Good day student welcome to mathgotserved.com
and this clip we're going to be going over
problems 1 to 5 of the multiple choice AP
calc exam 01 2
All right let's take a look at question number
1 E38 if Y equals X sinx then the derivative
dy/dx equals what so that if you want to keep
in mind for this problem is that you do not
distribute the derivative across a product
or quotient of functions okay if you have
a product or a quotient of functions you use
the product or quotient rule but you can distribute
the derivative across sums and differences
okay so let's all take a look at the formula
that will be guiding our problem solving process
for today See the formulas are first one is
the product rule which you should have mastered
before the AP exam the product rule is if
you're looking for the derivative UV Prime
of two functions you NV the derivative is
Vu prime plus UV Prime okay and then you need
to remember your trigonometric differentiation
rule the tree differentiation rule that we
are going to be using for this examples the
derivative of sine the derivative of sinx
is cosine X and lastly the power of row we're
going to be applying that's to the derivative
of x x Prime equals 1 okay now it's go ahead
and I'll solve the problem we have Y equals
X sinx so we have a product of two functions
delenia function X and the trigonometric function
sinx so we're going to do a label them you
and be okay so the first function is you and
the second function sinx is V okay so what
is dy/dx DUI DX is equal to X sign X sinx
Prime remember Prime end dy/dx are the same
thing okay so let's see what we have apply
in the product rule will have Vu Prime is
going to be Sinex Furby X you prime which
is X Prime plus you which is x times the prime
sign X Prime okay now is differentiate and
simplify sine X times the derivative of x
is just one plus x times of derivative of
sine is cosine X okay it would simplify will
have sinx + x cosine X as our final answer
and that is option letter b
now let's take a look at? To let F be the
function given by f of x equals 300 x minus
x to the Third on which of the following intervals
is a function is increasing one penny one
to keep in mind is the connection between
the sign of f Prime and the behavior of f
if a firm is positive on an interval then
it will be increasing on that interval okay
so let's keep that in mind now what's right
on the procedure for solving this problem
there are three parts to this process part
one involves finding the critical values okay
so you find the critical values what are the
critical values of the X values where the
derivative is equal to 0 or does not exist
okay after that we're going to set up a number
line sign chart okay so set up line sign chart
that enable us to see how the sign of f Prime
changes across the critical values and then
lastly we will find the intervals find intervals
where F Prime is positive okay let's do it
Part 1 we want to find the critical values
so find the critical values physically where
is f Prime is equal to zero or does not exist
start with the function f of x equals 300
x minus x to the third we're going to find
the first derivative of prime of X applying
the power rule the derivative of 300 X 300
in the derivative of x to the third is -3x
Square now to find the critical values we're
going to see wherever Prime of x equals 0
and wherever Prime of X does not exist okay
since we do not have a denominator here there
isn't an ex value where this function is going
to be on the find this is a polynomial function
it is smooth and continuous is differentiable
everywhere okay so are we just have to do
is find where the derivative is a pool to
0 this situation is not applicable so we'll
take the derivative and set it equal to zero
and let's solve this we can subtract 3x square
from both sides and flipped equation will
have 3 x squared equals 300 divide both sides
by 3 that you check square equals 100 and
then we are going to take the square root
of both sides not one thing you want to keep
in mind is that anytime you take the square
root of a square you have to introduce the
positive and negative roots Plus or minus
10 because when you square root when you squirt
end or -10 you end up with positive 100 okayNow
we have our critical values left list them
are critical values are x equals negative
10 and positive 10 we are done with part 1
of the procedure now it's Advanced 2 part
2 which involves creating a number line sign
chart let me partition my workspace here part
two and three are basically combined in one
okay so we have our number line sign chart
we're going to grab the critical values on
a number line this will be an open interval
because we do not have any cut strings on
the domain okay so we're going to grab a -10
and positive 10 now let's pick up some partitions
so we can have a visual distinction as to
what's going on where because we have two
critical values our sign charge will be divided
into three regions okay so in the first region
we're going to select an x value basically
an X Valley that's smaller than -10 let's
use x equals negative 11 and then in the second
interval is coldest one in the second interval
To pick next Value this between -10 and 10
x equal zero will be a good option because
it's easy to work with and then over here
we're going to pick x equals positive 11 4
into Vault 3 now what we want to do is we
want to use this time to determine the behavior
of f okay so we're going to be looking for
the sign of f Prime and use that to determine
the behavior of the function fAll right let's
start with the first interval so we are going
