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im not sure if this is a plug in chug later on. please help

 Mar 6, 2018

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 #1
avatar+22181 
+1

implicit differentiation

\(\begin{array}{|rcll|} \hline e^y &=& x \quad & | \quad \text{chain rule} \\ e^y\cdot y' &=& 1 \\ y' &=& \dfrac{1}{e^y} \quad & | \quad \dfrac{1}{e^y}= x \\ y' &=& \dfrac{1}{x} \\ \hline \end{array}\)

 

laugh

 Mar 7, 2018
 #1
avatar+22181 
+1
Best Answer

implicit differentiation

\(\begin{array}{|rcll|} \hline e^y &=& x \quad & | \quad \text{chain rule} \\ e^y\cdot y' &=& 1 \\ y' &=& \dfrac{1}{e^y} \quad & | \quad \dfrac{1}{e^y}= x \\ y' &=& \dfrac{1}{x} \\ \hline \end{array}\)

 

laugh

heureka Mar 7, 2018

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