Quotient rule

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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.[1][2][3]

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[edit]

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 ≠ 0) is:

\frac{\cos(x) x^2 - \sin(x)2x}{x^4}

Proof[edit]

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].
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. [4]

References[edit]

  1. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 0-495-01166-5. 
  2. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 0-547-16702-4. 
  3. ^ Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th ed.). Addison-Wesley. ISBN 0-321-58876-2. 
  4. ^ Berresford, Geoffrey; Rockett, Andrew (2008). Brief Applied Calculus (5th ed.). ISBN 0-547-16977-9.