In a simple function like #y=f(x)#, #y# is expressed 'explicitly' in terms of #x#. However, in many cases the variables involved in a function are not linked to each other in an explicit way and they are rather linked through an implicit formula like the one given here.

In such cases, we differentiate both sides of the equality, say w.e.t. #x#, using normal formulas of differentiation such as product, quotient or chain formula, but whenever say #y# is involved, its derivative is put as #(dy)/(dx)#.

The given implicit function is #5=x-1/(1-xy^2)# differentiating it with respect to #x# on each side, we get

#0=1-x-(-1)/((1-xy^2)^2)xxd/dx(1-xy^2)# or

#0=1-x-(-1)/((1-xy^2)^2)xx(-y^2-x xx2yxx(dy)/(dx))# or

#0=1-x-1/((1-xy^2)^2)xx(2xy(dy)/(dx)-y^2)#

multiplying both sides by #(1-xy^2)^2#

#0=(1-x)(1-xy^2)^2-(2xy(dy)/(dx)-y^2)# or

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

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

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