to look for the sign of f Prime what we're
looking for is f Prime of 11 so we'll simply
plug in -11 sorry who was simply plug a -11
into the derivative function right here so
that is 300 -3 times -11 square if we simplify
that we will have 300 -3 x 121 which gives
a 300 - 363 and then the difference is going
to be negative okay so we have -63 that tells
us that the function is going to be decrease
in on this interval right here now let's take
a look at interval 2 x equal 0 or 1 to look
for the sign of f Prime F Prime of 0 is going
to be 300 - 3 x 0 square is in the order of
operations we have 300 minus 3 times 0 which
is just see row and we can see that F Prime
is positive on this interval what does that
tell us about the behavior of f it tells is
that f is increasing on this interval the
last one is from 10 to Infinity we're using
x equals 11 - looking for f Prime of 11 which
is 300 - 3 X 11 Square if we simplify I will
have 300 -3 x 121 which equals 300 minus 363
and we'll end up with a negative result again
so on this interval the function is going
to be decreasing now what is our conclusion
our conclusion is that let's write it down
conclusion since the first derivative F Prime
is greater than 0 on the interval -10 - 10
then if is doing what f is increasing on that
interval 
okay so answer it is going to be option letter
B that is where the function is increasing
at
Now let's take a look at the problem number
3 where to find the integral of secant x 10
x now it's see if you want to keep in mind
it has to do with the reversal of known derivatives
now if you know the difference rule for a
function then you automatically know what
the related integration rule is the reversal
okay so the formula that we're going to use
it for this problem is as follows the derivative
of secant x is equal to secant x Tan x okay
now how does this help us with this problem
in this problem we are asked to find the integral
of secant x tan x what is the relationship
between the integral and derivative there
inverse operations right so since this is
a derivative are integrating this we're going
to end up with the antiderivative with you
II X2 when we integrate II X 10 X this is
a known derivative okay so by simply reversing
the differentiation rules up here we are going
to end up with secant x plus c answer is option
letter A
All right let's take a look at problem 4 it
reads if f of x equals 7 x -3 + the natural
logarithm of X then F Prime of 1 is equal
to what now caution that you want to keep
in mind has to do with evaluating derivatives
now whenever you're in the derivative at a
point you want to remember to differentiate
first before you substitute you never want
to substitute and ever I'll wait then differentiate
that's incorrect so you do your Calculus first
before you do your arithmetic now these are
the steps that I use highlight head in the
cautionary steps that we went over so the
steps are first you find the derivative find
F Prime and then secondly you evaluate the
derivative at the indicated point so ever
X is equal to 7 x -3 + the natural logarithm
of X Step 1 will find the derivative of prime
of x the derivative of 7 x is just seven the
derivative of 3 using the constant rule of
0 and the derivative of the natural log of
x is 1 / axe so if we simplify a prime of
X will have 7 plus 1 over X now the derivative
at 1 when I'll do end Step 2 okay the derivative
evaluated at x equals one is 7 + 1/1 which
is 7 + 1 and give those the answer 8 option
letter is the correct answer for this problem
All right let's take a look at this question
number 5 it reads the graph of the function
f is shown above so I have it right here which
of the following statements is false so one
thing you want to keep in mind is when the
double sided limit exists and when it does
not exist giving a graphical representation
as is the case with this problem the double
sided limits always exist at a point where
a function is continuous or a point where
a function has a removable discontinuity so
if you have a scenario like this out of continuous
a point to where the function is continuous
or cause that point or where you have a removable
discontinuity in these two cases the limits
will exist because the left and right hand
limits will be equal to exactly the same y
value okay the approaching the same y value
as you approach these two points from the
left or right so let's keep that in mind so
what we're going to do is take a look at options
a to d they all have to do with the existence
of limitsSince the problem asks us for which
statement is false live see if maybe one of
these limits is non-existent okay if we do
not find the answer here we can take a look
at option e and do the test for continuity
so the strategy that we're going to be using
let's write it down strategy is want to find
where the limit exist that's what option a
2d limit exist when the left hand limit is
equal to the right hand limit okay all right
let's take a look at auction will start by
identifying the points that we are approaching
look at the left hand limit and then the right
hand limit season the ground and then we'll
see if right hand limit is equal to the left
hand limit okay? Now for auction a we're approaching
to so what direction are we approach into
from there new directions so that automatically
tells us that we are approaching from the
left and the right okay left hand limit is
basically the limit as X approaches 2 from
the left of the function that what on Earth
is that given the graph we are approaching
to from the left okay so when given a graph
the graphical limit is physically the Y value
