BAM!!! Mr. Tarrou. In this lesson we are going
to use these six derivative rules for finding
derivatives of Inverse Trig Functions. We
are going to use these rules that we derivatives
in the previous video to work out three examples.
Now yes, ok. We now know how to derive these
derivative rules on our own but you certainly
want to take the time to memorize these. You
do not want to derive one of these rules every
time you need one. I am sure that you are
not going to finish your next quiz or chapter
test, I know my kids won't, if you don't have
these put to memory. If you just first started
learning these make some flash cards or something.
Whatever it takes to get them memorized. Now
I have a little bit of alternative notation
here. We have the derivative with respect
to x of arcsin of u, so I already have this
set up for U substitution is equal to u prime
over the square root of 1 minus u squared.
That is the derivative for arcsine. Well let's
not forget about of course arcsine is the
alternative notation for the inverse sine
of u. So your textbook might have this written
slightly different than mine does. I am just
going to include these notes as my textbook
uses them for my students. But for you again
arcsine is the same as inverse sine. Over
here we have f(x) is equal to t times the
arc cosecant of 3x. We have a constant out
here, out front, which we are going to just
let float down now. We are probably very used
to do doing that. We have an inside function
of 3x. So we are going to let U equal 3x,
then the derivative of u with respect to x,
or u prime, is going to be equal to 3. So
the f prime of x, the derivative of f(x)...
the derivative of f with respect to x, is
equal to 5 from that constant floating down
out front times... well ok... the arc cosecant...
the derivative with respect to x of arc cosecant
of u is negative u prime, or the opposite
of u', and u' is 3... so we are going to change
that sign and have it be negative three...
over the absolute value of u... well u is
our inside function of 3x... now times the
square root of u squared minus one. I don't
know why I am writing u, it is equal to 3x...
now squared. All right. A couple of things
here. This three can be taken out of the absolute
value function because really I have this
absolute value function wrapped around the
3x to make sure that the factor remains positive.
Well 3 is positive, it is just making sure
that x remains positive. So I can take that
out. I am also going to put this 5 over one
and we are going to multiply across the top.
What is going to happen here is this three
that can be taken out of the absolute value,
I think I am going to show the work here....
Let's set it up instead of just saying it.
We have -3 times 5 over 3 times the absolute
value of x times the square root of... 3 squared
is 9 x squared minus 1. Well, now the 3's
are going to cancel out of course because
we can cancel out factors, not terms... not
something being added or subtracted to something
else. We get a final answer of -5 over the
absolute value of x times the square root
of 9x^2-1. You know we are just doing straight
forward examples. We are doing three examples
of finding the derivative. We are not going
to be trying to find equations of tangent
lines at a point or something like that. We
are just going through three examples of working
out these basic derivatives. Sometimes they
are not so basic, but that is all we have
for this first example. nanananana... Back
again. We have f(x) is equal to arc cosine
of 2x over x. So we have an inside function,
a u which is equal to 2x, and u prime which
is equal to 2. And we have a quotient so we
have to use the Quotient Rule for finding
a derivative. We have f(x) is equal to what
we have here. That means that f prime of x
is going to be equal to... Well the Quotient
Rule says take the denominator and multiply
by the derivative of the numerator. Again
the derivative of arc cosine of u is the opposite
of u prime over the square root of 1-u^2.
The opposite of u', so we have -2 over the
square root of 1 minus u squared, which is
our 2x... Ok yeah. I forgot where I was. The
denominator times the derivative of the numerator
minus the numerator... that is arccos(2x)...
times the derivative of the denominator which
going to be the derivative of x with respect
to x is 1... all over the denominator squared.
Ok. Now this is going to clean up not a lot,
but I do want to clean this up a little bit.
So I am going to write, I am going to bring
this x which we can write as x... Let's pick
a different color here... x/1. We are going
to bring this x up to the numerator and do
1 times the square root. I am going to give
each of these terms this denominator. I am
going to split this up, this one fraction
up into two separate terms and give each terms
this denominator of x squared because it can
clean up a little bit with our first term.
That means that we are going to have -2x over
the square root of 1 minus 2x squared is 4x^2
all over x^2 minus arccos(2x)/x^2. Now I am
going to just leave that like that. Not much
we can worry about doing there. Over here
we have a fraction divided by a fraction.
So what I am going to do is a couple of things.
I am running out of room. We are going to
write the numerator and the denominator as
a fraction so that we can see I am going to
take this numerator and divide by this denominator.
That means when you want to change division
to multiplication, like when you are dividing
fractions you multiply by the reciprocal of
the second value, or the reciprocal of what
you are dividing by. We are going to move
this out of the way and we are also going
to write this as -2x over the square root
of 1-4x^2 times 1/x^2. So again we changed
that division into multiplication of the reciprocal.
Our factors of x and x squared, well that
one x can cancel with one of those two in
the denominator. That just leaves us with
x times the square root of 1-4x^2. Don't be
tempted to take this x and cancel with one
of these x's down here in this denominator.
