We can take the quotient of two functions and get another function, but we must restrict the domain to be sure that the second function (the denominator or divisor) cannot attain the value 0. The derivative of a quotient is not, in general, merely the quotient of the derivatives.
It’s a little more complicated than that.
Definition and notation
When we want to divide a function by second function, we divide the expression of the first function by the expression of the second one. We find the ratio of the expressions that define the functions. Consider
f(x ) = 8x 2 and
g (x ) = 4x In this case,
[ f (x )] / [g(x )] = (8x 2) / (4x ) = 2x provided x≠ 0. Here’s another example. Suppose we have
f(x ) =x 2−x− 20 and
g (x ) =x− 5 The quotient in this case, as long as x≠ 5, is
[ f (x )] / [g(x )] = (x 2−x− 20) / (x− 5) = x+ 4
The rule in brief
The quotient rule for differentiation tells us this:
• To find the derivative of the quotient of two differentiable functions, we multiply the derivative of the first function by the second function, then multiply the derivative of the second function by the first function, then subtract the second product from the first product, and finally divide by the square of the second function.
If we want to use this rule, then both functions, as well as their ratio, must be differentiable at every point that interests us. (Sometimes a derivative exists for the quotient of two functions even when this rule can’t be applied. We’ll see an example of this shortly.) In addition, the Quotient Rule 95
denominator function can’t attain the value 0 anywhere in its domain. Stated symbolically, if f and g are differentiable functions, then
(f /g)′ = ( f ′ g−g ′ f ) / g2 for all points where g does not become 0.
Example
Let’s see how this rule works, using the examples from above. First, consider f(x ) = 8x 2
and
g (x ) = 4x
When we differentiate the quotient of the functions, we get
(f /g)′ (x ) = d /dx [(8x 2) / (4x )] = d /dx (2x ) = 2 When we use the quotient rule, we get
( f /g)′ (x ) = {[ f ′ (x )][ g (x )] − [ g ′ (x )][ f (x )]} / [ g (x )]2 = {[d /dx (8x 2)](4x ) − [d /dx (4x )](8x 2)} / (4x )2
= [(16x )(4x ) − 4 × (8x 2)] / (16x 2) = (64x 2− 32x 2) / (16x 2) = (32x 2) / (16x 2) = 2
This works only as long as g (x ) ≠ 0. That means the derivative of the quotient function is defined for all real numbers except 0.
Another example
Now let’s look at these two functions:
f(x ) =x 2−x− 20 and
g (x ) =x− 5 When we differentiate the quotient of these, we get
(f /g)′ (x ) = d /dx [(x 2−x− 20) / (x− 5)] = d /dx (x+ 4) = 1
When we use the quotient rule, we get
(f /g)′ (x ) = {[ f ′ (x )][g (x )] − [g ′ (x )][ f (x )]} / [ g (x )]2
= {[d /dx (x 2−x− 20)](x− 5) − [d /dx (x− 5)]( x 2−x− 20)} / (x− 5)2
= [(2x− 1)(x− 5) − 1 × (x 2−x− 20)] / (x 2− 10x+ 25)
= [(2x 2− 11x+ 5) − (x 2−x− 20) / (x 2− 10x+ 25)
= (2x 2− 11x+ 5 −x 2+x+ 20) / (x 2− 10x+ 25)
= (x 2− 10x+ 25) / (x 2− 10x+ 25) = 1
This works provided g (x ) ≠ 0. That means the derivative of the quotient function is defined for all real numbers except 5. That value of x is the zero of
g (x ) =x− 5 which forms the denominator in the quotient.
Are you astute?
The quotient of two functions can be differentiable in general, even if one or both of them individually is not. At first you might think that this sort of thing can’t occur. But once in awhile it happens. Here’s an example.
Take a look at Fig. 6-1. The derivatives of f (x ) and g (x ) both abruptly change at the point where x= 0.
Specifically:
f ′ (x ) = −2 when x< 0 f ′ (x ) = 2 when x> 0 g ′ (x ) = −1 when x< 0
g ′ (x ) = 1 when x> 0
Neither derivative is defined at the point where x= 0, so neither function is differentiable in general. You can’t use the quotient rule to find [ f (x ) / g (x )]′ unless you leave x= 0 out of the domains.
Nevertheless, a derivative exists for f (x ) / g (x ) over the entire set of reals, including x= 0. Because g (x ) never becomes equal to 0, you know that
f (x ) / g (x ) = 2 ( | x|+ 1) / ( | x|+ 1) = 2 Therefore,
[ f (x ) / g (x )]′ =d /dx 2 = 0
for all real numbers x. The graph of [ f (x ) / g (x )]′ is a line with a slope of 0 at every point along its infinite length, and that passes through the point (0,2). There is no discontinuity at the point (0,2) or anywhere else.
Here’s a challenge!
Derive the quotient rule for differentiation from the other rules we’ve learned so far.
Quotient Rule 97
Solution
Whenever we want to divide a quantity by another, we can multiply the first quantity by the reciprocal of the second, as long as the second quantity is not equal to 0. Knowing that, it seems that we should be able to derive the quotient rule for differentiation from the product and reciprocal rules for differentiation.
Let’s try that approach and see if it works.
Suppose that we’re given two functions called f and g, and the domain is restricted so the value of g can never become 0. We can write the reciprocal of g as (1/g). Our knowledge of algebra tells us that
f / g=f × (1/g)
Let’s insert a “times sign” (×) if there is any risk denoting a function of a function when we mean to denote a function times a function. The reciprocal rule for differentiation tells us that
(1/g)′ = −g ′ / ( g 2 ) The product rule for differentiation allows us to write
[ f × (1/g)]′ =f ′ × (1/g)+ (1/g)′f
x
f(x) = 2 (|x| + 1 )
g(x) = |x| + 1 f(x) / g(x) = 2 Each axis increment is 1 unit
Value of function f
g
f/g
Figure 6-1 In this situation, the quotient of f (x ) andg (x ) is differentiable over the entire set of reals, but neither f (x ) nor g (x ) is differentiable at the point where x= 0.
When we substitute ( f/g)′ for [ f × (1/g)]′ on the left-hand side of this equation, and we also substitute [−g ′ / (g 2)] for (1/g)′ on the right-hand side, we get
(f/g)′ =f ′ × (1/g)+ [−g ′ / ( g 2 )]f We can simplify this to
(f/g)′ =f ′ /g−g ′f / g 2
Now let’s multiply the quotient immediately after the equals sign by the quantity ( g/g). We know that (g/g) is defined under our domain restriction—that is, g≠ 0—so it’s equal to 1. When we do that, the above equation becomes
(f /g)′ = ( f ′/g)(g/g)−g ′f / g 2
We can multiply out the term immediately to the right of the equals sign to get (f / g)′ =f ′g / g2−g ′f / g 2
Now we have a difference of two fractions with a common denominator on the right-hand side of the equation. Consolidating this into a single fraction, we get
(f / g)′ = (f ′g−g ′ f ) / g 2 That’s the quotient rule for differentiation!