Derivative of Inverse Functions Calculus 1 AB READ DESCRIPTION

Derivative of Inverse Functions Calculus 1 AB READ DESCRIPTION

Bam! Mr. Tarrou In this Calculus lesson we
are going to be doing a lot of review which will be nice. We are actually…. A lot of
what we are learning in Calculus is review. Like we first started talking about derivatives
and really just went all the way back to algebra 1 and started with finding slope. So we are
going to be reviewing inverse functions with you. We are going to be using the Definition
of Inverse Function to validate that two functions are inverses of themselves. Now I keep saying
function so we will be talking about domain and range possibly needing to restrict the
domain of these functions so that they are inverses functions and not inverse relations.
We are going to be… I am going to give you a somewhat advanced questions and we are going
to work through finding the inverse of this given function. Then we are ultimately going
to get into with that calculus finding the derivative of inverse functions walking through
2 examples. The first one kind of showing you how the formula works. And it will be
overly detailed. And then one last example to finish up the video. So the inverse functions.
Well we know that, or hopefully remember that, if we were given a function and asked to find
it’s inverse as we are going to do in our first big example is we take the x and y values..
we switch places… and we resolve for y. So we have just a linear function y=2x+4.
And just pretending like I don’t know how to graph a line with just a y intercept and
a slope, I have got a table of values. There is a reason for that. I have picked the values
of -1, 0, 1, and 2 and found our y values. I plotted my function right here in white.
We have a y intercept of 4 and a slope of 2. And I have picked out one particular point
of (1,6). Then I have taken my x and y variables and swapped places and then I have resolved
this second equation for y. So subtracting both sides by 4 and dividing by 2, we get
y with an exponent of -1… which when we are talking about inverse functions it does
not mean write 1/y. It is an inverse function. It works in the reverse order. You know…
like trig functions. You put an angle in and you get the sides of a triangle. With inverse
trig function, as you put in the sides of the triangle and you get out this angle measure.
So the inverse of y, with the y being the original function of 2x+4, is 1/2x-2. And
again I have done a table of values. My x has a multiplication of 1/2 so I am for convenience
purposes with my table using even values. So 2, 4, and 6, and getting the y values of
-1, 0, and 1. Then we have this, if you can see the color of this green line right here,
is the inverse. Now a couple of things I want to point out here just in case you forgot
it from algebra, is if you look at my table of values I have (-1,2) where with the inverse
function I have (2,-1). I have the original point of (0,4) changing to a point of (4,0).
Then finally (1,6) is of course being (6,1). So not only are we seeing that we take the
x and y variables and we switch places, but it happens with the actually coordinates in
the t-tables. So what was the domain of my original function is becoming the range of
my inverse function. And then graphically how that looks is we have this line of y=x.
It makes sense because we are taking those x and y values and switching places, that
there is a reflection on this line where y and x are equal. And again, that is allowing
the domain of the original function become the range of the inverse function. You get
this reflection over that line of y=x. That is the graphical way of identifying inverse
functions. So in summary inverses are reflections over the line y=x. We notice that the domain
and the range values switch places. And we want to put this up. This is what we are going
to focus on in the video. We can see here in this simple linear functions, the derivative
which of course is… at least the first derivative… is the slope of the original function… So
the derivative of function f at (c,d) is the reciprocal of the derivative of the inverse
function at (d,c), switching again those domain and range values. Now these are linear functions
so this slope of these functions is going to remain constant. So I am pointing out this
(1,6) and this (6,1) because I know where the end of the video is going. Indeed as you
look at these two linear functions we can see that the slope of the original function
is 2 and the slope of the inverse function is 1/2. Those are reciprocals. And for these
two linear functions the slopes are the reciprocals everywhere in the domain. But of course with
Calculus and the first derivative that has allowed us to find the slope of curves at
a given point. So I am just saying that if you are at a point in one function and you
take the domain and range and swap them to find the point on the inverse function, that
the slopes are going to be reciprocals of themselves. Again, that becomes a big deal
in our last two examples. The Definition of Inverse Functions: A function g is the inverse
of function f if f of g of x, so you do that composition of functions, is equal to x when
