How do you implicitly differentiate #y= xy^2 + x ^2 e^(x -2y) #?
3 Answers
Explanation:
Take the derivative of the equation except for every time you differentiate
Note:
Use the product rule again:
Distribute the x^2:
Separate the terms with
Factor out the
Divide both sides by
# dy/dx = (y^2+(x^2+2x)e^(x-2y))/(1+2x^2e^(x-2y)-2xy)#
Explanation:
Using the Implicit Function Theorem we have:
# dy/dx = -((partial f)/dx)/((partial f)/dy) #
Where:
# f(x,y) = 0 #
We have:
# \ \ \ \ \ y=xy^2 + x^2e^(x-2y) #
# :. y=xy^2 + x^2e^x e^(-2y) #
Let:
# f(x,y)= xy^2 + x^2e^x e^(-2y)-y #
Then the partial derivatives are:
# (partial f)/(partial x) = y^2+(x^2)(e^x e^(-2y))+(2x)(e^x e^(-2y)) #
# \ \ \ \ \ \ \ = y^2+(x^2)(e^(x-2y))+(2x)(e^(x-2y)) #
# \ \ \ \ \ \ \ = y^2+(x^2+2x)e^(x-2y) #
And:
# (partial f)/(partial y) = 2xy-2x^2e^x e^(-2y)-1 #
# \ \ \ \ \ \ \ = 2xy-2x^2e^(x-2y)-1 #
And so:
# dy/dx = -(y^2+(x^2+2x)e^(x-2y))/(2xy-2x^2e^(x-2y)-1)#
# \ \ \ \ \ \ = (y^2+(x^2+2x)e^(x-2y))/(1+2x^2e^(x-2y)-2xy)#