### Video Transcript

In is video, we will learn how to
differentiate composite functions by applying the chain rule. We will see how to apply this to
simple functions initially. And then, weβll consider more
complex functions such as trigonometric and reciprocal trigonometric functions.

Firstly, a reminder of what
composite functions are. They are essentially functions of a
function. Suppose we have two functions, π
of π₯ equals two π₯ plus five and π of π₯ equals π₯ cubed. The composite functions π of π of
π₯ and π of π of π₯ are what we get if we composed these two functions in either
order. We apply one, and then we apply the
other.

π of π of π₯ means we applied π
first, giving π₯ cubed. And then, we take this as our input
for the function π, which will give two π₯ cubed plus five. π of π of π₯, however, is the
composite function we would get if we apply π first to give two π₯ plus five and
then take this as our input to the function π, which would give two π₯ plus five
all cubed. And if we distribute the
parentheses and simplify, this gives eight π₯ cubed plus 60π₯ squared plus 150π₯
plus 125.

So, we know how to compose
functions. But what about finding their
derivatives? Well, in this case, if youβre asked
to find the derivatives of either π of π of π₯ or π of π of π₯, it wouldnβt be
too bad. Because we could compose the
functions first, manipulate them algebraically, and then differentiate the resulting
polynomial.

But suppose, instead, the power of
π₯ in the function π of π₯ had been 10 or 20 rather than just three, it would be
extremely time consuming and tedious to distribute all these parentheses in order to
give a polynomial. So, it would be much more helpful
for us to have a rule that allows us to differentiate a composite function. And indeed, there is one. Itβs called the chain rule.

Weβll illustrate the chain rule by
first finding the derivative of the composite function π of π of π₯. So, weβve defined the functions π
of π₯ and π of π₯ to be two π₯ plus five and π₯ cubed, respectively. And we saw that the composite
function π of π of π₯ was two π₯ plus five all cubed, which simplifies to eight π₯
cubed plus 60π₯ squared plus 150π₯ plus 125. Now letβs consider finding the
derivative of this function.

To do so, we need to recall the
power rule, which tells us that the derivative with respect to π₯ of π, thatβs a
constant, multiplied by π₯ to the power of π is πππ₯ to the power of π minus
one. And we recall also that in order to
find the derivative of a sum or difference, we can just differentiate each term
separately and then add them together.

So, differentiating π of π of π₯
then gives the derivative π of π of π₯ prime, which is 24π₯ squared plus 120π₯
plus 150. Remember, the derivative of a
constant is just zero. So, when we differentiate that term
of plus 125, it just gives zero. Now letβs see if we can manipulate
this derivative to see if we can identify any relationship with the derivatives of
π and π individually.

Weβll first take out a common
factor of six to give six multiplied by four π₯ squared plus 20π₯ plus 25. We may then notice that four π₯
squared plus 20π₯ plus 25 is actually a perfect square. Itβs equal to two π₯ plus five all
squared. And as two π₯ plus five is our
expression for π of π₯, this is actually equal to π of π₯ squared. But what about that six? Well, six is equal to two times
three. So, we can write this whole
derivative as two times three times π of π₯ all squared. But how does this help?

Well, to see this, we need to find
the derivatives of π and π. Applying the power rule, we see
that π prime of π₯ is equal to two and π prime of π₯ is equal to three π₯
squared. So, that two in our derivative of
the composite function is the same as π prime of π₯. Now three times π of π₯ all
squared is actually the derivative of π evaluated at π of π₯. π prime of π₯ is three π₯
squared. So, π prime of π of π₯ is three
π of π₯ squared.

So, what have we found? Well, for this example, we found
that the derivative of π of π of π₯ is equal to the derivative of π, thatβs the
derivative of the inner function, multiplied by the derivative of π, thatβs the
outer function, with the inner function still inside. Now this is an illustration of the
chain rule. Itβs not a proof but that is beyond
the scope of what weβre going to look at in this video.

So, the chain rule then, it tells
us that the derivative of the composite function π of π of π₯ is equal to π prime
of π₯ multiplied by π prime of π of π₯. We can also express the chain rule
using Leibnizβs notation. If π¦ is equal to π of π of π₯,
and we let π’ equal π of π₯ so that π¦ becomes π of π’, a function of π’, then dπ¦
by dπ₯ is equal to dπ¦ by dπ’ multiplied by dπ’ by dπ₯.

