Cho hàm số \(f\left(x\right)=-\dfrac{\cos x}{2\sin^2x}+\ln\sqrt{\dfrac{1+\cos x}{\sin x}}\). Tính \(f'\left(\frac{\pi}{4}\right)\) . 1 \(\frac{1}{2}\) \(\sqrt{2}\) \(\frac{1}{\sqrt{2}}\) Hướng dẫn giải: - Tính \(\left(\dfrac{\cos x}{\sin^2x}\right)'\) : đặt \(u=\cos x,v=\sin^2x\) thì \(u'=-\sin x,v'=2\sin x\cos x\) và \(u'v-uv'=-\sin^3x-2\sin x\cos^2x\). Do đó \(\left(\dfrac{\cos x}{\sin^2x}\right)'=\dfrac{-\sin^3x-2\sin x\cos^2x}{\sin^4x}=-\dfrac{\sin^2x+2\cos^2x}{\sin^3x}=-\dfrac{1+\cos^2x}{\sin^3x}\) - Tính \(\left(\ln\sqrt{\dfrac{1+\cos x}{\sin x}}\right)'\): Ta có \(\dfrac{1+\cos x}{\sin x}=\dfrac{2\cos^2\dfrac{x}{2}}{2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\cot\dfrac{x}{2}\) nên \(\ln\sqrt{\dfrac{1+\cos x}{\sin x}}=\dfrac{1}{2}\ln\dfrac{1+\cos x}{\sin x}=\dfrac{1}{2}\ln\cot\dfrac{x}{2}\), do đó \(\left(\ln\sqrt{\dfrac{1+\cos x}{\sin x}}\right)'=\dfrac{1}{2}\left(\ln\cot\dfrac{x}{2}\right)'=\dfrac{1}{2}.\dfrac{-\dfrac{1}{2}.\dfrac{1}{\sin^2\dfrac{x}{2}}}{\cot\dfrac{x}{2}}=\dfrac{-1}{4\sin^2\dfrac{x}{2}\cot\dfrac{x}{2}}=\dfrac{-1}{4\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\dfrac{-1}{2\sin x}\) Vì vậy \(f'\left(x\right)=\dfrac{1+\cos^2x}{2\sin^3x}-\dfrac{1}{2\sin x}=\dfrac{1+\cos^2x-\sin^2x}{2\sin^3x}=\dfrac{\cos^2x}{\sin^3x}=\dfrac{\cot^2x}{\sin x}\). Khi \(x=\dfrac{\pi}{4}\) thì \(\cot x=1,\sin x=\dfrac{1}{\sqrt{2}}\) do đó \(f'\left(\dfrac{\pi}{4}\right)=\sqrt{2}\).
Cho hàm số \(g\left(x\right)=-\dfrac{1}{2\sin^2x}+\ln\tan x\). Tính \(g'\left(\frac{\pi}{6}\right)\) . \(\frac{8}{\sqrt{3}}\) \(\frac{12}{\sqrt{3}}\) \(\frac{16}{\sqrt{3}}\) \(\frac{32}{\sqrt{3}}\) Hướng dẫn giải: Ta có \(\left(\ln\tan x\right)'=\dfrac{\left(\tan x\right)'}{\tan x}=\dfrac{1+\tan^2x}{\tan x}\) và \(\left(-\dfrac{1}{2\sin^2x}\right)'=\dfrac{\left(\sin^2x\right)'}{2\sin^4x}=\dfrac{2\sin x\cos x}{2\sin^4x}=\dfrac{\cos x}{\sin^3x}\) . Từ đó \(g'\left(x\right)=\dfrac{\cos x}{\sin^3x}+\dfrac{1+\tan^2x}{\tan x}\). Khi \(x=\dfrac{\pi}{6}\) thì \(\cos x=\dfrac{\sqrt{3}}{2},\sin x=\dfrac{1}{2},\tan x=\dfrac{1}{\sqrt{3}}\) nên \(g'\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}:\dfrac{1}{8}+\left(1+\dfrac{1}{3}\right):\dfrac{1}{\sqrt{3}}=4\sqrt{3}+\dfrac{4}{\sqrt{3}}=4\left(\sqrt{3}+\dfrac{1}{\sqrt{3}}\right)=\dfrac{16}{\sqrt{3}}\) Đáp số: \(\dfrac{16}{\sqrt{3}}\)
