Equivalence between derivatives

Solution 1:

The substitution in the last line looks really wrong. You can't handle index summing that way. If you write this using $\sum$ instead of Einstein's notation, you'll see that.

Using chain rule:

1. LHS: \begin{equation} x^\mu\frac{\partial f(\lambda x)}{\partial x^\mu}=x^\mu\frac{\partial f(\lambda x)}{\partial \lambda x^\nu}\frac{\partial \lambda x^\nu}{\partial x^\mu}=x^\mu\frac{\partial f(\lambda x)}{\partial \lambda x^\nu}\lambda\delta^\nu_\mu=\lambda x^\mu \frac{\partial f(\lambda x)}{\partial \lambda x^\mu} \end{equation}
2. RHS: \begin{equation} \lambda\frac{\partial f(\lambda x)}{\partial \lambda}=\lambda\frac{\partial f(\lambda x)}{\partial \lambda x^\mu}\frac{\partial \lambda x^\mu}{\partial \lambda}=\lambda\frac{\partial f(\lambda x)}{\partial \lambda x^\mu}x^\mu=\lambda x^\mu\frac{\partial f(\lambda x)}{\partial \lambda x^\mu} \end{equation} So LHS=RHS.