you are done simplifying and g(f(x)) is equal to x as well. So you set up the composition,
simplify, and if both times f(g(x)) and g(f(x)) come out to be equal to x for each x in the
domain of g and for each x in the domain of f then f & g are inverses. Just for clarity
purposes if you needed it, I went ahead and rewrote that expression as f of the inverse
of f of x. That inverse function is getting put into the original function and it simplifies
to x. Then the original function getting put into the inverse function also, also not or,
has to simplify to be x. So our first example is going to be just using the Definition of
Inverse Functions to test whether two given functions are inverses. We are going to do
that right now. Show that functions f and g are inverse functions both analytically
and graphically. Now I just erased part of my example there which I will put back up
in a second. The reason I wanted to do that is I want to start the problem graphically
and again emphasize this restricting, some times we need to restrict the domains of these
functions to that they are inverse functions and not just relations. So we are going to
focus on function g first for just a moment, even though I have kind of listed that as
the second function. But function g has got one squared terms and it is on the x. So it
is a parabola opening, with that positive coefficient, up. So this vertical parabola
has a vertical stretch of 4 and it has a shift down of 1. I know my vertex is going to be
at (0,-1). Then with the vertical stretch either I can talk about transformations which
I just hinted to, or let’s just plug in a value of 1. One squared is 1, 4 times 1 is
4, and then 4 minus 1 is equal to 3. So we have a point of (1,3). Then this being a basic
polynomial with the x having an even power is going to pass the test for being an even
function. That means that is symmetric to the y-axis meaning if I plug in opposite values
x and -x I get the same y values. Or I could just plug in -1, -1 squared is 1, 1 times
4 is 4, and 4-1 is again 3. So yes indeed there is some vertical symmetry, or at least
symmetry to the y axis. The parabola is going to look something like this. And a parabola
is a function. But I want to verify that these are inverse functions, that they are inverse
functions to themselves. So if I take just my graph right now and I reflect it over the
line y=x, instead of going along the y axis it is run along the x axis. It would fail
the vertical line test. So while my g(x) is a function, f(x) if it were its inverse it
would be an inverse relation. We want to restrict these to be inverse functions. So that part
that I erased on my problem said that for my parabola that the x has to be greater than
or equal to zero. That is not because of any issues with graphing the parabola, but to
force these to be inverse functions not just inverse relations. So there we go, there is
g(x). Understanding that the domain of one function becomes the range of the other, I
could just write these two points down or even maybe a third and just swap the x and
y coordinates and sketch my inverse function or f(x) immediately. But to talk about transformations
for a second. f(x) is equal to 1/2, I can pull that 1/2 out front, times the square
root of x+1. So now this function has a vertical compression of a magnitude of 1/2. This plus
one, now this +1 is inside the math function not at the tail end… a constant basically
all by itself, so that is going to be a shift to the left with a compression of 1/2. So
my graph of f(x) is going to look something like this. And we can see… I am going to
need to talk over the air conditioner for a second before I can turn that off…. that
f(x) and g(x) as good as I can draw this on a chalk board are showing us a reflection
over this line again of y=x. So that is our graphical verification that these are inverse
functions. Now graphs are just that, they are approximations. Maybe with the tables
and our calculators we can verify it more exactly. But let’s do this analytically which
is a bit more reliable. So I am going to do the first half of this check and you are going
to do the second half. Or you can pause the video and do both on your own. We have f(g(x)).
We are going to validate that this is equal to just simply x. So we are going to take
function g and put into function f. So we have the square root of something plus 1 over
2. That something that we are plugging in for x is g. So it is going to be 4x^2-1. Ok,
so now we have the composition set up. There is nothing in front of these parenthesis.
There are no exponents, or denominator… There is a denominator of 2, but I am talking
about with the grouping symbol of the square root symbol. So this negative one and positive
one are going to just simply cancel out. We have the square root, and a broken piece of
chalk… the square root of 4x^2. The 4 is a perfect square, the x^2 is a perfect square,
and this is just simply one term. So we can square root the term. The square root of 4
is 2. The square root of x^2 is equal to x, and then over 2. That simplifies to x. That
is the first half of the check. We need to go through as well and validate
does g(f(x)) equal x. Now you are going to
set that up on your own. One thing by the way. When you take an even power of a value
