VIRGE CORNELIUS: Hi, Mark.
Hi, everybody.
Thanks so much for tuning in.
This is our first session of--
what are we calling it this year, Mark, AP practice?
MARK KIRALY: AP Daily Practice.
VIRGE CORNELIUS: Oh, AP Daily Practice.
And these are going to be short videos, so you can definitely
get a lot of practice in a short amount of time.
I'm Virge Cornelius.
I teach at Lafayette High School in Oxford, Mississippi.
MARK KIRALY: And I am Mark Kiraly at Ryan High
School in Denton, Texas.
VIRGE CORNELIUS: And this session, we're
going to focus on some multiple choice with no calculator.
And there's going to be a link if you
want to print these problems out for yourself.
They're brand new problems.
Mark and I wrote everything that's
going to be in these 8 sessions.
So find that link, print them out, and then come back
and join us.
Maybe you've already worked them or maybe you're
just going to work them along with us.
It doesn't matter to us.
We're so happy you're here.
So here's our first one, very typical question.
It says, if y is equal to x squared tangent of x, then dy,
dx equals, and then we've got four answer choices.
2x secant squared x, negative 2x squared x, negative x
squared secant squared x plus 2x tangent x.
And then x squared secant squared 2x plus 2x tangent x.
And what can I think about in terms of calculus here?
Well, I'm trying to take a derivative,
and I see that I have a product.
X squared tangent x means x squared times tangent x.
So if I'm taking a derivative and I have a product,
I wonder what I would use.
MARK KIRALY: It's almost like you
should have a rule for that.
VIRGE CORNELIUS: Oh, is it a product rule?
Look, I even have a yellow sheet right here.
Product rule is such an important rule in calculus.
If you have two functions that are multiplied together,
then the derivative is going to be
the first times the derivative of the second fg
prime plus the second times the derivative of the first g times
f prime.
If you don't have a list of rules for yourself,
now is the time to definitely start building one.
So let's see what we're going to do here.
So dy dx, that's another way of saying
the derivative, like the y prime,
would be my first which would be x
squared times the derivative of tangent x.
Ah, I need to derivative of tangent x.
So Mark, do you remember what that is?
MARK KIRALY: I believe it's secant squared x.
VIRGE CORNELIUS: Yeah, and actually,
just based on my answer choices, that's not a surprise, right?
MARK KIRALY: Yes, nothing surprising.
VIRGE CORNELIUS: Nothing surprising.
MARK KIRALY: But non negative because the cofunctions
have negative derivatives, but the non-cofunctions all
have positive derivatives.
VIRGE CORNELIUS: OK, so what you're saying is,
if it was something like y is equal to the cotangent of x,
then y prime would have been--
MARK KIRALY: Negative cosecant squared x.
VIRGE CORNELIUS: Yeah, OK.
Great.
So any time you have a cofunction,
the derivative is going to be negative.
All right, so I had first times the derivative
of the second, and then plus second,
which is tangent x, times the derivative
of the first derivative of x squared is 2x.
All right, so do I see an answer choice?
It's looking like D. What do you think, Mark?
MARK KIRALY: The answer's like D.
VIRGE CORNELIUS: OK.
Is that what we said yesterday when we worked on these?
MARK KIRALY: I believe so, yes.
VIRGE CORNELIUS: Good.
And of course, the 2x got moved to the front.
No big deal.
So let's take a look at our second one.
So our second one says, given that f of x
is equal to the natural log of x divided by root x,
find f prime 4.
So in this case here, this function
looks like it's a quotient.
I guess there must be a rule for that too, right?
So--
MARK KIRALY: And surprisingly, quotient rule.
VIRGE CORNELIUS: Oh, quotient rule.
So if you have a quotient such as,
and I'm just being very basic here, such as f divided by g,
then your derivative y prime would be--
sorry, I have to sing that off.
MARK KIRALY: Oh, I feel a song coming on.
