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

Question

1585 views around the world

You can reuse this answer
Creative Commons License

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

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

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

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