Extremize using total differential

Question

So I was playing around with lagrange multipliers and came to following question: suppose one wants to find an extreme value of $f(x_i)$ under the condition $g(x_i)=0$. First I define $L(x_i,\lambda)\equiv f(x_i)+\lambda g(x_i)$. To find the extremum, I then set $\frac{\partial L}{\partial x_i}=0$ and $\frac{\partial L}{\partial \lambda}=0$. My question is, if I can also set $dL=0$. Quite obviously, if the partial derivatives are zero, then also the total differential is, but my question is if this is true also the other way.

Answer 1

It is true precisely because the partial derivatives are the components of the differential, i.e. $$\langle dL(x_i,\lambda), h \rangle = \frac{\partial L}{\partial x_i}(x_i,\lambda) h_1 + \frac{\partial L}{\partial\lambda}(x_i,\lambda)h_2,$$ with $h=(h_1,h_2)$. Therefore since $dL(x_i,\lambda)=0$ is equivalent to $\langle dL(x_i,\lambda),h\rangle=0 \; \forall h$, it follows that the partial derivatives are zero ($h=(1,0)^\top$ gives $\partial L/\partial x_i = 0$ and $h=(0,1)^\top$ gives $\partial L/\partial \lambda = 0$).