Hàm số nào sau đây có đạo hàm bằng \(\dfrac{1}{\cos x}\) ?
\(f\left(x\right)=\ln\dfrac{1}{\sin x}\).\(g\left(x\right)=\ln\dfrac{1+\sin x}{\cos x}\).\(h\left(x\right)=\ln\dfrac{1}{\cos x}\).\(k\left(x\right)=\ln\cos x\).Hướng dẫn giải:Ta thử:
\(f'\left(x\right)=\dfrac{\left(\dfrac{1}{\sin x}\right)'}{\dfrac{1}{\sin x}}=\dfrac{\dfrac{-\cos x}{\sin^2x}}{\frac{1}{\sin x}}=-\cot x\)
\(g'\left(x\right)=\dfrac{\left(\dfrac{1+\sin x}{\cos x}\right)'}{\dfrac{1+\sin x}{\cos x}}=\dfrac{\dfrac{1+\sin x}{\cos^2x}}{\dfrac{1+\sin x}{\cos x}}=\dfrac{1}{\cos x}\)
\(h'\left(x\right)=\dfrac{\left(\dfrac{1}{\cos x}\right)'}{\dfrac{1}{\cos x}}=\dfrac{\dfrac{\sin x}{\cos^2x}}{\dfrac{1}{\cos x}}=\dfrac{\sin x}{\cos x}\)
\(k'\left(x\right)=\dfrac{\left(\cos x\right)'}{\cos x}=\dfrac{-\sin x}{\cos x}=-\tan x\)
Đáp số: \(g\left(x\right)=\ln\dfrac{1+\sin x}{\cos x}\).