VIRGE CORNELIUS: So excuse me but--
(SINGING) low-D-high min high-D-low, oh, low-low.
OK, so that would be my derivative of the quotient.
So low times the derivative of the numerator minus high times
the derivative of the denominator,
and then over the denominator squared.
So let's see what we can do here.
So I'm going to take the derivative.
So I've got--
MARK KIRALY: These are fun ones.
VIRGE CORNELIUS: So yes, they are.
I worked them yesterday.
All right, so f1 of x is equal to low.
So that's root x.
D high, uh-oh, derivative of natural log of x.
So that would be what, Mark?
May be something else I should put on my yellow sheet, which
is not very organized.
MARK KIRALY: Yeah, derivative natural log of x is 1 over x.
VIRGE CORNELIUS: Yes, it is.
It's 1 over x.
Good.
I like to put little boxes around my stuff.
So low-d-high, min, that's my minus.
High which is the lnx, d low, so times derivative of root x.
Now, I kind of very quickly did my power rule right here.
Derivative of x squared was 2x.
But on this one with root x, I might take a little more time
with my power rule.
So I might say to myself, in fact, I think Mark,
you do a little cloud bubble sometimes off on the side.
If I have-- say it again?
MARK KIRALY: Because I love the thought bubble.
VIRGE CORNELIUS: The thought bubble, right.
So the square root of x is just the same as x to the 1/2.
So it is a power function.
And if I take the derivative of that, it's going to be 1/2 x
to the negative 1/2.
1 minus 1/2--
1/2 minus 1 and the exponent there is negative 1/2.
Now, me personally, and this Mark only heard this 47 times,
times 47, I always have this just memorized as 1 over 2 root
x.
It's just easier for when you're substituting
n, which is what we're going to have to do in this case.
And algebraically, you guys can talk about that on your own
about how that transfers over.
(SINGING) So low-D-high, min high-D-low,
oh, low-low over the denominator squared.
And even though I could, right now,
start cleaning it up a little bit,
because I have to find f prime of 4,
I would go ahead and substitute in the 4 right now.
MARK KIRALY: Yeah, I did the same thing this afternoon.
VIRGE CORNELIUS: Subbing in the 4?
MARK KIRALY: Yeah.
VIRGE CORNELIUS: Yeah.
MARK KIRALY: Subbing those numbers in there
just to make it more comfortable.
VIRGE CORNELIUS: Yeah, so root 4 times 1 over 4
minus natural log of 4 times 1 over 2 root 4 all divided
by root 4 squared.
So I have router 4, which is 2 times of 4
minus natural log of 4.
And then down here, I have square root of 4
is 2, 2 times 2 is 4.
So times another 4, and then root 4 times
root 4 is going to be 4.
Now, this is not really anything like our answers quite yet.
But algebraically, I could take numerator and denominator
and multiply by this denominator of 4.
And then probably have something exactly what I want.
So in other words, I'm not multiplying 2 times a 4
only because I've done this from 5 times.
I know that I probably want to just do 4 times a 4
and have that cancel out and get 2.
And then here, 4 times a 4, so I just have natural lag of 4.
So 2 minus natural log of 4 in the numerator, and then 4 times
4 is 16 in the denominator.
But I think I have an answer now.
All right, here comes number three.
So number three says, if x to the 4 minus 3x cosine of y
equals 10, then find dy dx.
So I can't really very easily solve for y here,
and this wasn't given to me in terms of exclusively x.
So to me, this looks like a function
that is an implicitly defined function.
It might not even be a function.
It might be a relation of some sort.
So I would use implicit differentiation on that.
And so that means I'm just going to take the derivative
of each term as I go along.
Which then derivative of x to the fourth would be--
MARK KIRALY: It sounds like a power rule.
VIRGE CORNELIUS: Sounds like a power rule.
4x cubed derivative of 3x cosine y
is going to be another product rule, because I've
got my two different--
I didn't label these before, but have essentially got f times g.
