The Quotient Rule (Level 1) In the previous videos we dealt with functions
that were formed by a product of two or more functions. Recall that the way we dealt with
those types of functions was to apply the product rule. In this video we will learn
how to deal with functions formed by a quotient of functions.
The quotient of two differentiable functions f and g is itself differentiable at all values
of x for which g(x) does not equal 0. Because division by zero is undefined.
One way in which we can obtain a formula for the derivative of a function formed by the
quotient of two functions is by making use of the product rule. Say we have a function
R(x) that is formed by the quotient of f(x) and g(x). Making sure that g(x) does not equal
0, if this is the case than we can multiply both sides of this expression by the denominator
g(x). Next we take the derivative of both sides of the equation as follows. We will
go over this technique of taking the derivative of both sides of the equation in greater detail
in a much later video; this technique is known as implicit differentiation. We are essentially
taking the derivative of both sides of the equation, notice that the left hand side of
the equation is formed by a product of two functions so we need to apply the product
rule, applying the product rule we obtain the following expression, next we take the
derivative of the right hand side of the equation, in this case the derivative of f(x) is going
to be equal to f prime of x. Having taken the derivative of both sides, we now need
to solve for R prime of x, so we subtract the product of R(x) and g prime of x from
both sides, and then we divide by g(x) on both sides of the equation. We are almost
done, next we need to replace R(x) with an expression in terms of f(x) and g(x) recall
that R(x) was equal to the quotient of these two functions so we go ahead substitute this
expression. Then it is just a matter of cleaning up this expression and rewrite it as a single
fraction. We accomplish this by adding the terms in the numerator by finding a common
denominator, as follows. Then we simplify the complex fraction, resulting
in the following expression for the derivative of R of x. This formula is known as the quotient
rule. Using prime notation we can write the quotient rule as follows.
The quotient rule says that the derivative of a quotient is the denominator times the
derivative of the numerator minus the numerator times the derivative of the denominator, all
divided by the square of the denominator. The quotient rule and the other differentiation
rules learned so far enable us to compute the derivative of most rational functions.
Now, unfortunately, you do have to memorize this rule, keep in mind that you can always
derive it by using the technique of implicit differentiation. Another way of remembering
this rule is by using the following mnemonic device, Let the function on the numerator
be the hi function or top function, and let g(x) be the low function or bottom function.
We can now remember the quotient rule as “Lo D-Hi minus Hi D-lo all over the square of
what’s below”. Once again “Lo D-Hi minus Hi D-lo all over the square of what’s below”,
here the letter D represents the derivative. Alright let’s practice this new rule by
going over a couple of examples. Find the derivative of: R(x) equals e^x over
the square root of x. I hope you recognize this function. We found
the derivative of this function during the first product rule video. It turns out that
we can also find the derivative by using the quotient rule. Similar to the product rule
we need to first identify f(x), this is going to be the function located on the numerator
or the hi function which ends up being the natural exponential function. In the same
manner the function in the denominator or low function will be our g(x) in this case
it will be represented by the radical function. Once you identified your f(x) and g(x) it
is just a matter of finding each functions derivative and substituting these expressions
into the quotient rule formula. In this case R prime of x is going to be equal to
x raised to the power of one half times the derivative of e to the x minus e to the x
times the derivative of x to the power of one half all over x to the power of one half raised to the power of 2.
Remember to rewrite radical expressions into fractional powers this way you have an expression that
is in a derivative friendly form. Next we go ahead and find the derivative of e to the
x, and x to the 1/2 . We also simplify the denominator, doing that we obtain the following.
For the most part, the last step usually requires you to clean up the expression by collecting
like terms and or reducing complex fractions. This will be the case for most quotient rule
problems. With that in mind we can rewrite this expressions by dividing both terms in
the numerator by the denominator doing that we obtain the following. We can
also rewrite the fractional powers back into radical expressions resulting in the final
expression equal to e^x over the square root of x minus e^x over 2 times the square root
of x cubed. This is the same answer that we obtained in the product rule video. Note that
it might be easier to use the product rule instead of the quotient rule, make sure you
keep on the lookout for functions formed by a quotient it might be easier to find the
derivative by rewriting the function and applying the product rule instead.
Alright let’s try the next example. Find the derivative of: R(x) equals 2x over
the quantity 4 plus x squared. Similar to the previous example let's first
identify our f(x) and g(x), in this example our f(x) or hi function is 2x, and our g(x)
or low function is the quantity 4 plus x squared. Let’s go ahead and substitute these functions
and their respective derivatives into the quotient rule formula, here I have the quotient
rule written in prime notation. Substituting the functions we obtain the following. Taking
the derivative of f(x) and g(x) yields the following expressions. Now it is just a matter
of cleaning up the expression by collecting like terms and simplifying whenever possible.
Simplifying the numerator we obtain the following, then we go ahead and collect like terms, resulting
in the final expression equal to the quantity 8 minus 2x squared over the quantity 4 plus
x squared raised to the power of 2. Notice that there is really no need to expand the denominator it is perfectly
ok to leave your answer in this form. We can also factor out the constant 2 from both terms
in the numerator, notice that we can also factor out a negative sign, all of these forms
are acceptable answers for the derivative. Alright let’s try a slightly more complicated
example. Find the derivative of: R(x) equals the quantity
x cubed plus x over the quantity x to the power of 4 minus 2.
Alright, we once again have a quotient of two functions, similar to the previous examples
let’s go ahead and use the quotient rule. Our f(x) or hi function will be the quantity
(x^3 + x) and our g(x) or low function will be the quantity (x^4 - 2), having identified
our functions we go ahead and substitute them into the quotient rule formula. Doing that
we obtain the following. Next we go ahead and find the derivative of f(x) and g(x),
at this point it is just a matter of collecting like terms and simplifying the expression
in the numerator, we first need to distribute or FOIL out the expressions in the numerator,
after that, we make sure we distribute the negative sign on the second polynomial, from
here we go ahead and collect like terms, doing that we obtain the following expression. Notice
that there is really no need to expand the denominator it is best to just leave this
expression as it is, we can also factor out a negative sign in the numerator to make all
the terms positive, in the end the final derivative is equal to negative times the quantity x
to the power of 6 plus 3x to the power of 4 plus 6x squared plus 2 over the quantity
x to the power of 4 minus 2 raised to the power of 2. Alright and this is the quotient rule in a
nutshell in our next video we will go over slightly more challenging examples.