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what is $${\frac{{\mathtt{dy}}}{{\mathtt{dx}}}}$$ when $${{\mathtt{e}}}^{\left({\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{y}}\right)} = {\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}}$$?

Guest Mar 19, 2015

Best Answer 

 #1
avatar+92365 
+10

exy  = x + y    

Using implicit differentiation, we have

yexy + xy'exy  = 1 + y'

y' ( xexy - 1) = 1 - yexy

y' = [ 1 - yexy  ] / [ xexy - 1]  

 

  

CPhill  Mar 19, 2015
 #1
avatar+92365 
+10
Best Answer

exy  = x + y    

Using implicit differentiation, we have

yexy + xy'exy  = 1 + y'

y' ( xexy - 1) = 1 - yexy

y' = [ 1 - yexy  ] / [ xexy - 1]  

 

  

CPhill  Mar 19, 2015
 #2
avatar+94081 
+5

Thanks Chris,

I really need practice at these so it is great when they come onto the forum.

I even got the same answer as you - isn't that great :)

 

Thanks anon for giving us this question :)

Melody  Mar 19, 2015

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