why does (1/x)*(dy/dx) - (y/(x^2)) equal (d/dx)*(y/x)?

Question

thanks for your help in advance

Answer 1

d/dx(y/x)
use the product rule:
[d/dx[(f(x)g(x)]=f(x)g'(x)+g(x)f'(x)]
pretend that y is a function of x and differentiate
=(1/x)*(d/dx)(y)+(y)(-1/(x^2))
by the "advanced" chain rule [d/dx(y)=dy/dx]
=(1/x)(dy/dx)-(y/(x^2))

Answer 2

This is just the quotient rule:
d/dx (y/x) = [(dy/dx) (x) - (y) (1)] / x^2
= (1/x) dy/dx - y / x^2

You can also look at it as the product rule for y (x^(-1)):
d/dx (y x^(-1)) = (dy/dx) (x^(-1)) + y (-1)x^(-2)
= (1/x) dy/dx - y / x^2.