Câu 1: A
Câu 2: D
Câu 3: C
Câu 4: D
Câu 5: B
Câu 6: B
Câu 7: \(y=\dfrac{2x}{x-1}\)
=>\(y'=\dfrac{\left(2x\right)'\left(x-1\right)-2x\cdot\left(x-1\right)'}{\left(x-1\right)^2}=\dfrac{2\left(x-1\right)-2x}{\left(x-1\right)^2}\)
\(=\dfrac{-2}{\left(x-1\right)^2}\)
=>Chọn C
Câu 8:
\(f\left(x\right)=\dfrac{3x+1}{\sqrt{x^2+4}}\)
=>\(f'\left(x\right)=\dfrac{\left(3x+1\right)'\cdot\sqrt{x^2+4}-\left(3x+1\right)\cdot\left(\sqrt{x^2+4}\right)'}{x^2+4}\)
=>\(f'\left(x\right)=\dfrac{3\sqrt{x^2+4}-\left(3x+1\right)\cdot\dfrac{\left(x^2+4\right)'}{2\sqrt{x^2+4}}}{x^2+4}\)
=>\(f'\left(x\right)=\dfrac{3\sqrt{x^2+4}-\left(3x+1\right)\cdot\dfrac{x}{\sqrt{x^2+4}}}{x^2+4}\)
=>\(f'\left(0\right)=\dfrac{3\sqrt{0^2+4}-\left(3\cdot0+1\right)\cdot\dfrac{0}{\sqrt{0^2+4}}}{0^2+4}=\dfrac{6}{4}=\dfrac{3}{2}\)
=>Chọn C
