### Video Transcript

The Taylor series representation of
the function π is given by π of π₯ is equal to two minus four multiplied by π₯
minus seven squared plus three multiplied by π₯ minus seven to the fourth power
minus π₯ minus seven to the sixth power divided by six. And the series keeps going and
going. Find the value of the fourth
derivative of π at π₯ is equal to seven. Find the value of the third
derivative of π at π₯ is equal to seven.

The question wants us to find the
fourth derivative of the function at π₯ is equal to seven and the third derivative
of the function π at π₯ is equal to seven. And to do this, it gives us a
Taylor series representation of our function π of π₯. Weβll recall that the Taylor series
for a function π about π is equal to the sum from π equals zero to infinity of
the πth derivative of π evaluated at π divided by π factorial multiplied by π₯
minus π to the πth power. And this is for all values of π₯
where the series converges.

Since we know the representation of
our function π of π₯ is a Taylor series representation. And we can see that each term in
this representation has π₯ minus seven raised to various powers. So this must be a Taylor series
about seven. So thatβs the π in the definition
of a Taylor series to be equal to seven.

So substituting the value of π
equals seven into our definition of a Taylor series gives us that π of π₯ is equal
to the sum from π equals zero to infinity of the πth derivative of π evaluated at
seven divided by π factorial multiplied by π₯ minus seven to the πth power. At this point, we can start writing
out the terms of this series by substituting in each value of π into our
summand.

This starts with π equals zero
goes to π equals one and carries on. Since the zeroth derivative of π
is just equal to π, zero factorial is equal to one, and raising π₯ minus seven to
the zeroth power just gives us one. The first term is just π evaluated
at seven. Similarly, we can write the second
term in our Taylor expansion as π prime of seven multiplied by π₯ minus seven. We could do this for the third
term. In fact, we could do this for as
many terms as we would like.

What we want to notice from this
Taylor expansion is the coefficient of π₯ minus seven to the πth power. What we want to notice from this is
that the πth derivative evaluated at seven divided by π factorial is equal to the
coefficient of π₯ minus seven raised to the πth power. What this means in particular is
that the fourth derivative of π evaluated at seven divided by four factorial is
equal to the coefficient of π₯ minus seven to the fourth power. And similarly, we will also have
that the third derivative of π evaluated at seven divided by three factorial is
equal to the coefficient of π₯ minus three all cubed.

From the question, we can see that,
in the Taylor series representation of our function π, the coefficient of π₯ minus
seven to the fourth power is three. So we must have that the fourth
derivative of π evaluated at seven divided by four factorial is equal to three. Multiplying both sides of this
equation by four factorial gives us that the fourth derivative of π evaluated at
seven is equal to three multiplied by four factorial, which is equal to 72.

Next, we know that the third
derivative of π evaluated at seven divided by three factorial will be equal to the
coefficient of π₯ minus seven all cubed in our Taylor series representation. However, when we look at the
question, we see that there is no π₯ minus seven all cubed term. If there is no π₯ minus seven all
cubed term, this is the same as saying its coefficient is zero.

So we must have that the third
derivative of π evaluated at seven divided by three factorial is equal to zero. We can then multiply both sides of
this equation by three factorial to get that the third derivative of π evaluated at
seven is equal to zero multiplied by three factorial, which is just equal to
zero.

So what weβve shown is that if the
Taylor series representation of a function π is given by two minus four multiplied
by π₯ minus seven squared. Plus three multiplied by π₯ minus
seven to the fourth power minus π₯ minus seven to the sixth power divided by six, et
cetera. Then the fourth derivative of our
function π evaluated at seven is equal to 72. And the third derivative of our
function π evaluated at seven is equal to zero.