 in calculus, the quotient rule is a method of finding the derivative of a function that is the quotient of two other functions for which derivatives exist. if the function one wishes to differentiate, f(x), can be written as :f(x) = \frac{g(x)}{h(x)} and h(x)\not=0, then the rule states that the derivative of g(x)/h(x) is :f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}. more precisely, if all x in some open set containing the number a satisfy h(x)\not=0, and g'(a) and h'(a) both exist, then f'(a) exists as well and :f'(a)=\frac{g'(a)h(a) - g(a)h'(a)}{[h(a)]^2}. and this can be extended to calculate the second derivative as well (you can prove this by taking the derivative of f(x)=g(x)(h(x))^{-1} twice). the result of this is: :f(x)=\frac{g(x)[h(x)]^2-2g'(x)h(x)h'(x)+g(x)[2[h'(x)]^2-h(x)h(x)]}{[h(x)]^3}. the quotient rule formula can be derived from the product rule and chain rule. == examples == the derivative of (4x - 2)/(x^2 + 1) is: :\begin{align}\frac{d}{dx}\left[\frac{(4x - 2)}{x^2 + 1}\right] &= \frac{(4)(x^2 + 1) - (4x - 2)(2x)}{(x^2 + 1)^2}\\ & = \frac{(4x^2 + 4) - (8x^2 - 4x)}{(x^2 + 1)^2} = \frac{-4x^2 + 4x + 4}{(x^2 + 1)^2}\end{align} in the example above, the choices :g(x) = 4x - 2 :h(x) = x^2 + 1 were made. analogously, the derivative of sin(x)/x2 (when x&nbsp;â &nbsp;0) is: :\frac{\cos(x) x^2 - \sin(x)2x}{x^4} == proof == :let some function f(x)=\frac{g(x)}{h(x)}. :we wish to find \frac{d}{dx}[f(x)], and this is equivalent to \frac{d}{dx}\left[\frac{g(x)}{h(x)}\right] (by the reciprocal rule and the product rule, f will be differentiable). :we also know that g(x)=f(x)\cdot h(x). :by the product rule, we can say that \frac{d}{dx}[g(x)]=h(x)\cdot\frac{d}{dx}[f(x)]+\frac{d}{dx}[h(x)]\cdot f(x). :from the equation above, \frac{d}{dx}[f(x)]=\frac{\frac{d}{dx}[g(x)]-\frac{d}{dx}[h(x)]\cdot f(x)}{h(x)} :because f(x)=\frac{g(x)}{h(x)}, the right side simplifies to \frac{\frac{d}{dx}[g(x)]-\frac{d}{dx}[h(x)]\cdot\frac{g(x)}{h(x)}}{h(x)}, which simplifies to \frac{d}{dx}[f(x)]=\frac{\frac{d}{dx}[g(x)]\cdot h(x)-\frac{d}{dx}[h(x)]\cdot g(x)}{[h(x)]^2} : hence proved. {{cite book | last=berresford | first=geoffrey | last2=rocket
Differentiation_rules