rt,在学矩阵微积分。 就是证明这个东西,网上查不到: df(x1(t),x2(t),⋯ ,xn(t))dt=∑k=1n∂f∂xkdxkdt\frac{\mathrm{d} f(x_1(t),x_2(t),\cdots,x_n(t))}{\mathrm{d} t}=\sum_{k=1}^n\frac{\partial f}{\partial x_k}\frac{\mathrm{d}x_k}{\mathrm{d}t}dtdf(x1(t),x2(t),⋯,xn(t))=∑k=1n∂xk∂fdtdxk