that is also raised to that same even power, you are supposed to need to wrap that in absolute
value symbols. So this could possible look like 2 divided by 2 is 1 and just have a final
answer of the absolute value of x. But we have a restriction where x must be positive.
So there is no need to write those absolute value symbols. We are given that it is positive
as we do the problem. I am going to reveal the answer, but run through it first and make
sure that you can do it yourself. nanananana… Alright. Checks out both ways. g(f(x)) and
f(g(x)), they both come out to be x so these are inverse functions. Let’s go on to talking
about the existence of inverse functions. Put into words what I have said verbally,
show you the steps of finding an inverse functions and do one example of here is a function,
now find its inverse. We will do it a step further than maybe we did in Algebra 2 or
Precalculus. It will involve a little bit of practice with derivatives to check for
that 1 to 1 nature that we need to have an inverse function. The Existence of an Inverse
Function. Well a function has to have…. For a function to have an inverse function
it must be 1 to 1. Now it has to first be a function, that is why I keep saying function:)
But a 1 to 1 function, hopefully you remember, and if not let’s review it. That means this
function here that is increasing as we go to the right would pass the vertical line
test. No matter where I put a vertical line it would only cross once indicating that it
is a function. Then the 1 to 1 part means that it would also pass the horizontal line
test. Where… Everywhere I could draw a horizontal line it would only cross once. So that is
1 to 1 at least as a graphical check which is what you would have done in Algebra or
Geometry. Now we are into Calculus right? So we are going to be doing this. We need
to know how to use our graphing calculators as well. So a visual inspection for a 1 to
1 check could certainly be sufficient. But we can check this with derivatives as well.
If f is strictly monotonic, that means basically increasing as you go to the right or always
decreasing. It does not change direction. If f is strictly monotonic on the entire domain…
Let’s stop there for a second… The previous problem that I just did about checking whether
these to functions are inverses both graphically and analytically. I first graphed that parabola
right. It was decreasing from negative infinity to zero where we found the vertex of the parabola,
then it started to increase. That was not a monotonic function. It was decreasing and
then there was a place where the graph changed directions and started to increase or have
a positive slope. So we had to restrict that domain from the vertex to just part of the
function in that particular case where it was increasing. Not that it always needs to
be just increasing, but it does need to be monotonic. So we restricted the domain of
that parabola such that it was monotonic. …then it is 1 to 1 and therefore it has
an inverse function. Now as far as finding out, or determining, if a function is monotonic,
we can type that function into our calculator and do a visual inspection. But to review
and strengthen our Calculus skills for the example I do with you, and you are going to
help me with by practicing the problem yourself, we are going to check that monotonic situation
by looking at the first derivative. Basically like doing a first derivative test. Steps
for finding an inverse function. Well, we need to check if it is 1 to 1 visually or
with your first derivative. We are going to swap the x and y variables again as you saw
me do with this linear function. We are going to resolve after swapping the x and y variables,
resolve for y and then label that as the inverse function… whether it is y^-1 or f^-1(x).
We read that as the inverse of f(x) or the inverse of y. Determine the domain of the
inverse function. You know determining the domain of a function is much easier quite
often than determining the range. So it says here we are going to determine the domain
of the inverse function which is the range of the original function. We can take that
original function f(x) identify its domain, then find the inverse and find the domain
of the inverse function which is the range of the original function. Now determining
domain, your trig functions are a little bit more complicated with maybe where the vertical
asymptotes. Say for tangent function, they may be a little bit hidden and not obvious
from the equation unless you are really know your unit circle and your trig functions,
of course we should. But if you are looking at a pure algebraic equation, we can be very
cautious of restrictions in the domain if the dependent variable… the domain… or
the x variable is inside an even root or in the denominator of a rational function. It
does not guarantee that there is a restriction… like if the variable is in the denominator…
but it probably will have one. Then of course if you want to take it that one extra step
and make sure that you answers are inverses of each other you can check your definition
of inverse functions and make sure that both of these expressions comes out to be X. We
are not going to do this last step in our example of finding an inverse function. But