This is not arc cosine times 2 times x. That
2x of course is in that math function so nothing
else is going to cancel and I am done with
this example. BAM! One more example. So for
our last example we have f(x) is equal to
cosine of arcsin(x) minus the natural log
of x-1 over x+1. So a fairly complicated problem
here at the end, unless you take in small
pieces and it is not too bad. We have an inside
function, our u for cosine is going to be
arc sine of x. And we have the natural log
of a quotient. We want to split this up. We
want to expand that natural log function.
So I have added one more derivative rule up
here for us. The derivative with respect to
x of the natural log of u is u prime over
u. So we have f(x) is equal to the cosine
of arc... Actually for the purpose of space
I am going to just write the cosine of the
inverse sine of x minus the natural log of
x minus one... now when you take the natural
log.... um. When you take the natural log
of a quotient and you want to expand it, you
need to remember... Let's see here. We basically
get exponents from logarithms. When do you
subtract exponents? When you are dividing
like bases and our like bases are our base
of e for this natural log function. So again
we subtract exponents when we are dividing
with like bases and logarithms are basically
an exponent. When you log a value you get
an exponent. So it is this natural log of
x-1 which we need to think of as an exponent
because I am logging something to a particular
base and you get exponents from logarithms.
So it is going to be the ln(x-1) minus ln(x+1).
Now I know that when I take this one term,
this natural log of this quotient and expand
it I am going to get two terms. That means
that I need to take this negative sign and
wrap this expansion inside the parenthesis
so we remember that we need to distribute
this negative sign. If you don't do that,
you are going to do a lot more work in this
problem and you are going to have it wrong.
So we have f(x) is equal to cosine of the
inverse sine of x minus ln(x-1), and negative
times negative is positive, plus ln(x+1).
Ok for our problem here we have... We know
the derivative of cosine is negative sine
of x, but we have an inside function. We have
u is equal to the inverse sine of x or the
arc sine of x. That means that u prime or
the derivative of u with respect to x is going
to be x... so that is simply going to be 1...
because our u substitution basically now I
am looking at this and I have given you derivative
rules for inverse trig functions set up already
in terms of u substitution. So our u is our
x. So it is that inside function, you know,
the derivative of it and the derivative of
x with respect to x is 1. Now 1 over the square
root of 1 minus u which is for this simple
inverse sine of x, or arc sine that we have,
is just x. Ok. So take this nice and slow.
Don't make any mistakes. f'(x) is equal to...
the derivative of cosine is equal to negative
sine... OK... So the derivative of a trig
function when it has an inside function means
you have to use the Chain Rule. So it is the
derivative that trig function, and I am leaving
the inside u alone, times u prime. It is going
to be times u prime. So we have the derivative
of cosine of u is negative sine of u times
u prime... now minus the derivative of a natural
log of x-1 is going to be... well this inside
function is x-1 and the derivative of x-1
with respect to x is 1... So this is going
to be minus 1 over x-1... Plus.. and the derivative
of this inside function, right the u'/u, so
here it sort of U sub 3 if you will is x+1
and the derivative of x+1 is one... so 1/x+1.
I am also have a lot more simplifying to do
with this problem and I am running out of
space. Once again, let's get this green board
out of the way. Ok. So we have a lot of simplifying
here that we can do. Let's not forget that
we are taking the sine of the inverse sine.
These are inverse math operations and hopefully
you don't need the review. But what does this
x really represent if this is inverse sine
of x? Remember that we put angles into trig
functions and you get out basically the sides
of a right triangle, the reference triangle.
So if you are thinking about your angles in
standard position, it is the sine of theta
is equal to y over r. So this x is the ratio
of the opposite side over the hypotenuse going
into this inverse trig function and an angle
is going to come out. You take the sine of
that angle and you going to go right back
in. So it is sort of like the inverse sine
of 1/2 is going to be equal to pi/6, and the
sine of pi/6 takes you right back to 1/2.
So these inverse trig functions will cancel
out. You just end up with negative x. So the
sine of the inverse sine of x is just simply
x, then you have the negative coming in. You
have f'(x) equals -x over the square root
of 1-x^2. I hope now by the time you are in
Calculus... My Algebra 2 students, my PreCalculus
students are always trying to say these are
both perfect squares so that simplifies to
1-x and no it does not. Hopefully we are not
making that mistake still. So this here is
our first term simplified. Now I want to combine
these terms over here. This 1/(x-1) and this
1/(x+1), but to simplify the process a little
bit I am going to let this negative just float
up to the top and change this to an addition
of the negative number. Plus negative one
over x-1 plus one over x+1. Now I want to
bring these two together and we will be done
with this video so we need to find a common
denominator. That means I am going to multiply
this fraction by x+1 and this fraction by
x-1 in the numerator and denominator. Yes,
I have that over there. That is going to give
us negative x over the square root of 1-x^2...
Actually you don't need to see me talk through
a couple of algebraic fractions. Let's speed
this up a little bit. nananananana... I took
the negative, distributed it, combined like
terms over that common denominator, and also
distributed these. Basically we are looking
at a difference of squares pattern of x-1
times x+1, which comes out to be x^2-1. So
we have -x or the opposite of x over the square
root of 1-x^2 plus -2 over x^2-1. That is
the end of my last example. I am Mr. Tarrou.
BAM!!! Go Do Your Homework:D
