If f(x) = \begin{vmatrix} cosx & x & 1 \ 2sinx & x^2 & 2x \ tanx & x & 1 \end{vmatrix}, then lim_{x→0} f'(x)/x.

f(x) = \begin{vmatrix} cosx & x & 1 \ 2sinx & x^2 & 2x \ tanx & x & 1 \end{vmatrix} = cosx(x^2 - 2x^2) - x(2sinx - 2x tanx) + 1(2xsinx - x^2 tanx) = -x^2 cosx - 2xsinx + 2x^2 tanx + 2xsinx - x^2 tanx = x^2 tanx - x^2 cosx = x^2(tanx - cosx)
f'(x) = 2x(tanx - cosx) + x^2(sec^2x + sinx)
\therefore lim_{x→0} f'(x)/x = lim_{x→0} \frac{2x(tanx - cosx) + x^2(sec^2x + sinx)}{x} = lim_{x→0} 2(tanx - cosx) + x(sec^2x + sinx) = 2(0 - 1) + 0 = -2
Hence, \therefore lim_{x→0} f'(x)/x = -2