we are going to do the other four steps. Maybe again a bit more in-depth then maybe you would.
It would be easier to have the aid of a calculator, but we are going to practice our skills of
derivatives as we do this. So let’s get to that example of finding an inverse function.
Find the inverse function of f and state the domain and range of the original function
f and the inverse of f. Our original function is f(x) is equal to 2x divided by the square
root of x^2+5. Now with use of our graphing calculators which is something definitely
need to practice we can type this into our calculator, analyze the graph, and by analyzing
the graph determine the domain and there is some horizontal asymptotes going on here.
So I would have to recognize that asymptotic behavior, but we should also be able to identify
the range. We should also be able to just look at the graph and determine if it is monotonic
increasing or decreasing. In other words is it 1 to 1 and thus has an inverse function.
But you know that is fairly straight forward. Hopefully if you know how to use your graphing
calculator. But I want allow you the opportunity to remind yourself, restrengthen your derivative
skills in finding horizontal asymptotes of irrational functions. So I am going to step
out and reveal these solutions one step at a time. Maybe pause the video, find that first
derivative, determine if it is monotonic increasing or decreasing based on the first derivative,
and then if you remember how to find those horizontal asymptotes find the range of this
function as well. We can again find the domain of the inverse function which is a little
bit easier which will be the range of this original function. But I just want to review
some skills with you as we go through the video. Just quickly run through this. We have
a domain… now our denominator… we have variables that are in the denominator. We
cannot divide by zero, nor can we take an even root of a negative number. So with a
combination of those two we set x^2+5 is greater than 0. We subtract both sides by 5. We have
x^2 is greater than -5. Well certainly that we raise to an even power is going to be positive
so this will automatically be greater than negative five. So the solution is all real
numbers. For testing whether the function is monotonic thus being 1 to 1 we can find
the first derivative here. With the quotient rule we have the denominator times the derivative
of the numerator minus the numerator times the derivative of the denominator, which required
the chain rule over the denominator squared… the 2 cancels out… we have a factor of x^2+5
to the 1/2 power and the -1/2 power. That same factor is both terms so we take out that
lowest power of -1/2. Both of these terms to the left and the right side of the minus
sign have a factor of 2, so that 2 can be factored out as well. A little simplification
and we have f'(x) is equal to 10 over x^2+5 raised to the 3/2 power. So what is going
on here is again our x has an even power, so whatever we put in for x becomes positive
with the even power, then we add by 5 keeping it positive, and we have a positive numerator
so this expression… take a positive and raise it to an odd power comes out to be positive…
the even root with no negative out front becomes out to be positive.. So the first derivative
is going to be greater than zero or positive everywhere within the domain which is all
real numbers. Thus this is monotonic increasing. So thus it us one to one and we will have
an inverse function. Now coming over here I am going to point something out at the end
of this that maybe would saved us a little bit of time if we had thought of it at first.
But horizontal asymptotes, this is an irrational function. There is a fraction bar so it is
a fraction, but then it has a square root in it so this is an irrational function. So
we need to go through this limit process. To find the limit as x approaches positive
infinity we multiply the numerator by 1/x and to get that into the square root we take
that x and raise it to a second degree and square root it. So we have 1 over the square
root of x^2 which simplifies to be x when x>0. So this is equivalent to multiplying
by 1. When we do that we get ultimately a limit of 2 so as x approaches infinity the
graph is going to approach the value of 2. And to approach negative infinity we need
to have some kind of sign change to indicate that in our limit process. We can either do
-1/x, or 1 over negative square root of x squared as I indicated. Which is nice because
then you just put in a negative in front of your radical and the limit as x approaches
negative infinity is going to be negative two. So we have… I didn’t mark off, maybe
I should have, the horizontal asymptotes. But if we were to have graphed the original
function and tried to find domain and range, and whether the function was 1 to 1 with a
graph or a graphing calculator we would have seen this kind of graph. Just to again review
a lot of concepts that we have learned in the past, if I were to take this function
and rotate it 180 degrees it would appear to have symmetry about the origin. Well that
means that it is an odd function. An odd function are where you plug in opposite values of x
