The Quotient Rule (Level 2) In this video we will go over more challenging
examples that require the use of the quotient rule. The following examples require you to
really understand the quotient rule and combine it with various algebra and calculus concepts
learned so far. With that said let's go ahead and jump straight into the first example.
Find the derivative of: R(x) equals 1 over the quantity x plus 5 times e to the x.
Notice that this function seems to be formed solely by the function in the denominator.
In order to find the derivative of this function we actually need to apply quotient rule. We
can think of the constant 1 as our f(x) the hi function and the quantity (x + 5*e^x) as
our g(x) the low function. Then the derivative will be equal to the following expression.
Here the derivative of a constant is just equal to 0, and the derivative of our low
function is going to be equal to the following expression. Resulting in the final derivative
equal to negative times the quantity 1 plus 5 times e to the x over the quantity x plus 5 times e to the
x, squared. Be on the lookout for these types of functions that seem to be formed by the
reciprocal of a function. These types of functions pop out every now and then and as a result
we can generalize the derivative of these types of functions as follows: the derivative
of 1 over g(x) is going to be equal to negative times g prime of x over g of x squared, this is commonly
known as the reciprocal rule there is really no need to memorize this rule because it can
be obtained by simply applying the quotient rule. It can be useful to find the derivative
like the one in this example. Alright let’s try a slightly more challenging example.
Find the derivative of: R(x) equals x over the quantity x plus 5 over x.
Alright this function looks pretty insane it seems like we can apply the quotient rule,
but, before we use the quotient rule let's rewrite this function into a quotient rule
friendly form. We have a complex fraction that can be simplified by rewriting the denominator
as a single fraction, and then we can simplify the complex fraction as one single quotient
Now we are ready to apply the quotient rule, we will use x^2 as our f(x)
and the quantity (x^2 + 5) as our g(x). Then it is just a matter of substituting these
expressions 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). The last step is to simplify the numerator
by distributing the monomials, and by collecting like terms doing that yields the final derivative
equal to 10x over the quantity x squared plus 5, squared.
Alright let’s try a slightly more challenging example.
Find the derivative of: R(x) equals x times the quantity 1 minus 4 over x
plus 3. It seems that we have a product of two functions
but also notice that we have a quotient of two functions inside the expression.
Before you attempt to find the derivative of any function it is important to rewrite
the expression into a derivative friendly form. For this function let’s go ahead and
distribute the variable x to each term. This way we obtain an expression that becomes slightly
easier to differentiate. In this way, we now have to find the derivative term by term.
The derivative of the first term is going to be equal to one and the derivative of the
second term is a perfect candidate for the quotient rule, we will let 4x be our high
function and the quantity (x + 3) will be our low function, applying the quotient rule we
obtain the following expression, then, we go ahead and distribute each of the terms in the
numerator, next we simplify the expression by collecting like terms, in the end the derivative
of R(x) is equal to 1 minus 12 over the quantity x plus 3 squared. Always try to write functions
into a derivative friendly form before applying specific differentiation rules.
Ok, let’s try another example. Find the derivative of: R(x) equals the quantity
x squared plus c squared over the quantity x squared minus c squared.
Here c^2 represents a constant so we will treat it as such when applying differentiation
rules. This function requires an application of the quotient rule, we will let the quantity
(x^2 + c^2) be our high function and the quantity (x^2 - c^2) be our low function. Applying the
quotient rule we obtain the following expression, keeping in mind that c^2 is a constant when
finding the derivative. Next we go ahead and distribute the terms in the numerator
as follows. The last step is to simplify the expression and collect like terms doing that
we obtain the final derivative equal to negative 4 times c squared times x all over the quantity x squared minus
c squared raise to the power of 2. Alright, the last thing you should be aware
of is that you don’t need to use the quotient rule every time you see a quotient. Not every
function formed by a quotient needs to be differentiated by the Quotient Rule. At times
it is more convenient to rewrite the quotient first then use either the product rule or
constant multiple rule along with the power rule. For example the function R(x) equals the quantity
2x minus 1 over x squared. May look like it requires an application of
the quotient rule, but it turns out that we can find the derivative by making use of the
product rule by rewriting the denominator as a negative exponent as follows.
Alternatively, we can also divide both terms in the numerator by the denominator and then
apply the difference rule along with the power rule to find the derivative. Regardless of
what method you use you will obtain an expression that represents the derivative of the function.
Depending on your preference you might want to find the derivative of this function by
applying something other than the quotient rule.
Similarly the function 5/x^3 can be solved by rewriting it as a negative exponent and
then applying the power rule, you can also use the quotient rule if you like but the
first method is easier and requires less steps.
Lastly the function x^3/5 might look like a potential candidate for the quotient rule
but noticing that this is nothing more than a power function times the constant 1/5, we
can find the derivative by applying the power rule.
Again not all quotients require you to apply the quotient rule. Make sure you correctly
identify the function before blindly applying a differentiation rule.
Alright now let’s go over examples that require us to find an expression for the derivative
in terms of the given functions. If p(x) and q(x) are differentiable functions,
find an expression for the derivative of the following function.
R(x) equals p of x over q of x.
Alright in these types of problems we need to find an expression for the derivative in
terms of the function p(x), q(x), and their respective derivatives. Keep in mind that
we only want to find an expression for the derivative. Notice that R(x) is formed by
the quotient of p(x) and q(x) this means that we need to make use of the quotient rule.
The derivative of R(x) is going to be equal to q of x times p prime of x
minus p of x times q prime of x over q of x squared, and this is our expression for
the derivative of R(x). If we knew what p(x) or q(x) were equal to we can now substitute
these functions into our expression that represents the derivative of R(x). Alright let’s try
another example. R(x) equals p of x over the quanity 1 plus q of x.
Similar to the last example let's go ahead and apply the quotient rule.
The high function is p(x) and the low function is the quantity 1 + q(x) applying the quotient
rule we would obtain the following the quantity 1 plus q of x times p prime of x minus p of
x times q prime of x over the quantity 1 plus q of x, squared, we can also distribute
p prime of x in the numerator and we would obtain the following. Either expression is
an acceptable answer for the derivative. Keep in mind that we only want an expression for
the derivative, given the functions in the numerator and denominator. Ok, let’s try
one more similar example.
R(x) equals x squared over q of x.
Here we have a function formed by a power function in the numerator and a general function
q(x) in the denominator. We once again have to apply the quotient rule, notice that we
can actually find the derivative of the numerator by applying the power rule, for the denominator
we have no other choice but to write its derivative as q prime of x. Applying the quotient rule
and simplifying we obtain the final derivative equal to the quantity 2x times q of x minus
x squared times q prime of x over q of x squared. Alright in our next video we will find the
value for the derivative of a given function by making use of a set of values and using
the graph of functions.