### Video Transcript

In this video, weβll learn how to
apply the product rule to find the derivative of a function which is the product of
two or more other functions. Weβll begin by learning what the
product rule actually is before applying the rule to help us find the derivative of
a number of functions, including those which are themselves are product of at least
three different functions. Weβll also briefly consider the
application of the product rule for calculating the coordinates of the critical
points or stationary points on a graph.

Consider the example of π₯ squared
plus three π₯ plus one times π₯ minus four π₯ to the power of five.

Letβs think how we could find the
derivative of this expression with respect to π₯. In the past, we might have looked
to distribute this parentheses before applying the normal rules for differentiation
to each of the resultant terms. In fact though, this expression is
the product of two separate functions. The first is the function π₯
squared plus three π₯ plus one. The second is the function π₯ minus
four π₯ to the power of five.

And thereβs a rule that will allow
us to differentiate any expression which is the product of two or more
functions. Funnily enough, this rule is called
the product rule. Now, the proof of the product rule
is somewhat long-winded and thatβs outside of the constraints of this video. Instead, weβll simply state the
product rule and consider its application to calculus.

The product rule states that the
derivative of the product of functions π and π is equal to π multiplied by the
derivative of π plus the derivative of π multiplied by π. And this is sometimes alternatively
written as the derivative of π’π£ with respect to π₯ is π’ times dπ£ by dπ₯ plus π£
times dπ’ by dπ₯, where here π’ and π£ are functions in π₯. Itβs very much a matter of personal
preference as to which you decide to use. But this rule does need to be
memorized. Weβll begin by considering a simple
example of the application of this rule.

Find the first derivative of the
function π¦ equals three π₯ to the power of five plus seven multiplied by seven
minus three π₯ to the power of five.

Here, we have an equation which is
the product of two functions. Since this is the case, we can find
the derivative of this function by applying the product rule. Remember this rule says that the
derivative of the product of π and π is equal to π times the derivative of π
plus the derivative of π times π. Itβs always sensible to write down
what we actually know about the expression weβre looking to differentiate. We can say that itβs the product of
two functions. Letβs call the first one π of π₯
three π₯ to the power of five plus seven. And weβll call the second π of
π₯. Itβs seven minus three π₯ to the
power of five. Weβre going to need to find the
first derivative of each of these functions.

Remember the derivative of a simple
algebraic expressions such as ππ₯ to the power of π, where π and π are real
constants, is π times ππ₯ to the power of π minus one. We multiply by the exponent and
then reduce that exponent by one. And what this does actually mean is
that the derivative of a constant is zero since a constant β letβs say the number
one β is actually one π₯ to the power of zero. When we multiply by the exponent,
we actually just end up with an answer of zero. This means the derivative of π of
π₯ is five times three π₯ to the power of four plus zero which is simply 15π₯ to the
power of four. Similarly, the derivative of π of
π₯ is zero minus five times three π₯ to the power of four which is negative 15π₯ to
the power of four.

Letβs substitute what we know into
the formula for the product rule. dπ¦ by dπ₯ is equal to three π₯ to the power of
five plus seven times negative 15π₯ to the power of four plus 15π₯ to the power of
four times seven minus three π₯ to the power of five. We then distribute these
parentheses by multiplying each term by the term outside. And we get negative 45π₯ to the
power of nine minus 105π₯ to the power of four plus 105π₯ to the power of four minus
45π₯ to the power of nine. Negative 105π₯ to the power of four
plus 105π₯ to the power of four is zero. And we can see that the first
derivative of our function is negative 90π₯ to the power of nine. Itβs also useful to know that we
can apply this rule to find the derivative at a given point.

Letβs consider an example of
this.

Find the first derivative of π of
π₯ equals nine π₯ squared minus π₯ minus seven times seven π₯ squared minus eight π₯
minus seven at negative one, 24.

Here, we have an equation which is
the product of two functions. Since this is the case, we can find
the derivative of this function by applying the product rule. Remember that says that we can find
the derivative of the product of π and π by finding π times the derivative of π
plus the derivative of π times π. Now, actually our equation is π of
π₯. So weβre going to split our two
functions into π of π₯ and β of π₯. We can say that π of π₯ is equal
to nine π₯ squared minus π₯ minus seven and β of π₯ is equal to seven π₯ squared
minus eight π₯ minus seven. And then, Iβve changed the product
rule. This time, it says the derivative
of π times β is π times the derivative of β plus the derivative of π times β. Letβs find the derivative of
π.

