I have the following equation:
$$Y=[(A_L L)^{\frac{\sigma-1}{\sigma}}+(A_H H)^{\frac{\sigma-1}{\sigma}}]^{{\frac{\sigma}{\sigma-1}}}$$
I have to compute $\frac{\partial Y}{\partial L}$.
What I got is: $\frac{\partial Y}{\partial L} = A_L^{\frac{\sigma-1}{\sigma}} [(A_L L)^{\frac{\sigma-1}{\sigma}}+(A_H H)^{\frac{\sigma-1}{\sigma}}]^{\frac{1}{\sigma -1}} L^{\frac{-1}{\sigma}} $
However, the author proposes the following rearrangement:
$\frac{\partial Y}{\partial L}=A_L^{\frac{\sigma-1}{\sigma}} [(A_L)^{\frac{\sigma-1}{\sigma}}+(A_H)^{\frac{\sigma-1}{\sigma}} (H/L)^{\frac{\sigma-1}{\sigma}}]^{\frac{1}{\sigma -1}} $
How to get the latter expression?
We will simplify the question: Let us substitute $$A_L^{\tfrac{\sigma-1}{\sigma}}=k, \frac{\sigma-1}{\sigma}=u, (A_HH)^\tfrac{\sigma-1}{\sigma}=m^u.$$ Then we get, $$\frac{\partial Y}{\partial L}=k\bigg(kL^u+m^u\bigg)^{\tfrac{1}{\sigma-1}}L^{\tfrac{-1}{\sigma}}$$ As I mentioned in the comment, we take $L^\tfrac{\sigma-1}{\sigma}$, i.e. $L^u$ common, so we get $$\frac{\partial Y}{\partial L}=k\bigg(kL^u+m^u\bigg)^{\tfrac{1}{\sigma-1}}L^{\tfrac{-1}{\sigma}}=k\bigg(k+\bigg(\frac mL\bigg)^u\bigg)^{\tfrac{1}{\sigma-1}}L^{\tfrac{u}{\sigma-1}}\cdot L^{\tfrac{-1}{\sigma}}$$ Now we note that $\dfrac{u}{\sigma-1}-\dfrac{1}{\sigma}=0$ so we have $$\frac{\partial Y}{\partial L}=k\bigg(k+\bigg(\frac mL\bigg)^u\bigg)^{\tfrac{1}{\sigma-1}}=A_L^{\tfrac{\sigma-1}{\sigma}}\left(A_L^{\tfrac{\sigma-1}{\sigma}}+(A_HH/L)^{\tfrac{\sigma-1}{\sigma}}\right)^{\tfrac{1}{\sigma-1}}.$$