and negative x and you get out opposite answers. Well if I plug in opposite numbers into the
denominator, the negative or the opposite will be cancelled out by the even power thus
the denominator is going to be the same… it is always going to be the same whether
you plug in a positive or negative x. But the numerator is just simply 2x, so yes when
you plug in opposite values into this function… when you plug in -x because the sign of the
numerator changes and the denominator does not, you are going to get opposite answers.
Thus this would pass the test for the being an odd function. Now we can see that whether
you graph this with a table, or more than likely with your graphing calculator. So we
have not found the inverse function yet, but we have done a lot of work of finding domain,
finding whether it is 1 to 1, the extra work of finding the range. This is me just again
reviewing concepts. We could have found the inverse first and found the domain of the
inverse which would still give us the same values between -2 and 2. Let’s go ahead and
find that inverse function now. Got everything cleaned up. Summarizing what we just found,
the domain, our range, and our sketch. We are going to take these x and y variables
and swap places. So instead of y we are going to have x is equal to 2y over the square root
of y squared plus five. Again, we want to now solve this for y. We are going to square
both sides to get rid of this even root and manipulate the equation to the point where
we get y in terms of x. That will be the inverse function. Again I am going to step out and
reveal the solution one step at a time. Got those x and y’s swapped of course. That is
how we started the problem. I squared both sides of the equation to get rid of the square
root. We multiplied both sides by y^2+5, distributed the x^2 into the parenthesis. I moved my four
y squared over to the left with subtraction, and the same moving 5x^2 to the right. When
I got down to this point I have y squared is equal to -5x^2 over x^2-4. Well if I go
to get rid of this power of 2 by taking the square root both sides with that x squared
being positive… with that x^2 term whatever I plug in for x is going to be positive because
of the even power… then I get multiplied by negative five. Now I cannot square root
a negative number. So in this format the numerator is guaranteed to be negative and thus I cannot
take the square root of it. So I multiply the numerator and denominator by -1 to get
rid of that negative that is in the numerator and also of course that is going to take the
binomial in the denominator and swap the order. x^2 times negative one is negative x squared,
and -4 times -1 is positive 4. Now this expression has a pretty tight restricted domain for where..
you know… it can be positive but it is not guaranteed to be negative like we had in the
numerator. So I go ahead and then take the square root of both sides. I identify it as
the inverse of y. I get plus or minus the square root of 5 x over the square root of
4-x^2. Now again you cannot take an even root of a negative number and you cannot divide
by zero, so four minus x squared needs to be greater than zero. We get solved for x
squared, we square root both sides, and with inequalities you need to change the direction
of the sign for the negative answer. So we have that negative two is less than x which
is less than 2. Right!? The domain, I just solved for the domain of the inverse function
and it is checking out to be the same as the range of the original function. It is all
checking out to be good. We need to deal with this plus or minus that we have. I didn’t
pick a real simple question for finding the inverse. We have a plus or minus here. Well,
it is not both ok. We are looking for the function which the reflection over the line
y=x which is something like this. This is just a quick sketch. I don’t know if this
really crosses or not. But we are looking for the reflection of this original function
f(x) over the line y=x. If you use your graphing calculator to graph both the positive and
the negative version of your answer, one of those answers… which is the negative one…
will come down and cross through origin this way. It will still have vertical asymptotes
of x=-2 and x=2, but it will be coming through like this which is not your reflection over
the line y=x. We are looking for a graph that looks like this. And the graph that is that
reflection over the line y=x is the positive version of our answer. So our inverse function
is the square root of 5 times x over the square root of 4-x^2. And you can check that if you
like with the Definition of Inverse Functions. I have graphical check right here so we are
going to leave it at that. We are going to move on to the derivative of in inverse function,
write out that theorem, and some formulas so we can get to our last two examples. … and
find my remote. Let’s talk about a few properties of an inverse function. If f is a function
whose domain is an interval I. If f has an inverse function, then the following statements
are true. If f is continuous on its domain then the inverse of f is continuous on its
domain which is the range of the original function. If f is increasing which means that
