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

2 Answers

See the answer below:
enter image source here

Apr 4, 2017

# 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 #y# explicitly as a function of #x#, only implicitly through the equation #F(x, y) = 0# which defines #y# as a function of #x, y = y(x)#. Therefore we can write #F(x, y) = 0# as #F(x, y(x)) = 0#. Differentiating both sides of this, using the partial chain rule gives us

# (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 # F(x,y) = e^(xy^2-xy)-x^3y^2+y-2 #; Then;

# (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)#