In this section we'll develop the quotient rule of derivatives then apply it in order to extend the power rule to negative-integer exponents.
The quotient rule is a prescription for finding the derivative of a quotient of functions. Here is the formula first, then we'll derive it below.
The derivative of a quotient of functions is the denominator function multiplied by the derivative of the numerator function, minus the numerator function multiplied by the denominator, and all divided by the square of the denominator function.
If f(x) is the numerator and g(x) is the denominator, then we can abbreviate that definition like this:
The quotient rule can easily be derived using the difference quotient and a simple trick. Start by writing out the difference quotient for, f(x)/g(x). This derivation is really just like that of the product rule in another section.
As usual, give these fractions a common denominator:
Now the that will help us pull out some recognizable derivatives: add and subtract f(x)g(x) to the numerator. The net result, of course, is doing nothing at all, but you'll see how it will help:
Now rearrange, factoring g(x) out of the first term of the numerator and f(x) out of the second:
... separate the terms and switch around the denominator to get closer to derivatives:
Use the properties of limits to isolate terms and identify the derivatives, f'(x) and g'(x):
Finally, in the limit as h approaches zero, g(x+h) is g(x), so long as g(x) is a continuous function. In other words, g(x) must have a value at x.
The quotient rule can be remembered using this mnemonic phrase:
Low-d-high minus High-d-low over the square of what's below.
It means take what's in the denominator (low), multiply it by the derivative (d) of the numerator (high), subtract the opposite pairing, then divide the whole thing by the square of the denominator.
A mnemonic device, like an acronym (ROYGBIV, NASA) or a phrase (Every Good Boy Does Fine) is a device you might use to jog your memory of something more complicated.
Our function is
Solution: To set up the quotient rule we multiply the denominator by the derivative of the numerator, subtract the product of the numerator and the derivative of the denominator, then divide the whole thing by the square of the denominator. We need only take two derivatives:
The result is
Note: These product-rule derivatives can be tricky to simplify. It's not always worth spending a lot of time on it. If you see something easy, go for it, but don't be surprised if the derivative doesn't simplify too much.
Our function is
Solution: We set up the quotient rule according to the formula to get:
After taking the simple derivatives, we get this function. Remember to keep expressions in parenthesis:
Now we can take a couple of algebra steps to simplify this one. First multiply the binomials of the numerator:
Gather terms. I've left the denominator alone here. It's about as simple as it will get.
Being a being with a small brain, I don't always use the quotient rule when I need to find the derivative of a quotient. I don't have a lot of extra space in there, so I don't like the clutter.
In fact, any function f(x)/g(x) can be written as the product f(x)[g(x)]-1. So the product rule will take you a long way in finding derivatives of quotients.
Still, the quotient rule does make a lot of derivatives much easier in the long run. So memorization is up to you, but it helps to have the memory device: low-d-hi minus hi-d-low over the square of what's below. Who could forget that?
Up to this point, it's unlikely that you've developed the power rule of derivatives,
for negative exponents, such as the derivative of f(x) = x-1. We can do that now very easily with the quotient rule. Here's the derivative:
Applying the quotient rule, we get:
Now notice that the derivative of a constant is zero, and for the second derivative, we can use the power rule with a positive exponent. Condensing the result gives:
Now we can use the laws of exponents (namely that 1/(2n) = 2-n, and xa·xb = xa+b, to get our result:
... which is pretty much what we would have guessed, I guess.
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