Tính đạo hàm của các hàm số sau: a) y= x sin x/ 1- tan x ;

Trả lời

a, $y^{\text{'}}=\frac{\left(x\sin x\right)\text{'}\left(1-\tan x\right)+\left(x\sin x\right){\left(1-\tan x\right)}^{\text{'}}}{{\left(1-\tan x\right)}^2}$

$$\begin{gathered} =\frac{\left(\sin x+x\cos x\right)\left(1-\tan x\right)+\left(x\sin x\right)\left(-\frac{1}{{\cos }^{2}x}\right)}{{\left(1-\tan x\right)}^2} \\ =\frac{\sin x+x\cos x-\sin x\tan x-x\sin x+\frac{x\sin x}{{\cos }^{2}x}}{{\left(1-\tan x\right)}^2} \\ =\frac{\sin x+x\cos x-\sin x\tan x+x\sin x\left(-1+\frac{1}{{\cos }^{2}x}\right)}{{\left(1-\tan x\right)}^2} \end{gathered}$$

$$\begin{gathered} =\frac{\sin x+x\cos x-\sin x\tan x+x\sin x\frac{1-{\cos }^{2}x}{{\cos }^{2}x}}{{\left(1-\tan x\right)}^2} \\ =\frac{\sin x+x\cos x-\sin x\tan x+x\sin x\frac{{\sin }^{2}x}{{\cos }^{2}x}}{{\left(1-\tan x\right)}^2} \\ =\frac{\sin x+x\cos x-\sin x\tan x+x\sin x{\tan }^{2}x}{{\left(1-\tan x\right)}^2} \end{gathered}$$