Now, stylistically, you can do what you want here,
but I tend to keep my negative 3 outside,
and then do my power rule on the inside.
Some people like to leave that negative 3 coefficient
with their x term.
So I'm taking--
MARK KIRALY: I'm just [INAUDIBLE]
that bracket out a little and make f negative 3x.
VIRGE CORNELIUS: Yep, you could do that too.
You could do that too.
I think I'm going to just let f equal x though.
Yeah, and keep that three outside.
But again, stylistically, with different practice,
whatever's most efficient for you
and yet maintaining your accuracy and speed.
So I'm going to have my negative 3
outside the bracket, and then f, which is x or my first times g
derivative of cosine y.
That's going to be negative sine y.
But then I have to take the derivative
of that inside function.
Here comes the implicit differentiation.
So that's dy dx.
And then plus, good thing for a reasonable pens,
cosine of y, my second, times the derivative of my first
and the derivative of just x is 1.
Notice how I still have that negative 3 on the outside.
And this equals derivative of 10, which is zero.
MARK KIRALY: And don't lose your concentration
when you hit that constant.
VIRGE CORNELIUS: This one.
I know.
Although luckily, they don't show up in the answers
that much.
But I don't know.
It could be wrong.
I don't know all the multiple choice questions that well
because they don't--
MARK KIRALY: Free response or something like that.
Definitely could sneak its way in.
VIRGE CORNELIUS: Yeah, so all I'm just doing here
is distributing, because I do need to solve for dy dx.
And I just have a single dy dx term,
so I'm not going to have to gather them together and factor
out that dy dx.
That might ring a bell for some of you.
MARK KIRALY: I also want to point out,
you didn't say it while you did it, but real nice job when
she distributed the negative 3 to the negative sign.
VIRGE CORNELIUS: Ah, negative times negative
is positive, yeah.
MARK KIRALY: In case people are wondering
where that negative went.
VIRGE CORNELIUS: OK, good.
And then I'm going to subtract this 4x cubed from both sides.
And depending on how comfortable you are,
you might also, at the same time,
know that you're going to be dividing by 3x sine y and then
just kind of glance up at your answers and notice,
oh, it's got to be this one right here because that 3x sine
y is in the denominator, and I've got my negative 4x cubed.
But just to finish it off, it would look like this.
If you have time, you should really practice multiple choice
in terms of not finishing them off.
OK, Mark, do you know where I'm going with this?
MARK KIRALY: Yeah.
VIRGE CORNELIUS: To save yourself time.
A lot of times, you can get about 2/3 to 3/4 of the way
through a multiple choice and kind of know
what the answer is if you've practiced enough.
And that way, you have more time on the harder questions
to have time to think about what's going on those.
So for the noncalculator multiple choice,
how many minutes do we have per question?
MARK KIRALY: Two minutes, 30 problems in one hour.
VIRGE CORNELIUS: Yeah, two minutes per question.
So these, they go pretty quickly,
but they can go even quicker if you want to kind of stop here.
MARK KIRALY: But don't let that intimidate you either.
If you ever think that's really short,
stare at a clock for 2 minutes if you can.
60 seconds is a lot of times.
VIRGE CORNELIUS: Oh, to me, it's a lot of time too.
MARK KIRALY: It's not luxurious, but it's enough.
VIRGE CORNELIUS: So we just did another derivative here.
We just did-- that was quotient rule.
We just did the derivative of cosine.
I'm just going to put it of x.
What was the derivative of cosine?
A negative sine of x.
There goes that cofunction again.
And we also did implicit differentiation.
So that's something to review for yourself,
especially related rates.
Related rates, big time with implicit differentiation.
You're going to need that for that dx dt
and things like that.
OK, last multiple choice right here.
This is d dx of 5 raised up to the x squared,
evaluated at x equals 3.
We have to take the derivative of this function, which
is exponential.
And there's also-- it looks like a chain rule going on.
So we've got a lot going on here.
Maybe better review a couple of things on my sheet.
