Derivative of exponential in index notation

Question

I'm looking to reproduce \begin{align} \partial_{j_m,j_n}\bigg|_{\bf{j}=0}\exp\left(\frac12\bf{j}^\top\bf{B}\bf{j}\right) = B_{mn} \end{align} where $B_{mn}=B_{nm}$ is a real, symmetric, positive-definite $N\times N$ matrix. I have tried the following, and I know this is incorrect due to the surplus of indices. \begin{align} \partial_{j_m,j_n}\bigg|_{\bf{j}=0}\exp\left(\frac12j_mB_{mn}j_n\right) &= \left[\partial_{j_m} \frac12B_{mn}j_m\exp\left(\frac12j_mB_{mn}j_n\right)\right]_{\bf{j}=0} \\ &= \left[\frac12B_{mn}\exp\left(\frac12j_mB_{mn}j_n\right) + \frac14B_{mn}j_mB_{mn}j_n\exp\left(\frac12j_mB_{mn}j_n\right)\right]_{\bf{j}=0} \\ &= \frac12B_{mn} \end{align}

I'm naively working with the indices it seems. Can someone clarify my mistakes?

(PS. Please edit the title to something more descriptive if you can think of something.)

With Jiaqi Li's answer, I think I've understood how to go about this: \begin{align} \partial_{j_m,j_n}\exp\left(\frac12j_rB_{rs}j_s\right) &= \frac12\partial_{j_m}\left[\left(B_{ns}j_s+j_rB_{rn}\right)\exp\left(\frac12j_rB_{rs}j_s\right) \right] \\ &= \frac12\left[\left(B_{nm}+B_{mn}\right)\exp\left(\frac12j_rB_{rs}j_s\right) \\ \qquad\qquad\qquad\qquad+ \frac12\left(B_{ns}j_s+j_rB_{rn}\right)\left(B_{ms}j_s+j_rB_{rm}\right)\exp\left(\frac12j_rB_{rs}j_s\right) \right] \\ \end{align}

This appears to evaluate to the desired result for $\bf{B}$ symemtric. The two major mistakes were the initial usage of the same indices inside the exponential and forgetting that $\frac{\partial x_i}{\partial x_j} = \delta_{ij}$.

Answer 1

Using doubled index sum notation the result is straightforward. $$ \left.\frac{\partial}{\partial x_m}\frac{\partial}{\partial x_n} e^{\frac12 x_iB_{ij}x_j}\right|_{x=0}\\ =\left.\frac{\partial}{\partial x_m} \frac 12 (B_{nq}x_q+x_pB_{pn})e^{\frac12 x_iB_{ij}x_j}\right|_{x=0}\\ =\left.(\frac 12 (B_{nm}+B_{mn})+\frac 14 (B_{nq'}x_{q'}+x_{p'}B_{p'n})(B_{mq''}x_{q''}+x_{p''} B_{p''m}))e^{\frac12 x_iB_{ij}x_j}\right|_{x=0}\\ =\frac 12 (B_{nm}+B_{mn})\\ $$ The given result holds when $B$ is symmetric.

Answer 2

The way you work with indices are incorrect. It should be $$\partial_{j_m,j_n}\bigg|_{\mathbf{j=0}}\exp\left(\frac12\bf{j}^\top\bf{B}\bf{j}\right)=\partial_{j_m,j_n}\bigg|_{\mathbf{j=0}}\exp\left(\frac12j_rB_{rs}j_s\right)$$ since when a index appears twice, it means summation over all possible values. See the wikipedia page Einstein notation. So in the $\exp()$ function, you use indices to imply summation. But when taking the partial derivative, you don't want summation over all possible values for $m,n$. Therefore, you need to use different indices for the summation and the partial derivative.