Giải bài 2 trang 176 – SGK môn Đại số và Giải tích 11
Tìm đạo hàm của các hàm số sau:
\(a)\,y=2\sqrt{x}\sin x-\dfrac{\cos x}{x};\)
\( b)\,y=\dfrac{3\cos x}{2x+1};\)
\(c)\,y=\dfrac{{{t}^{2}}+2{\cos t}}{\sin t};\)
\(d)\,y=\dfrac{2\cos \varphi -\sin \varphi }{3\sin \varphi +\cos \varphi };\)
\(e)\,y=\dfrac{\tan x}{\sin x+2};\)
\(f)\,y=\dfrac{\cot x}{2\sqrt{x}+1}.\)
Gợi ý:
Áp dụng công thức đạo hàm của hàm số lượng giác:
\((\cos x)'=-\sin x\\ (\sin x)'=\cos x\\ (\tan x)'=\dfrac 1 {\cos^2 x}\\ (\cot x)'=\dfrac 1 {\sin ^2 x}\)
a)
\(\begin{align} y'&=\dfrac{2}{2\sqrt{x}}\sin x+2\sqrt{x}\cos x-\dfrac{-\sin x.x-\cos x}{{{x}^{2}}} \\ & =\dfrac{\sin x}{\sqrt{x}}+2\sqrt{x}\cos x+\dfrac{x\sin x+\cos x}{{{x}^{2}}} \\ & =\dfrac{x\sqrt{x}\sin x+2{{x}^{2}}\sqrt{x}\cos x+x\sin x+\cos x}{{{x}^{2}}} \\ & =\dfrac{\left( \sqrt{x}+1 \right)x\sin x+\left( 2{{x}^{2}}\sqrt{x}+1 \right)\cos x}{{{x}^{2}}} \\ \end{align}\)
b)
\(\begin{align} y'&=\dfrac{-3\sin x\left( 2x+1 \right)-2.3\cos x}{{{\left( 2x+1 \right)}^{2}}} \\ & =\dfrac{-3\sin x\left( 2x+1 \right)-6\cos x}{{{\left( 2x+1 \right)}^{2}}} \\ \end{align}\)
c)
\(\begin{align} y'&=\dfrac{\left( 2t-2\operatorname{sint} \right)\sin t-\left( {{t}^{2}}+2\cos t \right)\operatorname{cost}}{{{\sin }^{2}}t} \\ & =\dfrac{2t\sin t-2{{\sin }^{2}}t-{{t}^{2}}\cos t-2{{\cot }^{2}}t}{{{\sin }^{2}}t} \\ & =\dfrac{2t\sin t-{{t}^{2}}\cos t-2}{{{\sin }^{2}}t} \\ \end{align}\)
d)
\(\begin{align} y'&=\dfrac{\left( -2\sin \varphi -\cos \varphi \right)\left( 3\sin \varphi +\cos \varphi \right)-\left( 2\cos \varphi -\sin \varphi \right)\left( 3\cos \varphi -\sin \varphi \right)}{{{\left( 3\sin \varphi +\cos \varphi \right)}^{2}}} \\ & =\dfrac{-7(\sin^2\varphi+\cos^2\varphi)}{{{\left( 3\sin \varphi +\cos \varphi \right)}^{2}}}\\&=\dfrac{-7}{{{\left( 3\sin \varphi +\cos \varphi \right)}^{2}}} \\ \end{align}\)
e)
\(\begin{align} y&'=\dfrac{\dfrac{1}{{{\cos }^{2}}x}\left( \sin x+2 \right)-\tan x\cos x}{{{\left( \sin x+2 \right)}^{2}}} \\ & =\dfrac{\sin x+2-\sin x{{\cos }^{2}}x}{{{\cos }^{2}}x{{\left( \sin x+2 \right)}^{2}}} \\ & =\dfrac{2+{{\sin }^{3}}x}{{{\cos }^{2}}x{{\left( \sin x+2 \right)}^{2}}} \\ \end{align} \)
f)
\(\begin{align} y'&=\dfrac{-\dfrac{1}{{{\sin }^{2}}x}\left( 2\sqrt{x}-1 \right)-\dfrac{1}{\sqrt{x}}\cot x}{{{\left( 2\sqrt{x}-1 \right)}^{2}}} \\ & =\dfrac{\sqrt{x}\left( 1-2\sqrt{x} \right)-\cot x{{\sin }^{2}}x}{\sqrt{x}\left( 2\sqrt{x}-1 \right)^2{{\sin }^{2}}x} \\ & =\dfrac{\sqrt{x}\left( 1-2\sqrt{x} \right)-\sin x\cos x}{\sqrt{x}\left( 2\sqrt{x}-1 \right)^2{{\sin }^{2}}x} \\ \end{align} \)