its first derivative is positive on its domain, then the inverse of f is increasing on its
domain… so its first derivative is going to be positive. If f is decreasing on its
domain, then the inverse function f^-1 is decreasing on its domain. Finally if f is
differentiable on an interval containing c and if f prime of c is not equal to zero,
so basically the slope of that original function is not equal to zero… because when you reflect
that over the line y=x, that horizontal tangent line is going to become vertical which will
be an issue with trying to find the derivative, so that derivative of f of c cannot equal
zero… then the inverse function or the inverse of f is differentiable at f(c). Now again
that domain and range is swapping right as you find inverses. So f is differentiable
at c, and the inverse is differentiable at f(c). f(c) is the y value of your original
function. But when you take the inverse and you start talking about inverse functions,
the y of the original function is the x of the inverse function. That is why you see
that f(c) there. Ok, let’s look at the rule for the derivative of an inverse function.
This rule that allows us to do this more quickly than actually having to find the inverse…
then having to find the first derivative to find this slope. BAM!!! This is what the whole
video is building up to, The Derivative of an Inverse Function. Let f be function that
is differentiable on an interval I. If f has an inverse function which we are going to
identify as g, then g is differentiable at any x for which the derivative of f, or f
prime, of g(x) is not equal to zero. This is how the formula is initially given in the
book that I teach out of in class. It says the derivative of g of x is equal to one divided
by the derivative of f, or f prime, of g of x. Ok. I find that initially can be confusing.
So I am going to take this and just write it in terms of the inverse function notation.
The derivative of the inverse of f of x is equal to one over the derivative of f of the
inverse of f of x. Even that is kind of confusing. So let’s see here. Derivative is slope. So
the slope of the inverse function at its x is equal to 1 divided by the slope… the
derivative… the slope of the original function and we are plugging in the inverse of f of
x. It is easy for me to say. Basically the y that we get from the inverse function. Well
why are we asking for y? Well, the x of the inverse function is the y of the original
function. Remember we talked about that domain and range swapping places when we find inverses.
So the y of the inverse function is the x of the original function. Let’s go back to
that original diagram that I started with this video, those two linear functions…
which slopes do not change but just for simplicity sakes. We have y=2x+4. The slope of the original
function was 2. We swapped the x and y, resolved for y, and we got the inverse of y was equal
to (1/2)x-2. Well yeah, these slopes are reciprocals of each other. These slopes are reciprocals
of each other. And this formula says that the slope of your inverse function at its
x is equal to the reciprocal… one over the slope of the original function… but you
are plugging in the y again of that language of the inverse… or the y of the inverse
which is 2 was the x of the original function. So the slope of the inverse function at its
x is equal to the reciprocal of the original at that inverse’s y value. That y value was
the x of the original function. So right here again 1 over f’ or the slope of the original
function where you are plugging in the y of the inverse function which was the x of the
original function. I am going to really hit on this really hard. Kind of actually do the
first example longer than I need to just to show you that swapping of the domain and range
and how those slopes are reciprocals of each other. Then we will do the last example very
streamlined allowing this formula to basically really short cut the process to help us find
derivatives at particular x values of the inverse function without finding the inverse.
Verify that f has an inverse. We are going to assume that we are talking about inverses
like we have all video. Then use the function f and the given real number ‘a’ to find the
derivate of the inverse of f of a. Ok. So normally I just tell my students if you really
get stuck on how to do the problem, just let the formula tell you what you need. Ok. First
thing it says is verify that f has an inverse function. It needs to be monotonic. We are
going to find the first derivative. f prime of x is equal to… that is a constant so
it is zero… and -3 times 3 is -9x^2. So no matter what x is, it is going to be squared
and therefore this factor will be positive… not times negative nine will be negative.
So the first derivative is always going to be less than zero so this is monotonic decreasing.
a is equal to 5. So the derivative of the inverse of f of a is equal to 1 divided by…
Well I need the derivative. Actually I just figured that out, right. It is -9 times something
squared. Then if you did not really catch what I was trying to point out to you in the
previous screen, then you might be going I need the inverse of f(a). Maybe we will burn
some calories and do that. And if I want the inverse function I am going to take the x
