#(d^2y)/(dx^2)# is simply asking for the double derivative or #f''(x)#
We have:
#x+y=xy# apply #d/dx# on both sides
#d/dx[x+y]=d/dx[xy]#
The product rule:
#d/dx(f(x)*g(x))=f'(x)*g(x)+f(x)*g'(x)#
The power rule:
#x^n=nx^(n-1)# where #n# is a constant.
#=>1*x^(1-1)+dy/dx=1*x^(1*1)*y+x*dy/dx#
#=>1+dy/dx=y+x*dy/dx#
#=>1+dy/dx-y=x*dy/dx#
#=>1-y=x*dy/dx-dy/dx#
#=>1-y=dy/dx(x-1)#
#=>(1-y)/(x-1)=dy/dx#
Apply #d/dx# to both sides again.
#=>d/dx[(1-y)/(x-1)]=d/dx[dy/dx]#
We use the division rule:
#d/dx[(f(x))/(g(x))]=(f'(x)*g(x)-f(x)*g'(x))/(g(x))^2#
#=>(d/dx(1-y)*(x-1)-(1-y)*d/dx(x-1))/(x-1)^2=(d^2y)/(dx^2)#
#=>(-dy/dx*(x-1)-(1-y)*1)/(x-1)^2=(d^2y)/(dx^2)#
#=>(-dy/dx*(x-1)-(1-y))/(x-1)^2=(d^2y)/(dx^2)#
We know what #dy/dx# is!
#=>(-(1-y)/cancel(x-1)*cancel(x-1)-(1-y))/(x-1)^2=(d^2y)/(dx^2)#
#=>(-1+y-1+y)/(x-1)^2=(d^2y)/(dx^2)#
#=>(2y-2)/(x^2-2x+1)=(d^2y)/(dx^2)#
or you may prefer
#(2(y-1))/(x-1)^2=(d^2y)/(dx^2)#
