Google
Câu hỏi: Hàm số $y=\log _2 \sqrt{x^2+x}$ có đạo hàm là
A. $y^{\prime}=\dfrac{(2 x+1) \ln 2}{2\left(x^2+x\right)}$.
B. $y^{\prime}=\dfrac{2 x+1}{\left(x^2+x\right)}$.
C. $y^{\prime}=\dfrac{2 x+1}{2\left(x^2+x\right) \ln 2}$.
D. $y^{\prime}=\dfrac{2 x+1}{\left(x^2+x\right) \ln 2}$.
A. $y^{\prime}=\dfrac{(2 x+1) \ln 2}{2\left(x^2+x\right)}$.
B. $y^{\prime}=\dfrac{2 x+1}{\left(x^2+x\right)}$.
C. $y^{\prime}=\dfrac{2 x+1}{2\left(x^2+x\right) \ln 2}$.
D. $y^{\prime}=\dfrac{2 x+1}{\left(x^2+x\right) \ln 2}$.
$\left(\log _2 \sqrt{x^2+x}\right)^{\prime}=\dfrac{\left(\sqrt{x^2+x}\right)^{\prime}}{\sqrt{x^2+x} \ln 2}=\dfrac{2 x+1}{2\left(x^2+1\right) \ln 2}$.
Đáp án C.