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)#.
Differentiating #sqrt(x)+sqrt(y)+sqrt(xy)=1#, we get
#1/(2sqrtx)+1/(2sqrty)xx(dy)/(dx)+1/(2sqrt(xy))(1xxy+x xx(dy)/(dx))=0#
Observe that as we have differentiated #sqrt(xy)# w.r.t. #xy#, we have used Chain rule again differentiated #xy# w.r.t. #x# using product formula. The above then becomes
#1/(2sqrtx)+1/(2sqrty)xx(dy)/(dx)+y/(2sqrt(xy))+x/(2sqrt(xy)) xx(dy)/(dx)=0#
or #(dy)/(dx)(1/(2sqrty)+x/(2sqrt(xy)))=-(1/(2sqrtx)+y/(2sqrt(xy)))#
or #(dy)/(dx)=-(1/(2sqrtx)+y/(2sqrt(xy)))/(1/(2sqrty)+x/(2sqrt(xy)))#
= #-(sqrty+y)/(sqrtx+x)# #-># multiplying numerator and denominator by #2sqrt(xy)#
