2}} + \frac{8}{x^{3}} - \frac{15}{x^{4}} + \frac{24}{7 x^{5}} c, y' = (\frac{3 x^{2} - 6 x + 7}{4 x}) = (\frac{3}{4} x - \frac{3}{2} + \frac{7}{4 x}) = (\frac{3}{4} x)' - (\frac{3}{2})' + (\frac{7}{4 x})' = \frac{3}{4} - 0 - \frac{7}{4 x} = \frac{3}{4} - \frac{7}{4 x^{2}}$

$d, y' = [(\frac{2}{x} + 3 \sqrt{x}) (\sqrt{x} - 1)]' = (\frac{2}{x} + 3 \sqrt{x})' (\sqrt{x} - 1) + (\frac{2}{x} + 3 \sqrt{x}) (\sqrt{x} - 1)' = (\frac{- 2}{x^{2}} + \frac{3}{2 \sqrt{x}}) (\sqrt{x} - 1) + (\frac{2}{x} + 3 \sqrt{x}) . \frac{1}{2 \sqrt{x}} = \frac{- 2}{x \sqrt{x}} + \frac{3}{2} + \frac{2}{x^{2}} - \frac{3}{2 \sqrt{x}} + \frac{1}{x \sqrt{x}} + \frac{3}{2} = \frac{- 1}{x \sqrt{x}} + \frac{2}{x^{2}} - \frac{3}{2 \sqrt{x}} + 3 e, y' = (\frac{1 + \sqrt{x}}{1 - \sqrt{x}})' = (\frac{2}{1 - \sqrt{x}} - 1)' = (\frac{2}{1 - \sqrt{x}})' = \frac{- 2 . (1 - \sqrt{x})}{{(1 - \sqrt{x})}^{2}} = \frac{- 2 . (\frac{- 1}{2 \sqrt{x}})}{{(1 - \sqrt{x})}^{2}} = \frac{1}{\sqrt{x} {(1 - \sqrt{x})}^{2}} f' y' = \frac{- 4 x^{2} - 10 x + 15}{{(x^{2} - 3 x)}^{2}}$