Cho hàm số \(f\left(x\right)=\dfrac{1}{\sqrt{3}}\ln\left(\dfrac{\tan\dfrac{x}{2}+2-\sqrt{3}}{\tan\dfrac{x}{2}+2+\sqrt{3}}\right)\) Tính \(f'\left(\frac{\pi}{2}\right)\). \(\frac{1}{3}\) \(\frac{2}{3}\) 1 \(\frac{4}{3}\) Hướng dẫn giải: Cách 1: Dùng MTCT Cách 2: Đặt \(u=\dfrac{\tan\dfrac{x}{2}+2-\sqrt{2}}{\tan\dfrac{x}{2}+2+\sqrt{3}}\) thì \(f\left(x\right)=\dfrac{1}{\sqrt{3}}\ln u\) nên \(f'\left(x\right)=\dfrac{1}{\sqrt{3}}.\dfrac{u'}{u}\). Tính \(u'\) : Đặt \(t=\tan\dfrac{x}{2}\) thì \(u=\dfrac{t+2-\sqrt{3}}{t+2+\sqrt{3}}\) và \(u'=\left(\dfrac{t+2-\sqrt{3}}{t+2+\sqrt{3}}\right)'.t'\). Vì \(t=\tan\dfrac{x}{2}\Rightarrow t'=\dfrac{1}{2}\left(1+\tan^2\dfrac{x}{2}\right)\) và vì \(\left(\dfrac{t+2-\sqrt{3}}{t+2+\sqrt{3}}\right)'=\dfrac{2\sqrt{3}}{\left(t+2+\sqrt{3}\right)^2}\) nên \(u'=\dfrac{2\sqrt{3}.\dfrac{1}{2}\left(1+\tan^2\dfrac{x}{2}\right)}{\left(\tan\dfrac{x}{2}+2+\sqrt{3}\right)^2}=\dfrac{\sqrt{3}\left(1+\tan^2\dfrac{x}{2}\right)}{\left(\tan\dfrac{x}{2}+2+\sqrt{3}\right)^2}\) Do đó \(f'\left(x\right)=\dfrac{1}{\sqrt{3}}.\dfrac{\sqrt{3}\left(1+\tan^2\dfrac{x}{2}\right)}{\left(\tan\dfrac{x}{2}+2+\sqrt{3}\right)^2}.\dfrac{\tan\dfrac{x}{2}+2+\sqrt{3}}{\tan\dfrac{x}{2}+2-\sqrt{3}}\)\(=\dfrac{1+\tan^2\dfrac{x}{2}}{\left(\tan\dfrac{x}{2}+2\right)^2-3}\) Chú ý rằng khi \(x=\dfrac{\pi}{2}\) thì \(\tan\dfrac{x}{2}=\tan\dfrac{\pi}{4}=1\) nên \(f'\left(\dfrac{\pi}{2}\right)=\dfrac{1+1^2}{\left(1+2\right)^2-3}=\dfrac{1}{3}\) Đáp số: \(\dfrac{1}{3}\)
Cho hàm số \(y=e^{-x}.\sin x\). Khẳng định nào trong các khẳng định sau đây đúng ? \(y'+2y"-2y=0\) \(y"+2y'+2y=0\) \(y"-2y'-2y=0\) \(y'-2y"+2y=0\) Hướng dẫn giải: Có \(y=e^{-x}.\sin x\Rightarrow y'=-e^{-x}\sin x+e^{-x}\cos x\) \(\Rightarrow y"=-y'+\left(e^{-x}\cos x\right)'=e^{-x}\sin x-2e^{-x}\cos x-e^{-x}\sin x=-2e^{-x}\cos x\) a) \(\left\{{}\begin{matrix}y'=e^{-x}\left(-\sin x+\cos x\right)\\2y"=-4e^{-x}\cos x\\-2y=-2e^{-x}\sin x\end{matrix}\right.