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

Question

Aurora Alvarado · Accepted Answer

# 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) )#

Ezekiel Alexander · Answer

undefined To differentiate the equation $ -3 = \frac{e^{x^2 - y^2}}{x^2 - 2y} $ with respect to both $ x $ and $ y $, you can use the quotient rule and the chain rule.

Differentiating with respect to $ x $, you get:

$$ 0 = \frac{d}{dx}\left(\frac{e^{x^2 - y^2}}{x^2 - 2y}\right) $$

Applying the quotient rule and chain rule, you get:

$$ 0 = \frac{(x^2 - 2y)\frac{d}{dx}(e^{x^2 - y^2}) - e^{x^2 - y^2}\frac{d}{dx}(x^2 - 2y)}{(x^2 - 2y)^2} $$

Similarly, differentiating with respect to $ y $, you get:

$$ 0 = \frac{(x^2 - 2y)\frac{d}{dy}(e^{x^2 - y^2}) - e^{x^2 - y^2}\frac{d}{dy}(x^2 - 2y)}{(x^2 - 2y)^2} $$

You can then solve these equations for $ \frac{d}{dx}(e^{x^2 - y^2}) $ and $ \frac{d}{dy}(e^{x^2 - y^2}) $ using the chain rule.