This may look quite complicated,
but itβs actually a relatively straightforward process, as weβll see in our
examples. Leibnizβs notation is really
helpful because it makes the chain rule a little bit more intuitive. Remember that finding derivatives
is all about small changes in π₯. So, letβs allow Ξπ’ to represent a
small change in π’ as a result of a small change in π₯.

In order to find the derivative of
π¦ with respect to π₯ dπ¦ by dπ₯, we consider the difference quotient Ξπ¦ by
Ξπ₯. We see that by multiplying both the
numerator and denominator by Ξπ’, which must be nonzero, and then reordering the
terms, we get Ξπ¦ over Ξπ’ multiplied by Ξπ’ by Ξπ₯. As Ξπ₯ tends to zero, so will both
Ξπ’ and Ξπ¦, giving dπ¦ by dπ₯ equals dπ¦ by dπ’ multiplied by dπ’ by dπ₯. That is the chain rule. The chain rule allows us to
differentiate a wide class of complex functions. Letβs look at some examples.

Find the first derivative of the
function π¦ equals five π₯ squared minus six to the power of six.

Now we see that this is an example
of a composite function. If we consider the first function
to be five π₯ squared minus six and the second to be π₯ to the power of six. We take five π₯ squared minus six
as the input to our second function, giving five π₯ squared minus six all to the
power of six. As this is a composite function, we
can apply the chain rule.

The chain rule tells us that if π¦
is a function of π’ and π’ is a function of π₯, then dπ¦ by dπ₯ is equal to dπ¦ by
dπ’ multiplied by dπ’ by dπ₯. So, we need to decide how weβre
going to define the function π’. Well, we take π’ to be our first
function. Itβs the part inside the
parentheses, π’ equals five π₯ squared minus six. π¦, therefore, becomes a function
of π’. π¦ equals π’ to the sixth, and π’
is a function of π₯.

We need to find both dπ¦ by dπ’ and
dπ’ by dπ₯, which we can do by applying the power rule. In the case of the dπ¦ by dπ’, we
just need to think of all the π₯βs in the power rule as being π’βs. We have then that dπ¦ by dπ’ is
equal to six π’ to the five, and dπ’ by dπ₯ is equal to 10π₯. We write down the chain rule and
then make the relevant substitutions, giving dπ¦ by dπ₯ is equal to six π’ to the
five multiplied by 10π₯.

Now hereβs a really important
point. That derivative of π¦ with respect
to π₯ must be in terms of π₯, and, at the moment, we still have the variable π’
involved. So, we must make sure that we
reverse our substitution. π’ is equal to five π₯ squared
minus six, so we have six multiplied by five π₯ squared minus six to the power of
five multiplied by 10π₯. Simplifying then, we have that the
first derivative of the function π¦ equals five π₯ squared minus six to the power of
six is 60π₯ multiplied by five π₯ squared minus six to the power of five.

Now this illustrates a really
powerful application of the chain rule, in fact, a general rule for finding the
derivative of a bracket raised to a power. If we express the derivative as
10π₯ multiplied by six multiplied by five π₯ squared minus six to the power of five,
then we see what we have is the derivative of the bracket, or the derivative of
whatβs inside the parentheses, thatβs 10π₯, multiplied by the original power, six,
multiplied by that bracket with the power reduced by one from what it was
originally.

This gives us the chain rule
extension to the power rule. This tells us that if we have a
function π of π₯ raised to a power, then the derivative is equal to π prime of π₯,
thatβs the derivative of whatβs inside the parentheses, multiplied by π and then
multiplied by π of π₯ with the power reduced by one, π of π₯ to the π minus
one. This is particularly useful if we
have really high powers. So, letβs see how we can apply this
rule to another example.

Determine the derivative of π¦
equals negative two π₯ squared minus three π₯ plus four to the power of 55.

Now this is where we really see
the importance of the chain rule. When we have an exponent as
high as 55, we certainly donβt want to attempt to distribute all the
parentheses. Instead, weβre going to use the
chain rule extension of the power rule, which tells us that the derivative of π
of π₯ to the π is π prime of π₯ multiplied by π multiplied by π of π₯ to the
π minus one.

So, π of π₯ will be that
function inside the parentheses, negative two π₯ squared minus three π₯ plus
four. We can apply the power rule to
differentiate π of π₯, giving negative four π₯ minus three. Now we can work out dπ¦ by
dπ₯. Itβs equal to π prime of π₯,
thatβs negative four π₯ minus three, multiplied by π, thatβs 55, multiplied by
π of π₯ to the power of π minus one, thatβs negative two π₯ squared minus
three π₯ plus four to the power of 54.

