How do you differentiate #-2=e^(x^2-y^2)/y#?

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Answer 1 · 2016-06-21T11:01:22

# y' = (2xy) / (1 + 2y^2)#

start by making life easier for yourself

this is #-2y = e^{x^2 - y^2} #

and then further simplification #ln(-2y) = x^2 - y^2#

and then we differentiate to get

#1/(-2y) (-2) y' = 2x - 2y y'#

#(1/y + 2y) y' = 2x#

# y' = (2xy) / (1 + 2y^2)#