求導數

熱心網友

y'={2^tan[6^(1/x)]}'+[lg(1-x^2)]'=ln2*{2^tan[6^(1/x)]}*{sec[6^(1/x)]}^2*ln6*(-1/x^2)+1/ln10*1/(1-x^2)*(-2x)=-ln2*ln6/x^2*{2^tan[6^(1/x)]}/[x*cos[6^(1/x)]}^2-2x/[ln10*(1-x^2)]