### Video Transcript

In this video, weβre going to see
how three rules of differentiation can be combined to allow us to differentiate
increasingly complex functions. We remind ourselves of these three
rules, the product rule, the quotient rule, and the chain rule. And then, weβll apply combinations
of these rules to a couple of different examples.

First, letβs recall these three
rules and their uses. The first rule is the product rule,
which allows us to differentiate products of functions. It tells us that the derivative of
the product ππ, thatβs ππ prime, is equal to ππ prime plus π prime π. What weβre doing is multiplying
each function by the derivative of the other and adding them together.

The second rule is the quotient
rule, which allows us to differentiate quotients for functions, thatβs π over
π. And it tells us that the derivative
of π over π is equal to ππ prime minus ππ prime over π squared. Now, itβs important to note that
unlike the product rule, which is symmetrical in π and π, the quotient rule is not
symmetrical in π and π. So, we must make sure that we
define π to be the function in the numerator and π to be the function in the
denominator.

The third rule is the chain rule,
which allows us to differentiate composite functions; that is, functions of other
functions. Here, we have the function π of π
of π₯, which means we apply π first and then apply π. Its derivative is given by π prime
of π₯ multiplied by π prime of π of π₯. Thatβs the derivative of the inner
function multiplied by the derivative of the outer function, with the inner function
still inside it.

It is important that weβre clear on
the different uses of notation. In the product rule, ππ means π
multiplied by π. Whereas in the chain rule, π of π
of π₯ means the composite function we get when we apply π first and then apply
π. We can also express each of these
rules using Leibnizβs notation. And itβs more common to use the
letters π’ and π£ when we do so, although it doesnβt actually matter.

The product rule tells us that the
derivative with respect to π₯ of π’π£ is equal to π’ dπ£ by dπ₯ plus π£ dπ’ by
dπ₯. The quotient rule tells us that the
derivative with respect to π₯ of π’ over π£ is equal to π£ dπ’ by dπ₯ minus π’ dπ£
by dπ₯ all over π£ squared. And finally, the chain rule tells
us that if π¦ is equal to the composite function π of π of π₯, and we define π’ to
be π of π₯ so that π¦ becomes a function of π’, π¦ equals π of π’. Then, dπ¦ by dπ₯ is equal to dπ¦ by
dπ’ multiplied by dπ’ by dπ₯. We differentiate π¦ with respect to
π’ and then multiply by the derivative of π’ with respect to π₯. Weβll now apply combinations of
these three rules to some examples.

Find the first derivative of π¦
equals π₯ minus five multiplied by π₯ minus two to the power of six at one, negative
four.

Letβs begin then by considering
this function that weβve been given. We can see that it is a product of
two functions, π₯ minus five and π₯ minus two to the power of six. This suggests that weβre going to
need to use the product rule in order to find this derivative. So, we can define one function to
be π and the other function to be π. Here is the product rule but as
itβs symmetrical in π and π, it doesnβt actually matter which way round we make
this definition. We have then that π of π₯ is equal
to π₯ minus five and π of π₯ is equal to π₯ minus two to the power of six.

And in order to apply the product
rule, we need to find the derivatives of π and π. π prime of π₯ is
straightforward. Itβs just one. But what about π prime? We have a power of six, and we
certainly donβt want to distribute all of the parentheses to give a polynomial. So, how are we going to find this
derivative? π is actually a composite
function, which means that we can apply the chain rule in order to find its
derivative, dπ¦ by dπ₯ equals dπ¦ by dπ’ multiplied by dπ’ by dπ₯. We can amend the π¦s in the chain
rule to πs. And then, we can allow π’ to equal
π₯ minus two, which makes π equal to π’ to the power of six.

The derivative of π’ with respect
to π₯ is one. And applying the power rule, the
derivative of π with respect to π’ is six π’ to the power of five. Applying the chain rule, dπ by dπ₯
is equal to six π’ to the power of five, thatβs dπ by dπ’, multiplied by one,
thatβs dπ’ by dπ₯. But of course, multiplying by one
has no effect. So, dπ by dπ₯ is equal to six π’
to the power of five.