The derivative of nine π₯ squared
is 18π₯ and the derivative of negative π₯ is negative one. So the derivative of π of π₯ is 18
π₯ minus one. Similarly, the derivative of β of
π₯ is 14π₯ minus eight. Letβs substitute what we know into
the formula for the product rule. The derivative of π with respect
to π₯ is therefore nine π₯ squared minus π₯ minus seven times 14π₯ minus eight plus
seven π₯ squared minus eight π₯ minus seven times 18π₯ minus one. We distribute the parentheses
carefully and collect like terms. And we see that the derivative of
π of π₯ is 252π₯ cubed minus 237π₯ squared minus 208π₯ plus 63.

But we arenβt quite finished. Weβve been asked to find the
derivative at the point negative one, 24. This is the point in our Cartesian
plane, where π₯ is equal to negative one and π¦ is equal to 24. So we can substitute π₯ is equal to
negative one into our expression for the derivative. Thatβs 252 times negative one cubed
minus 237 times negative one squared minus 208 times negative one plus 63, which is
negative 218. In fact, itβs useful to remember
that this actually tells us the gradient of the tangent to the curve at the point
negative one, 24.

In our next example, weβll consider
how to apply the product rule to differentiate an expression which is the product of
more than two functions.

The product rule says that the
derivative of ππ is equal to the derivative of π times π plus π times the
derivative of π. Use this to derive a formula for
the derivative of π times π times β.

In this question, weβve been given
the product rule and asked to use it to find a formula for the derivative of the
product of three functions. These are π, π, and β. Weβre going to begin by splitting
π times π times β up. Weβre going to write it as ππ
times β. Remember since multiplication is
commutative, we could have alternatively written it as π times πβ and we will get
the same answer either way. So we can say that the derivative
of ππβ is equal to the derivative of ππ times β.

And weβre now going to apply the
product rule. We can see that this is equal to
the derivative of ππ times β plus ππ times the derivative of β. And now, we spot that the first
term we have is the derivative of ππ. We know though by the definition of
the product rule that this is the same as the derivative of π times π plus π
times the derivative of π. So we replace this in our
formula. And weβre going to distribute these
parentheses.

When we do, we see that the formula
for the derivative of ππβ is the derivative of π times π times β plus π times
the derivative of π times β plus π times π times the derivative of β. You might also like to see if you
can apply this idea to help you find a formula for the derivative of the product of
four functions, say ππβπ.

Next, weβre going to look at how
this example can help us to find the derivative of an expression which is the
product of three functions.

Find the first derivative of π of
π₯ equals π₯ to the power of eight plus four times three π₯ root π₯ minus seven
times three π₯ root π₯ plus seven at π₯ equals negative one.

To answer this question, we have
two options. We could use the product rule twice
or we could recall the definition of the derivative of the product of three
functions. The derivative of π times π times
β is the derivative of π times π times β plus π times the derivative of π times
β plus π times π times the derivative of β. Now since our function is actually
π of π₯, Iβve changed the functions in this formula to be π’ π£ and π€. So letβs work out what the
functions π’, π£, and π€ actually are. We can say that π’ of π₯ is equal
to π₯ to the power of eight plus four. π£ of π₯ is equal to three π₯ root
π₯ minus seven. And π€ of π₯ is equal to three π₯
root π₯ plus seven.

We need to differentiate each of
these functions with respect to π₯ as per the formula for the product rule with
three functions. The derivative of π’ is fairly
straightforward. Itβs just eight π₯ to the power of
seven. But what about π£ and π€? Well, we could use the product
rule. But actually, we can simply rewrite
each of these expressions. We know that the square root of π₯
is the same as π₯ to power of one-half. And the laws of exponents tell us
we can simplify this expression by adding the powers.

And when we do, we see that π£ of
π₯ can be written as three π₯ to the power of three over two minus seven and π€ of
π₯ is three π₯ to the power of three over two plus seven. And this means the derivative of
the π£ is three over two times three π₯ to the power of one-half or nine over two π₯
to the power of one-half. And actually, the derivative of π€
is the same.

