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

1 Answer
Jul 27, 2017

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

Hence, #(dy)/(dx)=e^(xy)xx2(x-y)(1-(dy)/(dx))+(x-y)^2e^(xy)(x(dy)/(dx)+y)#

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

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

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

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