If f(x)=cos⁡xx12sinxx22xtan⁡xx1 then limx→0 f′(x)x=

f(x)=cos⁡xx2−2x2−x[2sin⁡x−2xtan⁡x]+12xsin⁡x−x2tan⁡x=−x2cos⁡x−2xsin⁡x+2x2tan⁡x+2xsin⁡x−x2tan⁡x=−x2cos⁡x+x2tan⁡xf1(x)=−2xcos⁡x+x2sin⁡x+2xtan⁡x+x2sec2⁡xlimx→0 f1(x)x=limx→0 −2xcos⁡x+x2sin⁡x+2xtan⁡x+x2sec2⁡xx=limx→0 −2cos⁡x+xsin⁡x+2tan⁡x+xsec2⁡x=−2+0+0+0=−2