Thereβs no need to expand the
parentheses. So, weβve found that dπ¦ by dπ₯
is equal to 55 multiplied by negative four π₯ minus three multiplied by negative
two π₯ squared minus three π₯ plus four to the power of 54. And weβve done this by applying
the chain rule extension to the power rule.

We can also apply the chain rule
more than once within the same problem. So, letβs consider an example of
this.

Find the first derivative of
the function π¦ equals the square root of eight π₯ minus sin of nine π₯ to the
power of eight.

Here we have π¦ is equal to the
square root of another function, so we have a composite function. Weβre, therefore, going to
apply the chain rule. Weβre going to define π’ to be
the function underneath the square root, so π’ is equal to eight π₯ minus sin of
nine π₯ to the power of eight. Then, π¦ is 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 tells us that
dπ¦ by dπ₯ is equal to dπ¦ by dπ’ multiplied by dπ’ by dπ₯. So, we need to find each of
these derivatives. dπ¦ by dπ’ is relatively straightforward. Using the power rule, we get
one-half π’ to the power of negative one-half. For dπ’ by dπ₯, the derivative
of eight π₯ is just eight. But what about the derivative
of sin of nine π₯ to the power of eight?

We actually need to apply the
chain rule again. We can let π equal this
function. And we can change the notation
a little to write it as sin nine π₯ to the power of eight. Itβs an equivalent notation,
but it might it make it a little clearer how weβre going to find the
derivative.

We recall the chain rule
extension to the power rule, which told us that if we had a function π of π₯
raised to a power π, then its derivative was π prime of π₯ multiplied by π
multiplied by π of π₯ to the power of π minus one. Here, we have a function, sin
of nine π₯ raised to a power eight, so we can apply the chain rule extension to
the power rule. We need to recall one more rule
which is that the derivative with respect to π₯ of sin ππ₯ is π cos ππ₯.

So, we begin. The derivative of the part
inside the parentheses is nine cos nine π₯. Then, we multiply by the power
eight. And then, we have the function
inside the parentheses written out again, but with the power reduced by one. Simplifying gives 72 cos nine
π₯ sin to the power of seven nine π₯. So, 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 a half π’ to power of negative a half multiplied by eight minus 72 cos
nine π₯ sin nine π₯ to the power of seven. We must also remember to
replace π’ in terms of π₯. So, π’ is equal to eight π₯
minus sin of nine π₯ to the power of eight. Weβll also simplify the
fractions. Dividing by that denominator of
two leaves coefficients of four and 36 in the numerator.

And we recall also that π’ to
the power of negative a half is equal to one over root π’. So, our derivative dπ¦ by dπ₯
simplifies to four minus 36 cos nine π₯ sin nine π₯ to the power of seven all
over the square root of eight π₯ minus sin nine π₯ to the power of eight.

So, weβve seen within this
question that we can apply the chain rule more than once within the same
problem. In fact, we can apply it as
many times as is necessary.

Letβs remind ourselves then of some
of the key points that weβve seen in this video. The chain rule is useful for
differentiating composite functions, thatβs functions of other functions. If π¦ is equal to the composite
function at π of π of π₯, then dπ¦ by dπ₯ is equal to π prime of π₯, thatβs the
derivative of the inner function, multiplied by π prime of π of π₯. Thatβs the derivative of the outer
function with the inner function still inside.

Weβve also seen the that if we make
the substitution π’ equals π of π₯, then π¦ becomes a function of π’. And the chain rule can be expressed
as dπ¦ by dπ₯ is equal to dπ¦ by dπ’ multiplied by dπ’ by dπ₯. We find the derivative of π¦ with
respect to π’ and multiply by the derivative of π’ with respect to π₯. We must make sure that we undo our
substitution at the end, so that dπ¦ by dπ₯ is in terms of π₯ only.

Weβve also seen the chain rule
extension to the power rule, which tells us that the derivative of a function π of
π₯ to the power of π is π prime of π₯ multiplied by π multiplied by π of π₯ to
the π minus one. Finally, we saw that we can apply
the chain rule as many times as we like within a particular problem. The chain rule is a really powerful
tool. And it opens up a really wide class
of functions which weβre able to differentiate.