Tính đạo hàm của hàm số \(y=\left(\frac{2x-1}{x+1}\right)^{\sqrt{2}}\)
\(\sqrt{2}.\left(\frac{2x-1}{x+1}\right)^{\sqrt{2}-1}\) \(-\frac{3\sqrt{2}}{\left(x+1\right)^2}.\left(\frac{2x-1}{x+1}\right)^{\sqrt{2}-1}\) \(\frac{3\sqrt{2}}{\left(x+1\right)^2}.\left(\frac{2x-1}{x+1}\right)^{\sqrt{2}-1}\) \(\left(\frac{3}{\left(x+1\right)^2}\right)^{\sqrt{2}}\) Hướng dẫn giải:Sử dụng \(\left(u^{\alpha}\right)'=\alpha u^{\alpha-1}u'\) và \(\left(\frac{u}{v}\right)'=\frac{v.u'-u.v'}{v^2}\)
\(y'=\sqrt{2}.\frac{\left(x+1\right).2-\left(2x-1\right).1}{\left(x+1\right)^2}.\left(\frac{2x-1}{x+1}\right)^{\sqrt{2}-1}\)
\(=\sqrt{2}.\frac{3}{\left(x+1\right)^2}.\left(\frac{2x-1}{x+1}\right)^{\sqrt{2}-1}\)
\(=\frac{3\sqrt{2}}{\left(x+1\right)^2}.\left(\frac{2x-1}{x+1}\right)^{\sqrt{2}-1}\)
