im not sure if this is a plug in chug later on. please help
implicit differentiation
\(\begin{array}{|rcll|} \hline e^y &=& x \quad & | \quad \text{chain rule} \\ e^y\cdot y' &=& 1 \\ y' &=& \dfrac{1}{e^y} \quad & | \quad \dfrac{1}{e^y}= x \\ y' &=& \dfrac{1}{x} \\ \hline \end{array}\)
