Giải bài 5.76 - 5.77 trang 210 - SBT Đại số và Giải tích lớp 11
5.76 Tìm đạo hàm của \(g\left( \varphi \right)=\dfrac{\cos \varphi +\sin \varphi }{1-\cos \varphi } \)
\(A.g'\left( \varphi \right)=\dfrac{\cos \varphi -\sin \varphi -1}{{{\left( 1-\cos \varphi \right)}^{2}}}\)
\(B.g'\left( \varphi \right)=\cos \varphi -\sin \varphi \)
\( C.g'\left( \varphi \right)=\dfrac{\sin \varphi +\cos \varphi -1}{{{\left( 1-\cos \varphi \right)}^{2}}}\)
\(D.g'\left( \varphi \right)=1-\sin \varphi -\cos \varphi \)
5.77. Cho hàm số \(y=\cot \sqrt{1+{{x}^{2}}} \).Tính \(y’(1)\)
\(A.\dfrac{1}{\sqrt{2}{{\sin }^{2}}\sqrt{3}}\)
\(B.-\dfrac{1}{\sqrt{2}{{\sin }^{2}}\sqrt{2}}\)
\(C.\dfrac{1}{2\sqrt{2+{{\sin }^{2}}\sqrt{2}}}\)
\(D.\sqrt{{{\sin }^{2}}\sqrt{2}}\)
5.76.
Ta có:
\(\begin{aligned} & g'\left( \varphi \right)=\dfrac{\left( \cos \varphi +\sin \varphi \right)'.\left( 1-\cos \varphi \right)-\left( \cos \varphi +\sin \varphi \right)\left( 1-\cos \varphi \right)'}{{{\left( 1-\cos \varphi \right)}^{2}}} \\ & =\dfrac{\left( \cos \varphi -\sin \varphi \right).\left( 1-\cos \varphi \right)-\left( \cos \varphi +\sin \varphi \right).\sin \varphi }{{{\left( 1-\cos \varphi \right)}^{2}}} \\ & =\dfrac{\cos \varphi -{{\cos }^{2}}\varphi -\sin \varphi +\sin \varphi \cos \varphi -\cos \varphi \sin \varphi -{{\sin }^{2}}\varphi }{{{\left( 1-\cos \varphi \right)}^{2}}} \\ & =\dfrac{\cos \varphi -\sin \varphi -1}{{{\left( 1-\cos \varphi \right)}^{2}}} \\ \end{aligned}\)
Chọn A.
5.77
Ta có:
\(\begin{aligned} & y'=\left( \cot \sqrt{1+{{x}^{2}}} \right)' \\ & =\left( \sqrt{1+{{x}^{2}}} \right)'.\left( -\dfrac{1}{{{\sin }^{2}}\sqrt{1+{{x}^{2}}}} \right) \\ & =\dfrac{2x}{2\sqrt{1+{{x}^{2}}}}.\left( -\dfrac{1}{{{\sin }^{2}}\sqrt{1+{{x}^{2}}}} \right) \\ & =\dfrac{-x}{\sqrt{1+{{x}^{2}}}.{{\sin }^{2}}\sqrt{1+{{x}^{2}}}} \\ \end{aligned} \)
Vậy
\(y'\left( 1 \right)=\dfrac{-1}{\sqrt{1+1}.{{\sin }^{2}}\sqrt{1+1}}=-\dfrac{1}{\sqrt{2}{{\sin }^{2}}\sqrt{2}} \)
Chọn B