However, we need this derivative to
be in terms of π₯, so we must reverse the substitution. π’ is equal to π₯ minus two, so dπ
by dπ₯ in terms of π₯ is equal to six multiplied by π₯ minus two to the power of
five. We could also have seen this by
applying the chain rule extension to the power rule. Now that weβve found each
derivative, we can begin substituting into the product rule to find the derivative
of π¦.

dπ¦ by dπ₯ will be equal to π,
thatβs π₯ minus five, multiplied by π prime, thatβs six multiplied by π₯ minus two
to the power of five. We then add π prime, thatβs just
one, multiplied by π, which is π₯ minus two to the power of six. We can simplify by taking a common
factor of π₯ minus two to the power of five out. And then, simplifying in that
second bracket gives π₯ minus two to the power of five multiplied by seven π₯ minus
32. So, weβve found the first
derivative of π¦.

But we are asked to evaluate this
derivative at a particular point, the point one, negative four. This means that we need to make a
substitution. We need to substitute the π₯-value
at this point into our gradient function, giving one minus two to the power of five
multiplied by seven multiplied by one minus 32. That gives negative one to the
power of five multiplied by negative 25, which is 25.

So in this question, we required a
combination of the product and chain rules in order to answer this problem.

Next, weβll consider an example
which requires a different combination of rules.

Determine the derivative of the
function π of π’ equals π’ squared plus five over π’ squared minus one all to the
power of four.

Letβs begin by considering the
rules that will be useful to us in this question. We can see that we have a quotient,
π’ squared plus five over π’ squared minus one. So, weβre going to need to use the
quotient rule. Iβve written the quotient rule out
here using πs and πs because weβve already got π’s and πs in the question. So, this will enable us to find the
derivative of this expression inside the parentheses.

But we still have that power of
four. Weβll allow π£ to equal the
expression inside the parentheses. Itβs equal to π’ squared plus five
over π’ squared minus one. Then, π will be equal to π£ to the
power of four. And we can apply the chain rule,
dπ by dπ’ is equal to dπ by dπ£ multiplied by dπ£ by dπ’.

dπ by dπ£ can be easily calculated
by applying the power rule. Itβs equal to four π£ cubed. But in order to calculate dπ£ by
dπ’, weβre going to need to apply the quotient rule. Weβll let π equal the numerator of
π£, thatβs π’ squared plus five, and π equals the denominator, thatβs π’ squared
minus one. dπ by dπ’ and dπ by dπ’ can be found by applying the power rule. Theyβre each equal to two π’. Now, we can substitute into the
quotient rule to find dπ£ by dπ’.

We have π, thatβs π’ squared minus
one, multiplied by π prime or dπ by dπ’, thatβs two π’, minus π, thatβs π’
squared plus five, multiplied by π prime or dπ by dπ’, thatβs two π’. Itβs all divided by π squared,
thatβs π’ squared minus one squared. Now, letβs simplify. Expanding the parentheses in the
numerator, we have two π’ cubed minus two π’ minus two π’ cubed minus 10π’. And then the denominator remains π’
squared minus one all squared. The two π’ cubeds will cancel each
other out, leaving negative 12π’ over π’ squared minus one squared.

Weβll now delete some of this
working out for the quotient rule to make some room on the page. We now have then that dπ by dπ£ is
equal to four π£ cubed and dπ£ by dπ’ is equal to negative 12π’ over π’ squared
minus one all squared. So, we can substitute into the
chain rule. dπ by dπ’ is equal to four π£ cubed multiplied by negative 12π’ over π’
squared minus one all squared. Now, remember that dπ by dπ’ must
be in terms of π’ only. So, we need to reverse our
substitution.

π£ is equal to π’ squared plus five
over π’ squared minus one. So, we have four π’ squared plus
five over π’ squared minus one all cubed multiplied by negative 12π’ over π’ squared
minus one all squared. Simplifying gives negative 48π’
multiplied by π’ squared plus five cubed all over π’ squared minus one to the power
of five.

In this question, we saw then that
we needed to apply a combination of the quotient and chain rules in order to find
the derivative of the function π of π’.

Letβs now look at an example
involving trigonometric functions.

Determine the derivative of the
function π of π‘ equals the square root of negative sin π‘ plus seven over negative
cos π‘ plus seven.

