Calculus – Find the derivative of inverse trigonometric functions

Calculus – Find the derivative of inverse trigonometric functions


Welcome to MySecretMathTutor
For this video we are going to work on finding the derivative of inverse trigonometric functions. Now these guys are kinda crazy looking, probably
have seen them before if you’ve taken Trigonometry, but I’m taking about things like inverse tangent,
inverse sine or arcsine. Those are the things we want to take a derivative
of. Now I’m going to jump into an example of these
just to give you a sense that these don’t quite look like other derivatives that you
are use to. I’m gong to do a problem like this, y=3x^2
+ cot^-1(x) just to see what we are up against. Alright, so off I go to take a derivative. These derivatives really play well the other
derivative rules so I’m going to take the derivative of 3x^2 and then worry about the
derivative of inverse cotangent. So first the derivative of 3x^2, that just
gives us 6x. No problem there. Now comes the really interesting one, what
is the derivative of inverse cotangent. Well its negative one, all divided by 1 +
x squared. Isn’t that crazy? We could clean this up a little bit, call
it 6x minus 1 over 1 plus x squared. But go ahead and look at that derivative. Notice how it doesn’t have any more trigonometric
functions in it. All of a sudden it just turns into this fraction,
and now x squared is on the bottom. That is not what your intuition says that
this probably turns into for derivatives. So now is a good time to take a look at some
of these rules, to see well what is that we should do when we run across these inverse
trigonometric functions. So there are six functions, and we have six
rules. Don’t worry, even though they look kind of
complicated, there are only three rules you really have to remember, and those are the
ones for inverse sine, inverse tangent, and inverse secant. As soon as you have these rules down, you
have them memorized. You really have the other ones memorized as
well. They are exactly the same, they are usually
just different by a negative sign. And the good way you can connect these two
is sine and its cofunction, cosine those are the ones that are the same, but cosine has
the negative part. Here is inverse tangent and its cofunction,
inverse cotangent. It has the negative sign for inverse cotangent. So as you can see with these rules, none of
them actually contain any trigonometric functions once you are done taking that derivative. Whether that is inverse sine or inverse tangent,
its derivative is usually a fraction, there is usually an x squared, or a square root
floating around in there. In a different video I’ll explain why these
are the correct rules by going over some proofs, but for now might just want to start memorizing
these and getting really familiar with them. Even though they do look a little bit more
complicated than your average derivative rule. You really do want to get to know these as
much as possible so that you can recognize them, actually for later on when you do integrals. Alright, so let’s just quickly cover these
two. If I take the derivative of inverse sine this
is one divided by the square root of one minus x squared. If I take the derivative of inverse tangent,
then I have one divided by one plus x squared. And if I have the derivative of inverse secant,
then I have one divided by x multiplied by the square root of x squared minus one. Now before we jump into two examples I want
to really make a note that these definitions actually may be different than your particular
math book. It all depends on how the math book has set
these functions up, how they’ve defined them. And you may end up with something slightly
different, so yours may actually end up with some restrictions in there, or may it has
absolute values. Double check your math book before you start
on your homework, and you start using these particular rules. Alright, let’s go ahead and give these guys
a try. See what we have for my examples, and of course,
do some derivatives. Now like I said before these things really
play nice with the rest of your derivative rules, so you still want to know your chain
rule, your product rule, quotient rule, all those things are important, as well as knowing
these new derivative rules. Alright let’s go ahead and start off with
this guy over here. So we want to take the derivative of sin^-1
and on the inside it looks like we have a 4x^2. Since these are a little more complicated,
what I like to do when finding their derivatives, is you know, kinda start off with a template
of the what the derivative of inverse sine would look like So the derivative for the outside is 1 divided
by, then we have that square root. If you look back at my rule sheet it says
1 – x^2. I’m not going to write that x in there, because
that x is really just a stand in for what ever the inside piece is, and in this case
its a 4x^2. So using that chain rule, the inside would
stay exactly the same. 4x^2 Multiplied by the derivative of the inside
I have an 8x. Now too bad, we have a little bit of cleaning
up we can do. Let’s go ahead and say that the derivative
of y, we’ll put that 8x on top, all divided by the square root of 1 minus, square the
four, we get sixteen. Square the x^2, we get x^4. Nice. And there is our derivative. Notice again, no trigonometric functions,
looks a little complicated, but sure enough that is our derivative. Alright let’s do a very similar one, definitely
want to keep in mind the derivative of inverse sine while we are doing this. We’ll also going to use a chain rule on this
one, but notice how this time the inside piece is the inverse sine. So let’s take care of that outside piece. We’ll bring down the power, reduce it by one. Now the inside will stay exactly the same,
so its just inverse sine of x. And now we’ll multiply this by the derivative
of inverse sine. Here is where we get to use that fraction. So let’s see the derivative of inverse sine,
we have, lets see, 1 divided by the square root of one minus x squared. Not bad, not bad at all. So this one of course we have some cleaning
up as well. Might as well combine everything into a single
fraction, so we put the 4 and the sine inverse of x on top. And we’ll just leave all that other square
root stuff in the basement. Alright, so let’s do a couple more examples,
just so we can get a little bit better at doing this. Let’s go ahead and throw in a few more of
those inverse trigonometric rules, so we can see what those look like. Alright. This next one we’ll go ahead and take care
of the derivative of e^x multiplied by tan^-1(x). Now in order to take care of this guy I really
want to point out that we are going to have to use a product rule. So here we have one function, call this f. And really this guy over here is a second
function, I’ll call that g. Using our product rule then we’ll say that
it is our first function, just as it is, no changes multiplied by the derivative of the
second function, so I need the derivative of tan^-1(x). Ok, we’ve seen this on our rule sheet. It’s kind of a nice one actually, because
this the one that doesn’t involve any square roots, its just 1 divided by 1 + x^2. Alright, continue on with our product rule. Now I’ll take my second function, unchanged,
don’t mess with it, don’t do anything, multiplied by the derivative of the first function. So e^x. So I just want to point out what I’ve done
here. So I have my first function, that’s just the
e^x. We have the derivative of the second, there
is all of that stuff. Now we have our second function, just as it
is and the derivative of the first. The derivative of a natural exponential of
course is just that same natural exponential, e^x.
Ok, so I have everything, looks good. Now we’ll just go ahead and clean it up a
bit. So y’ is equal to, let’s throw that e^x on
top, why not. 1 + x^2 on the bottom, plus I don’t know. I like to put that e^x out front of my inverse
tangent, just to make sure everyone knows that its not on the inside of the inverse
tangent. It’s really just being multiplied. Alright. On to the next one. This is y=csc^-1(e^x). Now I wanted to throw this one in here because
if you look at the rules for the derivative of inverse cosecant, there are actually two
little x pieces in there. So watch for that to show up when we take
its derivative. So the derivative of inverse cosecant. So negative, divided by, here is where one
of those x’s goes. The square root, here is where another one
of those x’s were, squared minus one. And that is really just setting up the derivative
of the outside. Now of course we don’t have an x, we have
something a little more complicated on the inside, we have e^x. So that would become the inside here. So e^x. e^x. Now we’ll multiply by the derivative of the
inside. The derivative of the inside in this case
is just e^x. Alright, now there is definitely some good cleaning up we can do here. In fact we have a common e^x in the bottom
and in the top, so we’ll go ahead and get rid of those. So we’ll say that the derivative of here is
a -1 all divided by the square root of e^2x minus 1. And now we are done with that derivative. So these rules they look a little bit complicated. I highly recommend you start memorizing these
as soon as possible, because you will want to recognize them in reverse when you do get
to integrals. And of course double check your math book
to make sure that your definitions match with what you are going to use on the homework. Alright, thanks for watching. Please go ahead and like this video if it
helped you out and go ahead subscribe to my YouTube channel. If you want to see some more videos, go ahead
and check out MySecretMathTutor.com

