How do you implicitly differentiate #2= e^(xy^2-xy)-y^2x^3+y #?
2 Answers
See the answer below:
# dy/dx = (3x^2y^2-e^(xy^2-xy)(y^2-y))/(e^(xy^2-xy)(2xy-x)-2x^3y+1)#
Explanation:
There is another (often faster) approach using partial derivatives. Suppose we cannot find
# (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y)) #
We have:
# e^(xy^2-xy)-x^3y^2+y=2 #
So Let
# (partial F)/(partial x) = e^(xy^2-xy)(y^2-y)-3x^2y^2 #
# (partial F)/(partial y) = e^(xy^2-xy)(2xy-x)-2x^3y+1 #
And so:
# dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y) #
# \ \ \ \ \ \ = - (e^(xy^2-xy)(y^2-y)-3x^2y^2)/(e^(xy^2-xy)(2xy-x)-2x^3y+1 )#
# \ \ \ \ \ \ = (3x^2y^2-e^(xy^2-xy)(y^2-y))/(e^(xy^2-xy)(2xy-x)-2x^3y+1)#