我想把三个符号表达式转换为矩阵，比较长，而且里面带导数，我想系数为导数，提出它的系数矩阵。什么办法都可以，请各位大神指教。
2*l1(t)*diff(l1(t), t) == 2*(tan(y(t))*z(t) + b*sin(x(t))*sin(y(t)))*(tan(y(t))*diff(z(t), t) + z(t)*(tan(y(t))^2 + 1)*diff(y(t), t) + b*cos(x(t))*sin(y(t))*diff(x(t), t) + b*cos(y(t))*sin(x(t))*diff(y(t), t)) + 2*(z(t) + b*cos(y(t))*sin(x(t)))*(diff(z(t), t) + b*cos(x(t))*cos(y(t))*diff(x(t)
就是像这样的方程，我想提出diff(x(t), t),diff(y(t), t),diff(z(t), t)前面的相对应的系数，拜托各位大神了。
2*l1(t)*diff(l1(t), t) == 2*(tan(y(t))*z(t) + b*sin(x(t))*sin(y(t)))*(tan(y(t))*diff(z(t), t) + z(t)*(tan(y(t))^2 + 1)*diff(y(t), t) + b*cos(x(t))*sin(y(t))*diff(x(t), t) + b*cos(y(t))*sin(x(t))*diff(y(t), t)) + 2*(z(t) + b*cos(y(t))*sin(x(t)))*(diff(z(t), t) + b*cos(x(t))*cos(y(t))*diff(x(t)
就是像这样的方程，我想提出diff(x(t), t),diff(y(t), t),diff(z(t), t)前面的相对应的系数，拜托各位大神了。
