$\text{ If }f\left(x\right)=\left|\begin{array}{lll}\cos x& x& 1\\ 2\sin & x^2& 2x\\ \tan x& x& 1\end{array}\right|\text{ then }\underset{x\to 0}{\lim }\frac{f^{\text{'}}\left(x\right)}{x}=$

$\begin{array}{c}f\left(x\right)=\left|\begin{array}{l}\cos xx1\\ 2\sin x^{2}2x\\ \tan xx1\end{array}\right|\\ =\cos x\left[x^2-2x^2\right]-x\left[2\sin x-2x\tan x\right]+1\left[2x\sin x-x^2\tan x\right]\\ =-x^2\cos x-2x\sin x+2x^2\tan x+2x\sin x-x^2\tan x\\ =-x^2\cos x+x^2\tan x\\ f^1\left(x\right)=-2x\cos x+x^2\sin x+2x\tan x+x^2{\sec }^{2}x\\ \underset{x\to 0}{\lim }\frac{f^1\left(x\right)}{x}=\underset{x\to 0}{\lim }\frac{-2x\cos x+x^2\sin x+2x\tan x+x^2{\sec }^{2}x}{x}\\ =\underset{x\to 0}{\lim }-2\cos x+x\sin x+2\tan x+x^{ }{\sec }^{2}x\\ =-2+0+0+0\\ =-2\end{array}$