30 Replies to “Calculus – Find the derivative of inverse trigonometric functions”

  1. Thank you sir.. Uploaded just 8 days ago but its just in time for my exams. keep the video coming cuz i might need you in the next semester too 🙂

  2. In the last question, we should use product method isn't it? Can someone explain me please just the last question I didn't get.

  3. this is what I am after. derivatives with negative exponents, radicals in the denominator etc. i would like to see them more simplified though.

  4. Thank you so much! Great explanation, simply and well said. My calc prof skips steps and expects us to make the jumps along with him, then says it's our fault if we didn't follow along… Very frustrating, but now I'm more confident for our exam! 🙂

  5. A great tip comes, get ready. The key to memorize them is using repeatedly. Just write random functions and derive them with your cheat sheet. After a short while, you will have memorized them. Just do it prior to your exam.
    Btw, these are great to understand trig sub. Trig subs are making you to integrate a derivative you know. Of course, if you know these ones.
    I'm a freshman in college and find calculus fascinating, so in harmony with nature and this is the very reason why i hate linear algebra, leaves a bitter taste in my mouth.
    Great video, enjoy this good stuff folks.

  6. Sir your speech seen on screen create problem in question solution so u crroection in this style I hope you understand my problem and your trick is very useful for us

  7. Shouldn’t the derivatives of arcsec(x) and arccsc(x) be 1/(abs(x)sqrt(x^2-1)) and -1/(abs(x)sqrt(x^2-1)) respectively? abs stands for absolute value and sqrt stands for square root.

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