Derivatives of Quotients

f '(x) g(x)

f(x) g '(x)

Example:

z = sin(y) / (e^{y + 1} - 1)

The first thing to notice when finding the derivative of this function is that it is a quotient, as shown below:

z	=	sin(y)

		e^{y + 1} - 1

The Derivative Rule for Quotients:

The derivative of a quotient is the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all divided by the denominator squared.
If

z	= (	f(x)	)

		g(x)

then the derivative of z is

So our example,

z	=	sin(y)

		e^{y + 1} - 1

we can think of as

z	=	f(y)

		g(y)

So the derivative is

z'	= (	f(y)	)'

		g(y)

=	f '(y)	g(y)	-	f(y)	g '(y)

	( g(y) )²
=	( sin(y) )'	( e^{y + 1} - 1 )	-	( sin(y) )	( e^{y + 1} - 1 )'

	( e^{y + 1} - 1 )²

and we just need to know each of the derivatives on the right-hand side of the equation. In this case these are

( sin(y) )'	=	cos(y)	(by the derivative rules for basic functions)
( e^{y + 1} - 1 )'	=	e^{y + 1} (1 + 0) - 0	(by the derivative rule for sums, chain rule, and the derivative rule for constants)

so the finished derivative is

additional explanation for the quotient rule
see another quotient rule example
practice gateway test
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Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose, University of Michigan Math Dept.