Bài 4 trang 169 SGK Đại số và Giải tích 11

Đề bài

Tìm đạo hàm của các hàm số sau:

$\begin{array}{l} a)\,\,y = \left( {9 - 2x} \right)\left( {2{x^3} - 9{x^2} + 1} \right)\\ b)\,\,y = \left( {6\sqrt x - \dfrac{1}{{{x^2}}}} \right)\left( {7x - 3} \right)\\ c)\,\,y = \left( {x - 2} \right)\sqrt {{x^2} + 1} \\ d)\,y = {\tan ^2}x - {\cot}{x^2}\\ e)\,\,y = \cos \dfrac{x}{{1 + x}} \end{array}$

Lời giải chi tiết

$\begin{array}{l} a)\,\,y = \left( {9 - 2x} \right)\left( {2{x^3} - 9{x^2} + 1} \right)\\y' = \left( {9 - 2x} \right)'\left( {2{x^3} - 9{x^2} + 1} \right) \\+ \left( {9 - 2x} \right)\left( {2{x^3} - 9{x^2} + 1} \right)'\\ = - 2\left( {2{x^3} - 9{x^2} + 1} \right) + \left( {9 - 2x} \right)\left( {6{x^2} - 18x} \right)\\ = - 4{x^3} + 18{x^2} - 2 + 54{x^2} - 162x - 12{x^3} + 36{x^2}\\ = - 16{x^3} + 108{x^2} - 162x - 2\\ b)\,\,y = \left( {6\sqrt x - \dfrac{1}{{{x^2}}}} \right)\left( {7x - 3} \right)\\y' = \left( {6\sqrt x  - \dfrac{1}{{{x^2}}}} \right)'\left( {7x - 3} \right) + \left( {6\sqrt x  - \dfrac{1}{{{x^2}}}} \right)\left( {7x - 3} \right)'\\  = \left( {6.\dfrac{1}{{2\sqrt x }} - \dfrac{{ - \left( {{x^2}} \right)'}}{{{{\left( {{x^2}} \right)}^2}}}} \right)\left( {7x - 3} \right) + \left( {6\sqrt x  - \dfrac{1}{{{x^2}}}} \right).7\\ = \left( {\dfrac{3}{{\sqrt x }} + \dfrac{{2x}}{{{x^4}}}} \right)\left( {7x - 3} \right) + 7\left( {6\sqrt x  - \dfrac{1}{{{x^2}}}} \right)\\= \left( {\dfrac{3}{{\sqrt x }} + \dfrac{2}{{{x^3}}}} \right)\left( {7x - 3} \right) + 7\left( {6\sqrt x - \dfrac{1}{{{x^2}}}} \right)\\ = 21\sqrt x - \dfrac{9}{{\sqrt x }} + \dfrac{{14}}{{{x^2}}} - \dfrac{6}{{{x^3}}} + 42\sqrt x - \dfrac{7}{{{x^2}}}\\ = \dfrac{{ - 6}}{{{x^3}}} + \dfrac{7}{{{x^2}}} + 63\sqrt x - \dfrac{9}{{\sqrt x }}\\ c)\,\,y = \left( {x - 2} \right)\sqrt {{x^2} + 1} \\y' = \left( {x - 2} \right)'\sqrt {{x^2} + 1}  + \left( {x - 2} \right)\left( {\sqrt {{x^2} + 1} } \right)'\\ = 1.\sqrt {{x^2} + 1}  + \left( {x - 2} \right).\dfrac{{\left( {{x^2} + 1} \right)'}}{{2\sqrt {{x^2} + 1} }} \\= \sqrt {{x^2} + 1}  + \left( {x - 2} \right).\dfrac{{2x}}{{2\sqrt {{x^2} + 1} }}\\  = \sqrt {{x^2} + 1} + \left( {x - 2} \right)\dfrac{x}{{\sqrt {{x^2} + 1} }}\\  = \dfrac{{{x^2} + 1 + {x^2} - 2x}}{{\sqrt {{x^2} + 1} }}\\ = \dfrac{{2{x^2} - 2x + 1}}{{\sqrt {{x^2} + 1} }}\\ d)\,y = {\tan ^2}x - \cot {x^2}\\y' = \left( {{{\tan }^2}x} \right)' - \left( {\cot {x^2}} \right)'\\ = 2\tan x.\left( {\tan x} \right)'  - \left( {{x^2}} \right)'.\dfrac{{ - 1}}{{\sin ^2 {x^2}}}\\ = 2\tan x.\dfrac{1}{{{{\cos }^2}x}} + \dfrac{{2x}}{{{{\sin }^2}x^2}}\\  = \dfrac{{2\sin x}}{{{{\cos }^3}x}} + \dfrac{{2x}}{{{{\sin }^2}x^2}}\\ e)y = \cos \dfrac{x}{{1 + x}}\\y' = \left( {\dfrac{x}{{x + 1}}} \right)'.\left( { - \sin \dfrac{x}{{x + 1}}} \right)\\ =  - \sin \left( {\dfrac{x}{{1 + x}}} \right).\dfrac{{\left( x \right)'\left( {1 + x} \right) - x.\left( {1 + x} \right)'}}{{{{\left( {1 + x} \right)}^2}}}\\ = - \sin \dfrac{x}{{1 + x}}.\left( {\dfrac{{1 + x - x}}{{{{\left( {1 + x} \right)}^2}}}} \right)\\ = - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}.\sin \dfrac{x}{{1 + x}} \end{array}$

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