and y and swap them. Let’s do this in orange. So we have x… this is the extra part that
you don’t really have to do. I am just trying to tie everything together and we will streamline
the last example. This y is going to become x. We are going to put a y here. We are going
to resolve this for y. We are going to subtract both sides by 2, divide by negative three.
I want to turn this around too. Then we are going to take the cube root of both sides.
Now that I have finished resolving for y, this is my inverse function. I am not going
to check it with the composition of f(g(x)) and g(f(x)). But this is my inverse function.
Now that I have the inverse of y, I can come up here and go… I was looking for the inverse
and that is the cube root
of x, which I am plugging in ‘a’ so it is going to be 5 minus 2 over negative 3 all
squared. This is going to come out to 1 over negative nine times… 5-2=3 and 3 divided
by -3 is -1 squared. And the cube root of -1 is -1. -1 squared is positive 1. Positive
one times nine gives us a final answer of -1/9. The derivative or the slope of my inverse
function at ‘a’ is equal to negative one-ninth. Now what I want to point out here and to show
what I was trying to point out in the previous screen, I took the time to find the inverse
function and then find that value of that inverse function at ‘a’ which came out to
be -1. Then we plugged that into the first derivative. Ok. I did not actually have to
find the inverse function. There will probably be problems in your homework where you really
will have a very difficult time finding that inverse function. So we can stream line this
if we remember that the ‘a’ again was effectively… see it going into the inverse function and
thus giving us a y value. Well that ‘a’ is the domain of the inverse and it is the range
of the original function. So I can say I am recognizing now that the 5 is actually the
y value for the original function. So if I take this original function, let me get this
out of the way, and say 5 is equal to 2-3x^3. Subtract both sides by 2. Divide both sides
by -3. We get -1 is equal to x^3. Then cube root both sides and get negative one is equal
to x. The x that is there from my original function becomes the y of my inverse function.
Look at what I have inside for this value of the inverse of f of a. We have negative
one. So I don’t need to find the inverse function to find the slope of the inverse function
at ‘a’. I just have to recognize this swap of the domain and range is going on, and say
ok… the x for the inverse function is actually the y of the original function. I can find
that inverse of f of ‘a’ a lot easier. Then I can just plug in the negative one right
here. I think I pointed to that line, I should have been pointing to this one. Plug in -1
here and get my answer a lot quicker. So let’s stream line that as our final answer and last
example of the video. As our last example f(x) is x+4 over x-3 with a restricted domain
where x is greater than 3. We have an ‘a’ value of 8. So we are going to find the slope
of the inverse function at the value of 8. To stream line this again we are going to
find the derivative of f(x) to verify that it is 1 to 1, thus the inverse function exists.
Plus we need it as a denominator down here. We are going to set f(x) equal to ‘a’, solve
it for x which is the y value of that inverse function… plug it in, get our answer and
we will be done much quicker than the last example. So f prime of x is equal to the denominator
times the derivative of the numerator minus the numerator times the derivative of the
denominator over the denominator squared. So when we get done cleaning all of this up…
this is getting multiplied by 1 so that is not going to do anything… distribute that
negative one. We have f prime of x is equal to x minus x which is zero… negative three
minus four which is negative seven… over x-3 squared. Now the denominator is always
going to be positive with that even power. The numerator is negative, so our first derivative
is going to be negative always so this is monotonic decreasing thus it is 1 to 1. Ok.
We need to find this y value here, the inverse of f of ‘a’. So we are going to let this function
equal 8. 8 is equal to x plus four over x minus three. Multiply both sides by x-3 and
we get 8x-24 after we distribute which is equal to x+4. Move this x over to the left
and we get 7x, add both sides by 24 and we get 28. Divide both sides by 7, and we get
x is equal to 4. So the slope of the inverse function at 8 is equal to 1 over… ok…
f prime, the slope of the original function. Well that is -7 over parenthesis something
minus three squared. And again the y of our inverse function, the f^-1(a), is the x of
our original function here which we found to be 4. It is going to be four minus three
squared. Well 4-3 is 1. 1 squared is 1. Negative seven over one is just negative seven. Our
final answer is -1/7. I started finalizing the numbers and I had a little interruption.
So what we have here is the slope of the original function at 4 was -7, so swapping the domain
and range the slope of our inverse function at 8 is the reciprocal which is -1/7. So I
am Mr. Tarrou. BAM! Go Do Your Homework… and I will find my remote.

