Derivatives of Quotients

f '(x) g(x)

f(x) g '(x)

Example:

y = ((-2)*s² + 3) / e^{s + 1}

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

y	=	(-2)*s² + 3

		e^{s + 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,

y	=	(-2)*s² + 3

		e^{s + 1}

we can think of as

y	=	f(s)

		g(s)

So the derivative is

y'	= (	f(s)	)'

		g(s)

=	f '(s)	g(s)	-	f(s)	g '(s)

	( g(s) )²
=	( (-2)*s² + 3 )'	( e^{s + 1} )	-	( (-2)*s² + 3 )	( e^{s + 1} )'

	( e^{s + 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

( (-2)*s² + 3 )'	=	(-2)2s + 0	(by the derivative rule for sums, rule for constant multiples, and the derivative rule for constants)
( e^{s + 1} )'	=	e^{s + 1} (1 + 0)	(by the chain rule)

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.