+0  
 
0
70
1
avatar+276 

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

Veteran  Mar 6, 2018

Best Answer 

 #1
avatar+19344 
+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

heureka  Mar 7, 2018
Sort: 

1+0 Answers

 #1
avatar+19344 
+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

28 Online Users

New Privacy Policy (May 2018)
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  Privacy Policy