For the function $B_i=A_{\text{i}\text{j}\text{k}}{x}_j{x}_k$ where $A_{\text{i}\text{j}\text{k}}$ is constant, show that

del$\frac{B_i}{d}el{x}_j=(A_{\text{i}\text{j}\text{k}}+A_{\text{i}\text{k}\text{j}})x_k.$ Simplify the derivative for the case $A_{\text{i}\text{j}\text{k}}=A_{\text{i}\text{k}\text{j}}.$