# The Quotient Rule

The quotient rule gives us the derivative of the quotient of two functions (one function divided by another function). We know that we can find the differential of a polynomial function by adding together the differentials of the individual terms of the polynomial, each of which is a function in its own right. It is therefore tempting to assume that, for a function that is the quotient of two functions, we can simply divide the derivative of the numerator by the derivative of the denominator. Unfortunately, things are not going to be that easy. So how do we go about finding the derivative? Let's look at an example. Suppose we need to find the derivative of the following expression:

 y  = x2 + 5 3x - 9

On the right-hand side of the equals sign, we have the quotient of two functions. Finding the derivative of each function on its own is relatively easy. To find the derivative of the quotient, however, we will need to use the quotient rule. The rule states that the derivative of the quotient of two functions is equal to the product of the denominator and the derivative of the numerator, minus the product of the numerator and the derivative of the denominator, all over the denominator squared. Let's state this algebraically. Suppose we have the expression:

 y  = u v

where u and v are both functions. The derivative is given by the following formula:

dy  =
 v du -  u dv dx dx
dxv2

Let's use the quotient rule formula to find the derivative of y = (x2 + 5) / (3x - 9). We'll start by finding the derivative of each function separately:

 d (x2 + 5)  =  2x dx
 d (3x - 9)  =  3 dx

Now we'll assign the functions to the appropriate variables in the formula:

u(x)  =  x2 + 5

v(x)  =  3x - 9

Substituting the above values into the quotient rule formula, we get:

dy  =
 v du -  u dv dx dx
=  (3x - 9)(2x)  -  (x2 + 5)(3)
dxv2(3x - 9)2
 dy = 6x2 - 18x - 3x2 - 15 dx 9x2 - 54x + 81
 dy = 3x2 - 18x - 15 dx 9x2 - 54x + 81
 dy = x2 - 6x - 5 dx 3x2 - 18x + 27
 dy = x2 - 6x - 5 dx 3(x2 - 6x + 9)

There are a couple of things to note here. The first is that the numerator in the quotient rule formula is almost the same as the formula for the product rule - the only difference is that the addition operator is replaced by the subtraction operator. This second thing to note is that, because subtraction is involved, the terms in the numerator must appear in the correct order. Let's look at another example. This time we'll find the derivative of y = (4x - 2) / (x2 + 1). Once again, we are looking for the derivative of the quotient of two functions. As before, we start by finding the derivative of each function separately:

 d (4x - 2)  =  4 dx
 d (x2 + 1)  =  2x dx

Next, we assign the functions to the formula variables:

u(x)  =  4x - 2

v(x)  =  x2 + 1

Substituting these values into the quotient rule formula, we get:

dy  =
 v du -  u dv dx dx
=  (x2 + 1)(4)  -  (4x - 2)(2x)
dxv2(x2 + 1)2
 dy = (4x2 + 4)  -  (8x2 - 4x) dx (x2 + 1)2
 dy = -4x2 + 4x + 4 dx (x2 + 1)2

Now let's suppose we have an expression like this:

 y  = x3 + 2x 7

Using the quotient rule formula to find the derivative, we get:

dy  =
 v du -  u dv dx dx
=  (7)(3x2 + 2)  -  (x3 + 2x)(0)
dxv249
 dy = 21x2 + 14 dx 49
 dy = 3x2 + 2 dx 7

Did we really need to use the quotient rule here? Supposing we simply rewrite the expression as:

 y  = 1 · (x3 + 2x) 7

Now we can simply apply the basic rules of differentiation, including the constant coefficient rule:

 d ( 1 (x3 + 2x) ) = 1 · d (x3 + 2x)  = 3x2 + 2 dx 7 7 dx 7

Whenever you need to differentiate an expression involving the quotient of two functions, it is always worth checking to see whether you can rewrite the expression as we did here. If so, you will be able to find the derivative without having to apply the quotient rule. There is one more scenario involving a quotient that we should look at. What happens if we need to find the reciprocal of a function? We still have to differentiate a fraction for which the denominator will be a function, but the numerator will always be one (1). Consider the following expression:

 y  = 1 x2 + 3

Using the quotient rule formula to find the derivative, we get:

dy  =
 v du -  u dv dx dx
=  (x2 + 3)(0)  -  (1)(2x)
dxv2(x2 + 3)2
 dy = -2x dx (x2 + 3)2

It turns out that we can obtain the derivative of the reciprocal of any function u using the following somewhat less complex formula:

d(1/u)  =
 - d u dx
dxu2

So, we could simply write:

 dy = -2x dx (x2 + 3)2

Following on from that, suppose we had the following expression:

 y  = 5 x2 + 2x + 3

Since this function is simply the reciprocal multiplied by five (5), we can treat the five in the numerator as a constant coefficient, and just use the formula for the derivative of the reciprocal as we did above:

dy  =  5 ·
 - d (x2 + 2x + 3) dx
dx(x2 + 2x + 3)2
 dy =  5 · -2(x + 1) dx (x2 + 2x + 3)2
 dy = -10(x + 1) dx (x2 + 2x + 3)2

There will inevitably be situations where you need to find the derivative of the quotient of two functions. If you are not dealing with a reciprocal (or some multiple of a reciprocal), and you cannot simplify the function so that it is not a quotient (for example if the denominator is a constant value), then you will probably need to use the quotient rule to find the derivative.