49 Replies to “Derivative of Inverse Functions Calculus 1 AB READ DESCRIPTION”

  1. You are very welcome:) I appreciate the nice feedback and your support. I wanted to review a lot of "old" concepts in this video, but when I was finished I was afraid it was too long. Thanks for watching!

  2. good review… I'm unclear about something you said… starting at 13:26

    " When you take an even power of a value that is also raised to that same even power you are supposed to wrap that in an absolute value symbol."

    In this example, what is the value that you're taking an even power of? Are you saying the square root is an even power? (sqrt(4x^2))?

  3. At the beginning of the sentence I should have said even root. Good catch:) I added an annotation in the video. At least I was pointing to the radical when I said the wrong word!

  4. Thank you! From an adult student 40+

    I have been searching for a simple and verbose presentation of inverse functions. The problem I was having was finding the (f)inverse(x) without finding the inverse function.
    When you said the the (f)inverse(x) was really the y of f(x) everything clicked.

    If I could buy you a beer I would!

  5. That's awesome, it's never too late to learn huh?! I'm glad my videos could make that "light" go on for you:)…and thanks for taking the time to seek additional help and choose my channel. I can't drink on the job but there is a "Tip the Teacher" button on my homepage after you graduate and get that high paying job:)…in the mean time, thanks for showing your support by liking and subscribing!

  6. once again you have imparted your skill and wisdom to help me understand, especially the last two examples is when the clouds parted and the light shown brightly.  keep on keeping on, love your explanations and keeps me motivated. BAM!! doing my homework!! 

  7. Awesome video, it helped a ton! I'm having trouble doing this with trinomials though. When I set the given value equal to the original function, I can't figure out how to get 1 x value. Would there be 2 x values?

  8. infinitely better than any help I've gotten at a learning center or from the grad student they assigned to teach my class! phenomenal work Mr. Tarrou! subscribed and recommended for sure.

  9. I've been subbed for a while but just started recommending your videos to my classmates. They are now glued to your videos. Seriously, thank you for making calculus super fun. 🙂

  10. Taking a 8 week summer course is a challenge, and i just wanted to say thank you for your dedication to your craft and educating people. Your videos are very helpful and it is much appreciated that you take the time out of your day to record these to help those that are struggling. fantastic job.

  11. I'm taking Calc 1 through an online university since it is my only option for now, and it is very difficult trying to understand a concept just from the text. For all intents and purposes, you are my Calc professor. Thanks!

  12. Amazing!  I have been struggling with this concept in Calc AB, but you helped to explain it very nicely.  It is still a bit shaky, but I at least understand it now.  Great job!  Thank you very much.

  13. 12:36
    "We have a square root and a broken piece of chalk"
    ProfRobBob never fails to make me laugh with his smoothness

  14. Mr. Tarrou, I swear you're a magical genie that makes complicated things easy. I won't say I completely understand the methodology you use, the whole replacing back and forth voodoo, but just memorizing the process and applying it makes things a whole lot easier. I'll work on understanding it later. THANK YOU for another wonderful video!! lots of love from this suffering calculus student!!

  15. So far, this was the only differentiation-related topic that I hadn't gotten; thanks again, Mr Tarrou! You're a life-saver! 😀

  16. GREAT! Thank you SO MUCH! I reached an epiphany while watching this video. I understand this topic so much better! Thanks!

  17. at 24:00 you multiplied numerator and denominator by 1/x and 1/sqrt(x^2). but 1/x is not equal to 1/sqrt(x^2) as sqrt(x^2) = |x| {Absolute Value of x}

  18. Just when I thought I was going to be mentally killed by this torture tactic called calculus I was saved by a Math Jesus. Thank you so much man.

  19. srsly… your videos sir helped ma alot.. i am reviewing for my finals tomorrow..
    and derivatives of trig func. are what i hate the most..
    the questions look very impossible to solve but they really are manageable as long as you keep on practicing.. i felt like my brain was about to pop.. but no pain no gain

  20. actually I like your videos more than patric videos , but the reason of why I prefer to watch patric videos is that they are very short and easy to watch and look for what I don't understand in mathematicise , so if you split your videos into specific topics of videos that would be better for me and any student . if you agree with me press like . thank you

  21. hi sir i am struggling to undersand what to do in order to prove that a function is Not one-to-one. what are the steps to prove that? thanks

  22. Just finished watching your video. I studied Calc I with your vids, and now studying Calc II :)) Thank you!

  23. Prof Bob why are those 4 statements 32:20 true? And why is the definition of the derivative of an inverse function true? Should I not be concerning myself with these things? I've always understood the reasonings behind everything that's been said in math up till now but lately it's just been making my head hurt. 🙁 so many words.

  24. I think there is an error at about 22 minutes. You have to distribute the multiplier of 2 to both the X^2 and the +5, leaving you with 2X^2 + 10 – X^2. So the final numerator is X^2 + 10, not 10.

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