of the points that you're approaching so this
is the point even though it's on the find
what is the word value of this point that
we're approaching the Y values 2 right hand
limits limit as X approaches 2 from the right
of the function to her approach into from
the right take a look at this point is undefined
point here or a removable discontinuity is
more appropriate because it's actually did
find those your proach this removable discontinuity
what is the y-coordinate of this point it's
too also and so is the right hand limit sequel
to the left her limits the answer is yes limits
as X approaches 2 of the function exists option
A is correct all right now we just talked
about earlier has these two being the only
cases where the limit exists we can just look
at the graph and we can easily determine what
the answer is if you look at 2 3 4 5 2 3 4
5 do we have any cases where we have a discontinuity
that's a jump or infinite or oscillating answer
is yes take a look at auction for we do not
have a continuous function or a removable
discontinuity there we have a jump discontinuity
which automatically means the limits will
not exist and the answer will be up in see
okay but let's go ahead and just show you
the rest so if you look at or should be nicer
to exercise a limit exit approach in 3 from
what's Direction no sign is indicated so it's
automatically left and right is double-sided
now the left hand limits we are going to be
approached in 3 from the left side so graphically
look at the points reproaching right which
is 3 we're approaching 3 from the left if
you're approaching 3 from the left what is
the y-coordinate of the point you're approaching
its 5 right so there goes the answer Now let's
take a look at the right hand the limit limit
as X approaches 3 from the right of the function
so approaching the same points because they
have a continuous function from the right
what is the y-coordinate of the point and
its 5 are the right and left hand limit sequel
yes what does that mean the limit exists and
is equal to 5 no option see we said that the
limit does not exist there because it's not
continuous and it's not a removable discontinuity
we have a jump discontinuity there so let
me show you why the limit does not exist option
see we have X as a protein for from which
direction left or right by Direction all right
under the left hand limit for a thing as X
approaches for from the left of the function
so I text equals four coming from the left
what is the Y value that the function is approaching
what is the weather Valley of the point that
the function is approaching it's kind of like
1.9 right a little bit below so we have 1.9
for the left hand limit now this approach
for from the right what is the right hand
limits limit as X approaches 4 from the right
of the function now you notice there's a junk
we have a different function now and I'm looking
at this functional looks like a problem what
is the Y value of the points that you're approaching
here at the Y value is 4 that's the right
hand limit so is the right and left hand limit
the same the answer is no sew option left
letter see if they feel this is one that fails
hands that is I will correct answer option
10 letter cNow even though we have the answer
let's take a look at some action DND okay
so far option DX is approaching 5 from the
left and the right since we do not have any
directions to 10 - x is approaching from the
left let's do the left hand limit first and
see what it looks like so we have this point
right here what is the y-coordinate of the
point that you are approaching this 46 Valley
the y coordinate that you're approaching is
six from the left so this is equal to 6 and
then from the right to limit as X approaches
5 from the right of the function is also 6
you're approaching the same point from the
left and from the right so that tells us that
the limits doesn't fact exist option is talking
about something different is talking about
continuity and ask if the function f is continuous
at x equals 3 super continuity let's right
down the approach for that for continuity
the left hand limit has to be equal to the
right hand limit and there is another condition
it has to be equal to the value of the function
at the specified x value okay so let's take
a look at 3 what's happening at 3 x equals
3 we are already over it is showing that the
left her limits and the right hand limits
are equal to 5 okay but what is the value
of the function there to 4 option I will looking
at um x equals 3 and Kay from the left on
the right so it let me guess copy that what
I wrote earlier the left hand limit the function
is equal to 5 the right hand limit as X approaches
3 from the right is also equal to 5 we need
another condition we need to find the value
of the function what is the value of the function
at x equals 3 so are simply doing his evaluate
the function at x equals 3 so what is M45
at x equals 3 what is that outputs we can
clearly see that at x equals 3 the output
is 5 and K so if of 5 is equal to 5 now since
the right hand limit is equal to the left
hand limit and is equal to the value of the
function is automatically implies that the
function is continuous at x value do we see
why option is correct it's true okay so one
thing you just want to keep in mind is even
do a function has a cost for a corner or a
vertical tangents the limit still exist okay
the only time when the limit feels as we have
a jump discontinuity like right here an infinite
discontinuity like an asymptote or an oscillating
discontinuity Thanks so much for taking the
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