If the function one wishes to differentiate, f(x), can be written as
and h(x) ≠ 0; then, the rule states that the derivative of g(x) / h(x) is equal to the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator:
Or more precisely; for all x in some open set containing the number a, with h(a) ≠ 0; and, such that g '(a) and h '(a) both exist; then, f '(a) exists as well:
The derivative of (4x  2) / (x^{2} + 1) = [(x^{2} + 1)(4)  (4x  2)(2x)] / (x^{2} + 1)^{2} = [(4x^{2} + 4)  (8x^{2}  4x)] / (x^{2} + 1)^{2} = [4x^{2} + 4x + 4] / (x^{2} + 1)^{2}
The derivative of [sin(x)] / x^{2} (when x ≠ 0) is ([cos(x)]x^{2}  [sin(x)](2x)) / x^{4}. For more information regarding the derivatives of trigonometric functions, see: derivative.
Informal Proof A proof of this rule can be derived from Newton's difference quotient: The derivative of [f(x)] / [g(x)] = (the limit as h approaches 0):
To turn this into a proper proof, one has to pick Δx so small that the denominators are all nonzero (and one has to argue that this is always possible because the involved functions are continuous).
Using only the product rule:
The rest is simple algebra to make f'(x) the only term on the left hand side of the equation and to remove f(x) from the right side of the equation.
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