If g is differentiable at x, and g(x) does not equal 0, the derivative of 1/g at x is -g′(x) / g(x)2.
The numerator of the difference quotient is 1/g(x+h) - 1/g(x), which we can rewrite, as though we were subtracting two fractions. Actually we need to be a bit careful here. Since g is continuous, it is nonzero through a neighborhood around x, not just at x. So as long as we keep h small, it's ok to have g(x+h) in the denominator. Now, where were we. Oh yes, subtract the two fractions and get this.
g(x)-g(x+h) over g(x+h)×g(x)
Again we need continuity, so that g(x+h) approaches g(x). Invoke the limit product/quotient theorem, and the difference quotient produces -g′/g2.
Combine this with the product rule to show the derivative of the quotient f/g is f′×g-g′×f over g2. This is the quotient rule for differentiation.