Дұрыс жауапДұрыс жауап: f′(x)=tgxxln10+lgxcos2xf'(x) = \frac{tgx}{xln10} + \frac{lgx}{cos^{2}x}Жауап нұсқаларыA.f′(x)=cosx⋅lgx+lgxxf'(x) = cosx \cdot lgx + \frac{lgx}{x}B.f′(x)=cosx⋅lgx+lgxcos2xf'(x) = cosx \cdot lgx + \frac{lgx}{cos^{2}x}C.f′(x)=tgxxln10+lgxcos2xf'(x) = \frac{tgx}{xln10} + \frac{lgx}{cos^{2}x}дұрысD.f′(x)=sinx⋅lgx+lgxxf'(x) = sinx \cdot lgx + \frac{lgx}{x}E.f′(x)=tgx⋅lgx+cosxxf'(x) = tgx \cdot lgx + \frac{cosx}{x}