For chain rule, how do you usually write chain rule?
Mark, do you write it with like an f of g of x or?
MARK KIRALY: The f of g of x is how I introduce it.
And then later on, I'll change it to f of--
just I want to have of x.
I could just f of g, or f of v, or whatever
I want to do it as, but of u, or something like that.
VIRGE CORNELIUS: Or you could even
do it, yeah, like with some differentials even, right?
MARK KIRALY: Yes, I introduce it.
VIRGE CORNELIUS: Yeah, so for chain rule,
if you have a composite function, one function that's
tucked inside another function, which really looks similar,
unfortunately to a product.
You'll see as we do some later sessions how it can really
get tangled up and they look very, very similar.
But what you would do for the derivative
is you would take the derivative of the outside function
with the inside function still in there.
And then multiply by the derivative
of that inside function.
So that's going to be important in this problem
that we do chain rule.
But then also important is that we
know that the derivative of some exponential function--
and I'm just going to say y is equal to a raised up to the x.
Yeah, well, our favorite-- this is actually
truly one of my favorite functions
is y is equal to e to the x.
Mark, what's the derivative of that?
MARK KIRALY: E to the x.
VIRGE CORNELIUS: Like the Phoenix
rising from its own ashes.
So too, e to the x.
But I was thinking about this one where the base wasn't e.
So what would the derivative here be?
MARK KIRALY: It's going to start the same.
You say, you start with a to the x again.
Start with that same exponential,
but then a natural log of a, which is hiding there
in the other one, right?
Because what's the natural log of e?
VIRGE CORNELIUS: That's a great thing
to put on your formula sheet.
The natural log of e is 1.
So here, when you multiply by the natural log of that base e,
you're just multiplying by 1, which is why you don't see it.
So here-- so you need to see it.
MARK KIRALY: The rule for a of x works
when a is e, if we want to just move some vowels around.
VIRGE CORNELIUS: (SINGING) E-i-e-i-o, yeah.
[LAUGHTER]
All right, I think we have enough.
MARK KIRALY: I wonder if it's a 5.
I wonder if it's a 5.
VIRGE CORNELIUS: What if it's a 5?
No big deal.
OK, so if it's a 5, we're going to say that this derivative is
equal to 5 raised up to the x squared
because the derivative of an exponential function is itself.
Oh, but wait a minute.
We have a base of 5 instead of a base of e.
So we have to multiply by the natural log of 5.
Oh, but wait a minute.
We've got an x squared, so we have
to take the derivative of x squared.
And we're evaluating all this at x equals 3.
That vertical bar is telling us that we're
evaluating it at x equals 3.
So we have 2 times 3 is 6.
So natural log of 5 times 6, 3 squared is 9.
So woo, do we have that?
Is that a?
MARK KIRALY: We do.
The answer is a.
VIRGE CORNELIUS: Yay!
MARK KIRALY: So we have some takeaways today, Virge.
VIRGE CORNELIUS: Yes.
You must know your differentiation rules
and when to use them.
If you haven't built yourself a formula sheet yet or flashcards
or something, now is the time.
And my students have a yellow sheet,
but it wouldn't be a bad idea for them
to recopy it since it's been a little bit since they've
done calculus.
That was in December of 2022.
So you must know them.
And we did four problems, but we covered a lot of rules,
like a lot of rules.
There's still more, though.
MARK KIRALY: At this point in the year,
you actually can do a large portion, if not
all of the problems on the exam once you know
what you're being asked to do.
And so that's what we're here for,
for the next seven sessions after this for these two weeks.
And if you go back and look at the videos that are still
available from 2021, 2022, a lot of it
is just understanding, what am I being asked,
so you know what to do.
And then I always say, keep your pencils moving.
Don't stare at it for too long.
If you're going to stare at it for very long,
move on or keep moving.
Well, thank you.
And hopefully, we will see you in the next video.
VIRGE CORNELIUS: Yeah, looking forward to it.