### Video Transcript

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

So, if we take a look at what we’re
trying to differentiate here, we’ve got 𝑦 equals 𝑥 minus five multiplied by 𝑥
minus two and this is to the power of six. Well, this is in the form 𝑦 equals
𝑢𝑣. So therefore, we can use the
product rule. And what the product rule tells us
is that if we have 𝑦 equals 𝑢𝑣, then d𝑦 d𝑥, or the derivative, is equal to
𝑢d𝑣 d𝑥 plus 𝑣d𝑢 d𝑥, where this means that you’ve got 𝑢 multiplied by the
derivative of 𝑣 plus 𝑣 multiplied by the derivative of 𝑢.

So, in our function, we’ve got 𝑥
minus five is our 𝑢 and 𝑥 minus two to the power of six is our 𝑣. And therefore, if we’ve got 𝑢 is
equal to 𝑥 minus five, we can find d𝑢 d𝑥 by differentiating with respect to
𝑥. You just get one. And this is because if you
differentiate 𝑥, if you multiply the exponent by the coefficient, it’d be one
multiplied by one. And then you reduce the power of
𝑥. Well, that’d be 𝑥 to the power of
zero. So, we’re just left with one.

And then, if you differentiate
negative five you just get zero. So, d𝑢 d𝑥, or the derivative of
𝑢, is one. And then, if you got 𝑣 is equal to
𝑥 minus two all to the power of six, then the derivative of this is gonna be equal
to six multiplied by 𝑥 minus two to the power of five.

And to calculate this, so to
differentiate 𝑥 minus two all to the power of six, what we use is something called
the chain rule. And the chain rule tells us that
d𝑦 d𝑥 is equal to d𝑦 d𝑡 multiplied by d𝑡 d𝑥. And I’ll show you what that means
and how we use it. So, if we have 𝑦 equals 𝑥 minus
two all to the power of six, and then we say that 𝑡 is equal to 𝑥 minus two, then,
therefore, 𝑦 must be equal to 𝑡 to the power of six.

So then, first of all, what we’re
gonna do is work out d𝑦 d𝑡. And to work out d𝑦 d𝑡, we
differentiate. So, we’ve got 𝑡 to the power of
six. And if we differentiate that, we
get six 𝑡 to the power of five. And that’s again, we multiplied the
exponent by the coefficient, so six by one which gives us six. And then, we reduced the exponent
by one. So, we’ve got 𝑡 to the power of
five.

And then, if we differentiate 𝑡
with respect to 𝑥, we just get one. So d𝑡 d𝑥 is just equal to
one. And we’ve already explained why
that would be. So therefore, if we substitute this
back into our chain rule, we’re gonna get the derivative, or d𝑦 d𝑥, is equal to
six 𝑡 to the power of five, because that was our d𝑦 d𝑡, multiplied by one,
because one was our d𝑡 d𝑥.

So then, if we substitute back in
our value for 𝑡, which was 𝑥 minus two, we get d𝑦 d𝑥 is equal to six multiplied
by 𝑥 minus two all to the power of five. So, that’s what we’ve got when we
go back to what we were doing with the original derivative.

Okay, great, so, if we go back to
what we were doing, we’ve now got 𝑢, d𝑢 d𝑥, 𝑣, and d𝑣 d𝑥. So, now what we can do is we can
use our product rule. So therefore, we get d𝑦 d𝑥 is
equal to our 𝑢 d𝑣 d𝑥, which is 𝑥 minus five multiplied by six 𝑥 minus two to
the power of five, then plus our 𝑣 d𝑢 d𝑥, which is 𝑥 minus two all to the power
of six multiplied by one. So, now we’re gonna rewrite this to
make this easier to simplify.

So, when we rewrite it, we get six
multiplied by 𝑥 minus five multiplied by 𝑥 minus two all to the of power five plus
𝑥 minus two all to the power of six. Okay, so, usually we’d look to
simplify this further, maybe expand the parentheses. However, we don’t need to do that
in this problem because in this problem we’ve been given a value to substitute in to
our derivative.

And that value is 𝑥 is equal to
one because what we’re trying to do is find the first derivative of our function at
one, negative four. So therefore, if we substitute in
𝑥 equals one, we’re gonna get six multiplied by one minus five multiplied by one
minus two all to the power of five plus one minus two all to the power of six. So, this is gonna give us six
multiplied by negative four multiplied by then we’ve got negative one to the power
of five then plus negative one to the power of six.

So, this is gonna give us 24 plus
one. And we get 24 from the first part
because if you have six multiplied by negative four, it’s negative 24. Then, if you multiply this by
negative one, it will give us 24. And it’s negative one because if
you raise negative one to the power of five, this is an odd exponent. And if it’s an odd exponent, we get
a negative answer, if it’s already negative. Because the sign will remain the
same. However, then we add on one because
we have negative one to the power of six, which we’ve got an even exponent. The negative one becomes
positive. So, it’s just one.

So therefore, we can say that the
first derivative of 𝑦 equals 𝑥 minus five multiplied by 𝑥 minus two all to the
power of six at the point one, negative four is going to be 25.