How do you differentiate #-3=e^(x^2-y^2)/(x^2-2y)#?

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Answer 1 · 2017-01-25T15:00:09

# y' = (2x ( e^(x^2-y^2) + 3))/(6 + 2y e^(x^2-y^2) )#

Firstly, simplify by re-writing as:

#-3(x^2-2y)=e^(x^2-y^2)#

Then implicitly differentiate wrt x.

#-6x +6y' =(2x - 2y y')e^(x^2-y^2)#

And re-order:
# 6y' + 2y y'e^(x^2-y^2) =2x e^(x^2-y^2) + 6x#

# y' (6 + 2y e^(x^2-y^2) ) =2x ( e^(x^2-y^2) + 3)#

#implies y' = (2x ( e^(x^2-y^2) + 3))/(6 + 2y e^(x^2-y^2) )#