Now, we can see straightaway in
this question that we have a quotient. So, weβre going to need to apply
the quotient rule at some point. But are we going to need to do
anything else? Well, we donβt just have this
quotient. We have the square root of this
quotient, which means we have a composite function. And so, weβre also going to need to
apply the chain rule. Weβll begin by allowing π’ to be
that quotient underneath the square root. π’ is equal to negative sin π‘ plus
seven over negative cos π‘ plus seven. Then, π becomes a function of
π’. Itβs equal to the square root of
π’, which we can express using index notation as π’ to the power of one-half.

The chain rule, using the letters
π , π‘, and π’ as we have in this question, tells us that the derivative of π with
the respect to π‘ is equal to dπ by dπ’ multiplied by dπ’ by dπ‘. Applying the power rule, we see
that dπ by dπ’ is equal to one-half π’ to the power of negative a half. But in order to find dπ’ by dπ‘,
weβre going to need to apply the quotient rule.

Weβll define π to be the function
in the numerator, thatβs negative sin π‘ plus seven, and π to be the function in
the denominator, negative cos π‘ plus seven. In order to find the derivatives of
these two functions, we need to recall how we differentiate sin and cos. There is a helpful little cycle
that we can remember. The derivative of sin π‘ is cos
π‘. The derivative of cos π‘ is
negative sin π‘. The derivative of negative sin π‘
is negative cos π‘. And the derivative of negative cos
π‘ is sin π‘. And then we go around the cycle
again.

Remembering that the derivative of
a constant is just zero, we have that π prime is equal to negative cos π‘ and π
prime is equal to sin π‘. Now, we can substitute into the
quotient rule to find dπ’ by dπ‘. Itβs π, thatβs negative cos π‘
plus seven, multiplied by π prime, negative cos π‘, minus π, thatβs negative sin
π‘ plus seven, multiplied by π prime, thatβs sin π‘, all over π squared. Now, we need to do some
simplification. So, weβll expand the parentheses in
the numerator, which gives cos squared π‘ minus seven cos π‘ plus sin squared π‘
minus seven sin π‘ over negative cos π‘ plus seven squared.

We can recall, at this point, one
of our trigonometric identities, cos squared π‘ plus sin squared π‘ is equal to
one. So, this simplifies to one minus
seven cos π‘ minus seven sin π‘ over negative cos π‘ plus seven squared. Now that we found both dπ’ by dπ‘
and dπ by dπ’, we can substitute into the chain rule. We have then that dπ by dπ‘ is
equal to dπ by dπ’, thatβs a half π’ to the power of negative a half, multiplied by
dπ’ by dπ‘, thatβs one minus seven cos π‘ minus seven sin π‘ over negative cos π‘
plus seven squared.

Remember, though, that dπ by dπ‘
must be in terms of π‘ only. So, we need to reverse our
substitution. We have negative sin π‘ plus seven
over negative cos π‘ plus seven to the power of negative a half multiplied by one
minus seven cos π‘ minus seven sin π‘ over two multiplied by negative cos π‘ plus
seven squared. That power of negative a half means
a reciprocal, so we can deal with that by inverting the fraction. And the first part becomes negative
cos π‘ plus seven over negative sin π‘ plus seven to the power of positive
one-half.

We can then simplify the
powers. We have negative cos π‘ plus seven
to the power of a half in the numerator and then negative cos π‘ plus seven to the
power of two in the denominator. Which will lead to a power of
negative three over two overall. Thatβs a power of three over two in
the denominator. This leads us to one minus seven
cos π‘ minus seven sin π‘ in the numerator. And in the denominator two times
the square root of negative sin π‘ plus seven. Thatβs negative sin π‘ plus seven
to the power of a half multiplied by negative cos π‘ plus seven to the power of
three over two.

In this question then, weβve seen
that we can apply both the quotient and chain rule to a problem involving the
derivatives of trigonometric functions.

Letβs now summarize what weβve seen
in this video. Weβve reminded ourselves of these
three key rules, the product rule, the quotient rule, and the chain rule, each
expressed here using Leibnizβs notation. Weβve seen that these three rules
can be used together in order to find the derivatives of more complex functions.

And although we havenβt done an
example in this video, we can, of course, apply each rule multiple times if the
problem requires it. These three rules are incredibly
powerful. And by combining them or using the
same rule in succession, this opens up a wide class of complex functions whose
derivatives weβre now able to find.