\) \(\Rightarrow y'+2y"-2y=-3e^{-x}\left(\sin x+\cos x\right)\ne0\) b) \(\left\{{}\begin{matrix}y"=-2e^{-x}\cos x\\2y'=-2e^{-x}\sin x+2e^{-x}\cos x\\2y=2e^{-x}\sin x\end{matrix}\right.\) \(\Rightarrow y"+2y'+2y=0\) c) \(\left\{{}\begin{matrix}y"=-2e^{-x}\cos x\\-2y'=2e^{-x}\sin x-2e^{-x}\cos x\\-2y=-2e^{-x}\sin x\end{matrix}\right.\) \(\Rightarrow y"-2y'-2y=-4e^{-x}\cos x\ne0\) d) \(\left\{{}\begin{matrix}y'=e^{-x}\left(-\sin x+\cos x\right)\\-2y"=4e^{-x}\cos x\\2y=2e^{-x}\sin x\end{matrix}\right.\) \(\Rightarrow y'-2y"+2y=e^{-x}\sin x+5e^{-x}\cos x\ne0\) Đáp số: \(y"+2y'+2y=0\)
Cho hàm số \(y=x.\sin x\). Khẳng định nào trong các khẳng định sau đây đúng ? \(xy-2\left(y'-\sin x\right)+xy"=0\) \(xy'-2\left(y-\sin x\right)+xy"=0\) \(xy-2\left(y"-\sin x\right)+xy'=0\) \(xy'+2\left(y'+\sin x\right)-xy"=0\) Hướng dẫn giải: Ta có \(y=x.\sin x\Rightarrow y'=\sin x+x\cos x\Rightarrow y"=\cos x+\cos x-x\sin x=2\cos x-x\sin x\) a) \(\left\{{}\begin{matrix}xy=x^2\sin x\\-2\left(y'-\sin x\right)=-2x\cos x\\xy"=2x\cos x-x^2\sin x\end{matrix}\right.\) \(\Rightarrow xy-2\left(y'-\sin x\right)+xy"=0\) b) \(\left\{{}\begin{matrix}xy'=x\sin x+x^2\cos x\\-2\left(y-\sin x\right)=-2x\sin x+2\sin x\\xy"=2x\cos x-x^2\sin x\end{matrix}\right.\) \(\Rightarrow xy'-2\left(y-\sin x\right)+xy"=x^2\left(\cos x-\sin x\right)+x\left(-\sin x+2\cos x\right)+2\sin x\ne0\) c) \(\left\{{}\begin{matrix}xy=x^2\sin x\\-2\left(y"-\sin x\right)=-2\left(2\cos x-x\sin x\right)+2\sin x\\xy'=x\sin x+x^2\cos x\end{matrix}\right.\) \(\Rightarrow xy-2\left(y"-\sin x\right)+xy'=x^2\left(\sin x+\cos x\right)+3x\sin x-4\cos x+2\sin x\ne0\) d)\(\left\{{}\begin{matrix}xy'=x\sin x+x^2\cos x\\2\left(y'+\sin x\right)=2\left(2\sin x+x\cos x\right)\\-xy"=-x\left(2\cos x-x\sin x\right)\end{matrix}\right.\) \(\Rightarrow xy'+2\left(y'+\sin x\right)-xy"=x^2\left(\cos x+\sin x\right)+x\sin x+4\sin x\ne0\) Đáp số: \(xy-2\left(y'-\sin x\right)+xy"=0\)
