Differentiating power series | Series | AP Calculus BC | Khan Academy

- [Instructor] We're told here that

f(x) is equal to this infinite series,

and we need to figure out what is

the third derivative of f,
evaluated at x equals zero.

And like always, pause this
video and see if you can

work it out on your own
before we do it together.

Alright, so there's two
ways to approach this.

One is we could just take the derivative

of this expression while
it's in sigma notation.

The other way you do it is we could just

expand out f(x) and take
the derivative three times,

and see if we get an answer
that, I guess, makes sense.

Let me do it the second way first.

Let me just expand it out.

F(x) is equal to, let's see,
when n is equal to zero,

this is negative one to the
zero, which is just one,

times x, times x to the zero plus three,

so it's gonna be x to the third

over two times zero, so that's
zero plus one factorial,

so that's just going to be over one.

Then the next term,
when n is equal to one,

well now it's gonna be
negative one to the one,

so now we're just gonna
have a negative out front.

Negative, and it's gonna be
two times one plus three,

so that's gonna be x to the fifth power

over two times one plus
one, so it's gonna be

two plus one is three factorial.

So it's gonna be x to the fifth over six.

And then when x is equal to two,

this is going to be positive again,

and it's gonna be x to the seventh power

over five factorial.

Is that right? Yeah.

Five factorial, and five factorial...

Actually, let me just write
that out as five factorial.

Five factorial would be,
what, it would be 120.

It'd be five times four
times six, so it'd be 120.

But we could just keep going minus plus,

and it goes on and on and on forever.

Well now let's just take the derivatives.

F'(x) is going to be equal to,

we're still applying the power rule here,

it's going to be three x squared

minus 5/6x to the fourth, plus seven

over five factorial x to the sixth,

I'm just applying the power rule,

minus plus, we just keep going
on and on and on forever.

The second derivative, f''(x) is going

to be equal to, apply
the power rule again.

It's going to be six x to the
first minus four times five

over six, I'll just write
that as 20/6x to the third,

plus six times seven, so
it's 42 over five factorial

x to the fifth, and we're
just gonna keep on going,

minus plus, keep going,
or alternate between minus

something then plus something,
on and on and on forever.

Then we get to the third derivative.

The third derivative is equal to,

let's see, the derivative of six x is six,

and then we have minus
20 times three is 60/6,

which of course is 10, x squared,

plus five times 42 is what,
210 over five factorial

times x to the fourth power, minus plus

over and over and over again, and then

we just evaluate this at zero.

F'''(0), well, when x is equal to zero,

all of these terms with
xes are gonna go to zero,

and you're just gonna be
left with this six here.

So f'', the third derivative evaluated

at zero is just equal to six.

Now another way that we
could've tackled this

is just kept it in this sigma notation.

We could've said that f'(x) is equal

to the infinite sum, and
actually, let me line them up.

So this is where we
did f'(x) expanded out,

but we could've said
f'(x) is equal to the sum

from n equals zero to infinity,

and you take the derivative here,

you're gonna get, and
you're taking the derivative

with respect to x, so for that purpose,

you assume everything else is,

well, the n is just gonna tell us,

is gonna tell us how we
change from term to term,

so if we take the derivative
with respect to x here,

use the power rule, bring the
two n plus three out front,

so it's gonna be negative one to the n

times two n plus three,
times x to the decrement,

the exponent, two n plus two over

two n plus one factorial.

And then if you wanna take
the second derivative,

and this is the same thing as this.

If you take the second derivative, f''(x),

well now we're taking the sum from zero

to infinity of negative one to the n.

Let me move over to the right a little bit

so we have some space.

And now, we take this exponent out front,

so you're gonna have two n plus three

times two n plus two,
all of that's going to be

over two n plus one factorial,
and this is gonna be

times x to the two n plus one.

All I'm doing every time,
it seems really complicated,

I'm just taking the exponent out front,

multiplying it out front,
and then decrementing it.

So two n plus two minus
one is two n plus one.

And then if I wanna take
the third derivative,

the third derivative is
the sum where n equals zero

to infinity, negative one to the n.

We take this, bring it, multiply it,

so we're gonna have two n plus three

times two n plus two,
times two n plus one,

all of that over two n plus one factorial,

and then that is going to be times

x to the two n power.

Now let's now evaluate this
thing when x is equal to zero.

F'''(0) is gonna be the
sum from n equals zero

to infinity of negative
one to the nth power.

This is gonna be interesting.

We're gonna have all of this business,

two n plus three times two n plus two

times two n plus one, all of that

over two n plus one factorial,

times zero to the two n power.

You might be tempted to say, well, hey,

if zero to these different powers,

maybe everything's gonna be zero,

but remember, we're
starting at n equals zero,

so for any n that's not equal to zero,

this zero to that power
is just gonna be zero

and that term's gonna disappear,

kinda like what we saw
when we expanded it out.

And so the only term that matters

here is when n is equal to zero.

So this is just going to be equal to,

because for n equals one,
two, three, four, five,

all the way to infinity,
this thing is gonna dominate.

It's just gonna multiply,
it's gonna be zero.

You're just gonna zero everything out.

And so this is just gonna
reduce to the first term,

when n equals zero,
and when n equals zero,

it's gonna be negative one to the zero.

This is gonna be, which is just one.

Let me just write that as one.

Times, this is three times two times one

over one factorial, and
then times zero to the zero,

which is equal to one.

So this is equal to one,
and so this is equal to six.

Either way, I think
the first way we did it

was a little bit more straightforward,

a little bit more
intuitive, closer to what

you might have seen
before, but it's important

to realize that we did
the same thing both times,

we just kept it in the sigma notation

on this time to the right.

This technique is useful
because you'll see it a lot

in math, where you might
wanna do things a little

bit more of a general way,
and so it might be helpful

to take the derivatives while you stay

in that sigma notation.