So I have to find the partial derivative $\frac{d}{dRL}$ which means that all variables except RL are constants. In the end I get $\frac {E^2((Rs+RL)^2-2RL)}{(Rs+RL)^3}$ and the answer should be $\frac {E^2*(Rs-RL)}{(Rs+RL)^3}$
I arrived at that answer by using $\frac {f(x)}{g(x)}$ -> $\frac {g*f' - f*g'}{g^2}$
Could someone please help me because I've been stuck on this for an hour
I hope this can help:
\begin{equation} \frac{\partial}{\partial RL} (\dfrac{E^{2}RL}{(Rs+RL)^{2}})= \frac{\partial}{\partial RL} (E^{2}RL\cdot\dfrac{1}{(Rs+RL)^{2}}) \end{equation}
For convenience, I'm going to apply the product rule $(f\cdot g)'=f'\cdot g+f\cdot g'$ \begin{equation} =\frac{\partial (E^{2}RL)}{\partial RL}\cdot(Rs+RL)^{-2}+E^{2}RL\cdot\frac{\partial ((Rs+RL)^{-2})}{\partial RL} \end{equation}
\begin{equation} =E^{2}\cdot(Rs+RL)^{-2}-2E^{2}RL\cdot(Rs+RL)^{-3} \end{equation}
\begin{equation} =\frac{E^{2}}{(Rs+RL)^{2}}-\frac{2E^{2}RL}{(Rs+RL)^{3}} \end{equation}
\begin{equation} =\frac{E^{2}(Rs+RL)-2E^{2}RL}{(Rs+RL)^{3}} \end{equation}
\begin{equation} =\frac{E^{2}Rs+E^{2}RL-2E^{2}RL}{(Rs+RL)^{3}} \end{equation}
\begin{equation} =\frac{E^{2}Rs-E^{2}RL}{(Rs+RL)^{3}} \end{equation}
\begin{equation} =\frac{E^{2}(Rs-RL)}{(Rs+RL)^{3}} \end{equation}