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# what is dy/dx when e^xy=x+y?

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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

#1
+80935
+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
Sort:

#1
+80935
+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
#2
+91435
+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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