Cho hàm số \(y=e^{\cos x}\). Hãy chọn hệ thức đúng ? \(y'.\cos x+y.\sin x+y"=0\) \(y'.\sin x+y.\cos x+y"=0\) \(y'.\sin x+y''.\cos x+y'=0\) \(y'.\cos x-y.\sin x-y"=0\) Hướng dẫn giải: \(y=e^{\cos x}\Rightarrow y'=-\sin xe^{\cos x}\Rightarrow y"=-\cos xe^{\cos x}+\sin^2xe^{\cos x}\) a) \(y'\cos x=-\sin x\cos xe^{\cos x},y\sin x=\sin xe^{\cos x}\) nên \(y'.\cos x+y.\sin x+y"=e^{\cos x}\left(-\sin x\cos x+\sin x-\cos x+\sin^2x\right)=e^{\cos x}\left(\sin x-\cos x\right)\left(1+\sin x\right)\ne0\) b) \(y'\sin x=-\sin^2xe^{\cos x},y\cos x=\cos x.e^{\cos x}\) nên \(y'.\sin x+y.\cos x+y"=e^{\cos x}\left(-\sin^2x+\cos x-\cos x+\sin^2x\right)=0\) c) \(y'\sin x=-\sin^2xe^{\cos x},y"\cos x=\cos x\left(-\cos x+\sin^2x\right)e^{\cos x},y'=-\sin xe^{\cos x}\) nên \(y'.\sin x+y''.\cos x+y'=e^{\cos x}\left(-\sin^2x+\cos x\left(-\cos x+\sin^2x\right)-\sin x\right)=e^{\cos x}\left(-1-\sin x+\cos x\sin^2x\right)\ne0\) d) \(y'\cos x=-\sin x\cos xe^{\cos x},y\sin x=\sin xe^{\cos x},y"=-\cos xe^{\cos x}+\sin^2xe^{\cos x}\) nên \(y'.\cos x-y.\sin x-y"=e^{\cos x}\left(-\sin x\cos x-\sin x+\cos x-\sin^2x\right)\ne0\) Đáp số: \(y'.\sin x+y.\cos x+y"=0\)
Cho hàm số \(y=\sin\left(\ln x\right)+\cos\left(\ln x\right)\). Khẳng định nào trong các khẳng định sau đây đúng ? \(xy"-x^2y'+y=0\) \(x^2y"-xy'-y=0\) \(x^2y"+xy'+y=0\) \(x^2y"-xy'-y=0\) Hướng dẫn giải: Có \(\left(\sin\ln x\right)'=\dfrac{1}{x}\cos\ln x,\left(\cos\ln x\right)'=-\dfrac{1}{x}\sin\ln x\) nên \(y'=\dfrac{1}{x}\left(\cos\ln x-\sin\ln x\right)\Rightarrow xy'=\cos\ln x-\sin\ln x,x^2y'=x\cos\ln x-x\sin\ln x\) \(y"=-\dfrac{1}{x^2}\left(\cos\ln x-\sin\ln x\right)+\dfrac{1}{x}\left(-\dfrac{1}{x}\sin\ln x-\dfrac{1}{x}\cos\ln x\right)=-\dfrac{1}{x^2}\left(2\cos\ln x\right)\Rightarrow x^2y"=-2\cos\ln x,xy"=\dfrac{-2\cos\ln x}{x}\) a) \(xy"-x^2y'+y=\dfrac{-2\cos\ln x}{x}-\left(x\cos\ln x-x\sin\ln x\right)+\left(\sin\ln x+\cos\ln x\right)\)\(=\left(-\dfrac{2}{x}-x+1\right)\cos\ln x+\left(x+1\right)\sin\ln x\ne0\) b) \(x^2y"-xy'-y=\left(-2\cos\ln x\right)-\left(\cos\ln x-\sin\ln x\right)-\left(\sin\ln x+\cos\ln x\right)=-4\cos\ln x\ne0\) c) \(x^2y"+xy'+y=\left(-2\cos\ln x\right)+\left(\cos\ln x-\sin\ln x\right)+\left(\sin\ln x+\cos\ln x\right)=0\) d) \(x^2y"+xy'-y=\left(-2\cos\ln x\right)+\left(\cos\ln x-\sin\ln x\right)-\left(\sin\ln x+\cos\ln x\right)=-2\left(\cos\ln x+\sin\ln x\right)\ne0\) Đáp số: \(x^2y"+xy'+y=0\)