We now have everything we need to
substitute this into our formula for the product rule. Now, at this point, you might be
tempted to jump straight into substituting π₯ is equal to negative one into the
derivative. However, we have some roots here
and that might cause issues. Instead, we carefully distribute
each set of parentheses and simplify fully. And when we do, we see that the
derivative of π of π₯ is 99π₯ to the power of 10 minus 392π₯ to the power of seven
plus 108π₯ to the power of two.

And we can now evaluate this at π₯
is equal to negative one. Itβs 99 times negative one to the
power of 10 minus 392 times negative one to the power of seven plus 108 times
negative one squared which is equal to 599.

In our final example, weβll
consider how to use the product rule to solve problems involving critical
points.

Find the coordinates of the
critical points on the curve with equation π¦ equals π₯ over 16 plus π₯ squared.

The critical points of a curve are
found when the derivative is made equal to zero. Weβll begin then by differentiating
our equation π¦ in terms of π₯. We can begin by rewriting our
equation as π₯ times 16 plus π₯ squared to the power of negative one. And we can now differentiate this
using the product rule. Weβll let our first function β Iβve
called that π’ β be equal to π₯. And weβll say our second function
is 16 plus π₯ squared to the power of negative one. Itβs quite straightforward to find
the derivative of π’. We just get one. But weβre going to need to use the
chain rule to differentiate 16 plus π₯ squared to the power of negative one.

Weβre going to let π‘ be equal to
16 plus π₯ squared. Then, differentiating π‘ with
respect to π₯ gives us two π₯. And we can now say that π£ is equal
to π‘ to the power of negative one. We need to differentiate π£ with
respect to π‘. And when we do, we see that itβs
negative π‘ to the power of negative two. The derivative of our function π£
with respect to π₯ is equal to dπ£ by dπ‘ times dπ‘ by dπ₯. At the moment, thatβs negative π‘
to the power of negative two times two π₯. We can now replace π‘ with 16 plus
π₯ squared. And weβve differentiated π£ with
respect to π₯.

We can now clear some space and
apply the product rule. Itβs π’ multiplied by the
derivative of π£ plus π£ multiplied by the derivative of π’. We then rewrite it slightly. We say that the derivative of π¦
with respect to π₯ is negative two π₯ squared over 16 plus π₯ squared all squared
plus one over 16 plus π₯ squared. Weβre going to simplify this
expression by multiplying the denominator and the numerator of our second fraction
by 16 plus π₯ squared. And we see that we now have
negative two π₯ squared over 16 plus π₯ squared squared plus 16 plus π₯ squared over
16 plus π₯ squared squared.

By adding the numerators, we see
weβve successfully found the derivative. Itβs negative π₯ squared plus 16
over 16 plus π₯ squared all squared. Now earlier, we said that the
coordinates of the critical points could be found by letting the derivative be equal
to zero. So weβre going to say our fraction
negative π₯ squared plus 16 over 16 plus π₯ squared squared is equal to zero. And then, we consider what needs to
be true for this to be the case. Well, it doesnβt actually matter
what the denominator of our fraction is. If the numerator of our fraction is
zero, then the entire fraction must be equal to zero.

So we say that negative π₯ squared
plus 16 is equal to zero and we solve this. We add π₯ squared to both sides
which gives us π₯ squared equals 16. And then, we find the square root
of both sides of the equation, remembering to take both the positive and negative
square roots of 16. And we end up with two values for
π₯: positive and negative four. So the critical points of our curve
occur when π₯ is equal to plus or minus four. Weβre going to need to substitute
this back into the original equation to find the coordinates of the critical
points.

When π₯ is equal to four, π¦ is
four over 16 plus four squared which is one-eighth. And when π₯ is equal to negative
four, π¦ is negative four over 16 plus negative four squared which is negative
one-eighth. So the coordinates of the critical
points on the curve with equation π¦ equals π₯ over 16 plus π₯ squared are four,
one-eighth and negative four, negative one-eighth.

In this video, weβve learned that
we can apply the product rule to find the derivative of the product of two
functions. Weβve learned that for two
functions π and π, the derivative of their product is π times the derivative of
π plus the derivative of π times π. And finally, weβve seen that whilst
we can apply the product rule in succession, we can also find the product of three
functions by using the given formula.