==+2xsinx−x2tanx)=⇒∴=== So, x→0limxf′(x)=−2f(x)=cosx2sinxtanxxx2x12x1cosx(x2−2x2)−x(2sinx−2xtanx)−x2cosx−2xsinx+2x2tanxx2tanx−x2tanxf′(x)=2x(tanx−cosx)+x2(sec2x+sinx)x→0limxf′(x)x→0limx2x(tanx−cosx)+x2(sec2x+sinx)x→0lim(tanx−cosx)+x(sec2x+sinx)2(0−1)+0=−2