Cho hàm số \(f\left(x\right)=\dfrac{\cos x}{1-\sin x}\). Tính \(f'\left(\frac{\pi}{6}\right)-f'\left(-\frac{\pi}{6}\right)\) \(\frac{4}{3}\) \(\frac{4}{9}\) \(\frac{8}{9}\) \(\dfrac{2}{5}\) Hướng dẫn giải: Đặt \(u=\cos x,v=1-\sin x\) thì \(u'=-\sin x,v'=-\cos x,u'v-uv'=-\sin x\left(1-\sin x\right)-\cos x.\left(-\cos x\right)=1-\sin x\) \(f'\left(x\right)=\dfrac{u'v-uv'}{v^2}=\dfrac{1-\sin x}{\left(1-\sin x\right)^2}=\dfrac{1}{1-\sin x}\) \(f'\left(\dfrac{\pi}{6}\right)-f'\left(-\dfrac{\pi}{6}\right)=\dfrac{1}{1-\dfrac{1}{2}}-\dfrac{1}{1+\dfrac{1}{2}}=2-\dfrac{2}{3}=\dfrac{4}{3}\)
Cho hàm số \(y=f\left(x\right)=\frac{\cos x}{1+2\sin x}\). Khẳng định nào trong các khẳng định sau là sai? \(f'\left(\frac{\pi}{2}\right)=-\frac{1}{3}\) \(f'\left(-\frac{\pi}{2}\right)=-1\) \(f'\left(\frac{\pi}{6}\right)=-\frac{5}{4}\) \(f'\left(0\right)=-2\) Hướng dẫn giải: \(f'\left(0\right)=-2\)
Viết điều kiện với \(a,b\) để hàm số \(f\left(x\right)=\dfrac{b\cos x+a\sin x+1}{\cos x-\sin x+1}\) có \(f'\left(-\frac{\pi}{4}\right)=0\) . \(a+b=0\) \(a-b=0\) \(2a+b=0\) \(2b+a=0\) Hướng dẫn giải: Đặt \(u=a\sin x+b\cos x+1,v=\cos x-\sin x+1\) thì \(u'=a\cos x-b\sin x,v'=-\sin x-\cos x\). \(u'v-uv'=\left(a\cos x-b\sin x\right)\left(\cos x-\sin x+1\right)-\left(a\sin x+b\cos x+1\right)\left(-\sin x-\cos x\right)\) \(=a\cos^2x+b\sin^2x-\left(a+b\right)\sin x\cos x+a\cos x-b\sin x-\left(-a\sin^2x-b\cos^2x-\left(a+b\right)\sin x\cos x-\sin x-\cos x\right)\) \(=a+b+\left(a+1\right)\cos x-\left(b-1\right)\sin x\) \(f'\left(x\right)=\dfrac{u'v-uv'}{v^2}=\dfrac{a+b+\left(a+1\right)\cos x-\left(b-1\right)\sin x}{\left(\cos x-\sin x+1\right)^2}\), \(f'\left(-\dfrac{\pi}{4}\right)=\dfrac{a+b+\dfrac{\left(a+1\right)}{\sqrt{2}}+\dfrac{\left(b-1\right)}{\sqrt{2}}}{\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}+1\right)^2}\) Điều kiện được thực hiện khi \(a+b+\dfrac{a+1+b-1}{\sqrt{2}}=0\Leftrightarrow\left(a+b\right)\left(1+\dfrac{1}{\sqrt{2}}\right)=0\Leftrightarrow a+b=0\)
