### Video Transcript

Find the Taylor polynomials of the
third degree approximating the function π of π₯ equals the square root of π₯ at the
point π equals nine.

Letβs start this question by
writing out the general form for Taylor polynomials up to degree π. And letβs begin by substituting in
the point where weβre centering our approximation, at π equals nine. Also note that weβve also been
asked to find the Taylor polynomials up to the third degree. From here, we can see that weβre
going to need to find π of nine, π prime of nine, π double prime of nine, and π
triple prime of nine. To do this, weβre going to need to
find π prime of π₯, π double prime of π₯, and π triple prime of π₯.

Well, we know from the question
that π of π₯ equals the square root of π₯, which we know is the same as π₯ to the
half power. Writing it this way is going to
help us differentiate it to find the first derivative. Using the power rule, which tells
us to multiply by the power and then subtract one from the power, gives us the first
derivative. π prime of π₯ equals one over two
π₯ to the power of negative one-half.

And we then differentiate this to
get the second derivative of π. Applying the power rule again gives
us that the second derivative of π is negative one over two multiplied by one over
two π₯ to the power of negative three over two. But we can just simplify this to
negative one over four π₯ to the power of negative three over two.

And we differentiate once more to
get the third derivative of π. This gives us negative three over
two multiplied by negative one over four π₯ to the power of negative five over
two. But again, we can simplify this to
give us three over eight π₯ to the power of negative five over two.

But what we actually want here is
π of nine, π prime of nine, π double prime of nine, and π triple prime of
nine. So we substitute nine into each of
these functions. And then we substitute the values
that we found back into our working. And from here, we just need to do
some simplification to achieve our final answer.

If we start here with one over six
over one factorial and we recall that π factorial is the product of π and all the
integers below it down to one. Then one factorial is just one. So one over six over one factorial
is just one over six over one, which is just one over six.

Now letβs simplify negative one
over 108 divided by two factorial. Well, two factorial is just two
multiplied by one, which is two. So this is just negative one over
108 over two. And we know from fraction laws that
this is just negative one over 108 multiplied by two, which is negative one over
216. And finally, we need to simplify
one over 648 divided by three factorial. Well, three factorial is three
multiplied by two multiplied by one, which is six. So this is just one over 648
multiplied by six, which gives us one over 3888. So that gives us our final
answer. The Taylor polynomials of the third
degree approximating the function π of π₯ equals the square root of π₯ at the point
π equals nine.