Đạo hàm của hàm số \(y=\dfrac{\cos x-1}{9^{2x}}\) là
\(y'=\dfrac{\sin x-4\left(\cos x-1\right)\ln3}{3^{4x}}\). \(y'=\dfrac{\sin x-2\left(\cos x-1\right)\ln3}{3^{4x}}\). \(y'=-\dfrac{\sin x+4\left(\cos x-1\right)\ln3}{3^{4x}}\). \(y'=-\dfrac{\sin x+2\left(\cos x-1\right)\ln3}{3^{4x}}\). Hướng dẫn giải: Áp dụng \(\left(\dfrac{u}{v}\right)'=\dfrac{v.u'-u.v'}{v^2}\) Ta có: \(y'=\dfrac{9^{2x}.\left(\cos x-1\right)'-\left(\cos x-1\right)\left(9^{2x}\right)'}{\left(9^{2x}\right)^2}\) \(=\dfrac{-9^{2x}\sin x-\left(\cos x-1\right)2.9^{2x}\ln9}{9^{4x}}\) \(=\dfrac{9^{2x}\left[-\sin x-2\left(\cos x-1\right)\ln9\right]}{9^{4x}}\) \(=-\dfrac{\sin x+2\left(\cos x-1\right).\ln3^2}{9^{2x}}\) \(=-\dfrac{\sin x+4\left(\cos x-1\right).\ln3}{3^{4x}}\).