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Current time:0:00Total duration:2:58

Part C. Find the value
of the sixth derivative of f evaluated at 0. So you could imagine
if you just tried to find the sixth derivative of
f, that would take you forever. And then to evaluate it at 0,
because this is x squared here. And you'd have to keep doing
the product rule over and over again, and the chain
rule and all the rest. It would become very,
very, very messy. But we have a big clue here. The fact that they made
us find the first four terms of the Taylor series
of f about x equals 0 tells us that there
might be a simpler way to do this, as
opposed to just taking the sixth derivative of
this and evaluating at 0. The simplest way to do
this is to just go back. In the last problem,
we were able to come up with the first four
non-zero terms of the Taylor series of f. And if you look at your
definition of the Taylor series right here-- and we go into
depth on this in another Khan Academy video where we talk
about why this makes sense-- you see that each degree
term of the Taylor series, its coefficient
is that derivative. And this Taylor Series
is centered around 0, and that's what we care about
in terms of this problem. We see the coefficient is
that derivative divided by that degrees, that
derivative evaluated at 0 divided by that
degrees factorials. So the second degree term,
it's the second derivative of f evaluated at 0
divided by 2 factorial. The fourth degree term
is the fourth derivative of f evaluated at 0
divided by 4 factorial. So the sixth degree term--
let's remind ourselves what we're even
trying to figure out-- so they want us to figure
out the sixth derivative of f evaluated at 0. That's what they want
us to figure out. Well, if you think about
the Taylor series centered at 0, or at 0, or
approximated around 0, the sixth degree term in the
Taylor series approximation of f is going to be f prime
of the sixth derivative of f evaluated at 0 times x to
the sixth over 6 factorial. This is going to
be the sixth degree term in Taylor approximation,
in Taylor series. And we have that term
sitting right over here. This is the sixth degree term. We figured it out
in the last problem. This right here is
the sixth degree term. So you have x to the sixth over
here, x to the sixth over here, you have 6 factorial over
here, 6 factorial over here. So this negative 121 must
be the sixth derivative of f evaluated at 0. So that's our answer. This is equal to negative 121. And we're done.

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