### Video Transcript

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

One thing we could do is to take
this five π₯ squared minus six all to the power of six and expand it binomially β
that is using Pascalβs triangle. Having done this, we would get a
polynomial in π₯, which we could then differentiate term by term in the normal
way. But this will take a long time both
to binomially expand the expression and to differentiate all the terms you get.

Surely, thereβs a better way. Luckily, there is a better way. Let π§ equal five π₯ squared minus
six β that is the expression inside the parenthesis. Then, simply, by a substituting π§
for five π₯ squared minus six, we see that π¦ is equal to π§ to the sixth.

How does this help us find the
derivative of the function with respect to π₯ though? Well, we can use the chain rule,
which says that the derivative of π¦ with respect to π₯ is the derivative of π¦ with
respect to some other variable π§ times the derivative of π§ with respect to π₯. Itβs almost like the derivatives
ππ¦ by ππ§ and ππ§ by ππ₯ are fractions and the ππ§s cancel.

Applying the chain rule, we now
need to find ππ¦ by ππ§ and ππ§ by ππ₯ and multiply them together. Letβs start with ππ¦ by ππ§. π¦ is equal to π§ to the sixth. And so differentiating π¦ with
respect to π§, we get six times π§ to the five. Here, weβve used the fact that just
like the derivative of a power of π₯ with respect to π₯, you find the derivative of
a power of π§ with respect to π§ by taking a copy of the exponents down to the front
and multiplying by it and then reducing the exponent by one.

It doesnβt matter what the variable
is β whether itβs π₯ or π§ or even capital π. Having found ππ¦ by ππ§, we now
need to find ππ§ by ππ₯. Well, we have that π§ is equal to
five π₯ squared minus six. And differentiating this, we get
that ππ§ by ππ₯ is 10π₯. Hence, we see that ππ¦ by ππ₯ is
~~seven~~ [six] π§ to the five times 10π₯, which is 60π₯π§ to the five. Weβre almost done here. The last thing to say is that we
want ππ¦ by ππ₯ in terms of π₯ alone. And we can do this by substituting
our expression for π§ in terms of π₯. π§ is five π₯ squared minus
six. And so 60π₯π§ to the five is 60π₯
five π₯ squared minus six to the five.

And this is our final answer. Well, as we said at the beginning
of the video, we can solve this problem without using the chain rule. Using the chain rule is much more
efficient. For many differentiation problems,
thereβs no way of avoiding the chain rule. So itβs really worth
remembering.