QUOTIENT RULE
Quotients of Functions
We can build new functions from old ones by forming quotients.
Given functions f and g with target sets consisting of real numbers, the quotient of f and g is the new function f/g defined as follows.
(f/g)(x) = f(x)/g(x)
We must be a little careful in determining the domain of the quotient f/g: a real number x lies in the domain of f/g if and only if it is in the domain of both f and g AND g(x) is not equal to zero. Division by zero is never allowed. Not now. Not ever.
A special type of quotient is the reciprocal of a function.
The domain of the reciprocal is the set of all points x in the domain of g for which g(x) is not zero.
Example. The cosecant function is the reciprocal of the sine function.
csc(x) = 1/sin(x)
Example. Power functions with negative exponents also arise as reciprocals.
x -n = 1/xn
The reciprocal of a differentiable function is differentiable. Our next general differentiation rule shows us how to differentiate the reciprocal of a differentiable function.
Why is THAT true?
A quotient of differentiable functions is differentiable. The quotient rule shows us how to differentiate a quotient of differentiable functions. The quotient rule is another basic tool that must be memorized.
Quotient Rule
|
|
( |
|
) |
= |
| f '(x)g(x) - f(x)g'(x) |
|
| g(x)2 |
|
|
The quotient rule says that the derivative of the quotient is "the derivative of the top times the bottom, minus the top times the derivative of the bottom, all divided by the bottom squared".....At least, that's one way to remember it. Be sure to get the order of the terms in the numerator correct. The numerator in the quotient rule involves SUBTRACTION, so order makes a difference!!
It is actually quite simple to derive the quotient rule from the reciprocal rule and the product rule. We simply recall that the quotient f/g is the product of f and the reciprocal of g.
Proof of the Quotient Rule
| Dx(f(x)/g(x)) |
= |
Dx((f(x))(1/g(x))) |
PRODUCT
RULE--> |
= |
(Dx(f(x)))(1/g(x)) + f(x)(Dx(1/g(x))) |
RECIPROCAL
RULE--> |
= |
|
COMMON
DENOM.--> |
= |
| f '(x)g(x) - f(x)g'(x) |
|
| g(x)2 |
|
Example. Differentiate y = (2x + 1)/(x2 - 3x + 4).
|
|
= |
|
(Dx(2x+1))(x2 - 3x + 4) - (2x + 1)(Dx(x2 - 3x + 4))
|
|
| (x2 - 3x + 4)2 |
|
|
= |
|
2(x2 - 3x + 4) - (2x + 1)(2x - 3)
|
|
| (x2 - 3x + 4)2 |
|
|
Our job in this type of problem is simply to express the derivative in terms of the given independent variable.
|
Note that this typical differentiation problem did NOT ask us to simplify our answer in any algebraic sense. In other situations you will find it convenient to simplify an answer such as this one (for example, by multiplying out the numerator and collecting like terms). But you do not need to do this unless you have a specific reason.
Example. We can now differentiate power funcitons having negative integers as exponents. Each of the following involves differentiation of a reciprocal y = 1/g(x).
g(x) = x:
Dx(x -1) = Dx(1/x) = (-Dx(x))/x2 = -1/x2 = -x -2
g(x) = x2:
Dx(x -2) = Dx(1/x2) = (-Dx(x2))/(x2)2 = -2x/x4 = -2x -3
|
See also: Derivatives of Trigonometric Functions. |
Example. There are six basic trigonometric functions; all six are defined as quotients of the sine and cosine functions. For example, the cosecant function is the reciprocal of the sine function. The derivative of the cosecant function is computed using the reciprocal rule as follows.
Dx(csc(x)) = Dx(1/sin(x)) = -Dx(sin(x))/sin2(x) = -cos(x)/sin2(x) = -csc(x)cot(x)
What's Next: Now we move on to the chain rule, which considers the problem of how to differentiate the composition of two functions.
/Stage6/Lesson/quotientRule.html
© CalculusQuestTM
Version 1
All rights reserved---1996
William A. Bogley
Robby Robson