# 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)#