Tìm đạo hàm của hàm số \(y=\dfrac{1}{\sqrt{x^2+1}}\).
![]() | \(y'=\dfrac{x}{\left(x^2+1\right)\sqrt{x^2+1}}\) |
![]() | \(y'=\dfrac{-x}{\left(x^2+1\right)\sqrt{x^2+1}}\) |
![]() | \(y'=\dfrac{x}{2\left(x^2+1\right)\sqrt{x^2+1}}\) |
![]() | \(y'=-\dfrac{x\left(x^2+1\right)}{\sqrt{x^2+1}}\) |
Chọn phương án B.
\(\begin{aligned}
y'&=-\dfrac{\sqrt{x^2+1}'}{\sqrt{x^2+1}^2}\\
&=-\dfrac{\left(x^2+1\right)'}{\left(x^2+1\right)2\sqrt{x^2+1}}\\
&=-\dfrac{2x}{2\left(x^2+1\right)\sqrt{x^2+1}}\\
&=\dfrac{-x}{\left(x^2+1\right)\sqrt{x^2+1}}.
\end{aligned}\)