View Question - 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, 20150 users composing answers..
Best Answer
#1exy = x + y
Using implicit differentiation, we have
yexy + xy'exy = 1 + y'
y' ( xexy - 1) = 1 - yexy
y' = [ 1 - yexy ] / [ xexy - 1]
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2+0 Answers
#1exy = x + y
Using implicit differentiation, we have
yexy + xy'exy = 1 + y'
y' ( xexy - 1) = 1 - yexy
y' = [ 1 - yexy ] / [ xexy - 1]
![]()
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 Post New Answer1 Online Users
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