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

1 Answer
Jul 2, 2017

#(dy)/(dx)=(e^(xy)+y(x-y^2)e^(xy)-y)/(1+x+2ye^(xy)-x(x-y^2)e^(xy))#

Explanation:

Implicit differentiation is a special case of the chain rule for derivatives. Generally differentiation problems involve functions i.e. #y=f(x)# - written explicitly as functions of #x#. However, some functions y are written implicitly as functions of #x#. So what we do is to treat #y# as #y=y(x)# and use chain rule. This means differentiating #y# w.r.t. #y#, but as we have to derive w.r.t. #x#, as per chain rule, we multiply it by #(dy)/(dx)#.

Here we have #y=(x-y^2)e^(xy)-xy#

#:.(dy)/(dx)=(1-2y(dy)/(dx))e^(xy)+(x-y^2)e^(xy)xx(y+x(dy)/(dx))-(y+x(dy)/(dx))#

#=e^(xy)-2ye^(xy)(dy)/(dx)+y(x-y^2)e^(xy)+x(x-y^2)e^(xy)(dy)/(dx)-y-x(dy)/(dx)#

or #(dy)/(dx)+2ye^(xy)(dy)/(dx)-x(x-y^2)e^(xy)(dy)/(dx)+x(dy)/(dx)=e^(xy)+y(x-y^2)e^(xy)-y#

or #(dy)/(dx)=(e^(xy)+y(x-y^2)e^(xy)-y)/(1+x+2ye^(xy)-x(x-y^2)e^(xy))#