Bài 3 trang 169 SGK Đại số và Giải tích 11

Đề bài

Tìm đạo hàm của các hàm số sau:

$\begin{array}{l}a)\,\,y = 5\sin x - 3\cos x\\b)\,\,y = \dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}\\c)\,\,y = x\cot x\\d)\,\,y = \dfrac{{\sin x}}{x} + \dfrac{x}{{\sin x}}\\e)\,\,y = \sqrt {1 + 2\tan x} \\f)\,\,y = \sin \sqrt {1 + {x^2}} \end{array}$

Lời giải chi tiết

$\begin{array}{l} a)\,\,y = 5\sin x - 3\cos x\\ \Rightarrow y' = 5\left( {\sin x} \right)' - 3\left( {\cos x} \right)'\\y' = 5\cos x - 3.\left( { - \sin x} \right)\\y' = 5\cos x + 3\sin x\\ b)\,\,y = \dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}\\ \Rightarrow y' = \dfrac{{\left( {\sin x + \cos x} \right)'\left( {\sin x - \cos x} \right) - \left( {\sin x + \cos x} \right)\left( {\sin x - \cos x} \right)'}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\y' = \dfrac{{\left( {\cos x - \sin x} \right)\left( {\sin x - \cos x} \right) - \left( {\sin x + \cos x} \right)\left( {\cos x + \sin x} \right)}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\ y' = \dfrac{{2\sin x\cos x - 1 - 1 - 2\sin x\cos x}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\ y' = \dfrac{{ - 2}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\ c)\,\,y = x\cot x\\ \Rightarrow y' = \left( x \right)'.\cot x + x.\left( {\cot x} \right)'\\y' = \cot x + x.\left( { - \dfrac{1}{{{{\sin }^2}x}}} \right)\\y' = \cot x - \dfrac{x}{{{{\sin }^2}x}}\\ d)\,\,y = \dfrac{{\sin x}}{x} + \dfrac{x}{{\sin x}}\\ \Rightarrow y' = \left( {\dfrac{{\sin x}}{x}} \right)' + \left( {\dfrac{x}{{\sin x}}} \right)'\\y' = \dfrac{{\left( {\sin x} \right)'.x - \sin x.\left( {x} \right)'}}{{{x^2}}} + \dfrac{{\left( x \right)'.\sin x - x.\left( {\sin x} \right)'}}{{{{\sin }^2}x}}\\y' = \dfrac{{x\cos x - \sin x}}{{{x^2}}} + \dfrac{{\sin x - x\cos x}}{{{{\sin }^2}x}}\\ y' = \left( {x\cos x - \sin x} \right)\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{{{\sin }^2}x}}} \right)\\ e)\,\,y = \sqrt {1 + 2\tan x} \\ \Rightarrow y' = \dfrac{{\left( {1 + 2\tan x} \right)'}}{{2\sqrt {1 + 2\tan x} }}\\y' = \dfrac{{2\left( {\tan x} \right)'}}{{2\sqrt {1 + 2\tan x} }}\\y' = \dfrac{{\left( {\tan x} \right)'}}{{\sqrt {1 + 2\tan x} }}\\y' = \dfrac{{\dfrac{1}{{{{\cos }^2}x}}}}{{\sqrt {1 + 2\tan x} }}\\ y' = \dfrac{1}{{{{\cos }^2}x.\sqrt {1 + 2\tan x} }}\\ f)\,\,y = \sin \sqrt {1 + {x^2}} \\ \Rightarrow y' = \cos \sqrt {1 + {x^2}} .\left( {\sqrt {1 + {x^2}} } \right)'\\y' = \cos \sqrt {1 + {x^2}} .\dfrac{{\left( {1 + {x^2}} \right)'}}{{2\sqrt {1 + {x^2}} }}\\y' = \cos \sqrt {1 + {x^2}} .\dfrac{{2x}}{{2\sqrt {1 + {x^2}} }}\\ y' = \dfrac{x}{{\sqrt {1 + {x^2}} }}\cos \sqrt {1 + {x^2}} \end{array}$

Loigiaihay.com