When I'm done with school, I'm gonna start math from the beginning (not really the beginning, but from where I'm comfortable) just because you make so much sense of it all. I love the way you teach. Update: I have learnt a great deal. I've recently completed ODEs, particularly wronskian theorem and uniqueness theorem. Ive gotten over my fear of trigonometry and statistics. It's been great ever since i started working in insurance, especially since I now have some money to play the stock market. Keep at it, guys. KA doesn't have everything, but it will, at the very least, position you to better understand published texts without tuition.
I used to hate maths in highschool, but it was just because I didn't know how to study maths. I did like it very much, but I wasn't disciplined enough to really learn it, so I thought I hated it. So I went through the entirety of KA's fundamentals and now I'm a Computer Engineering student.
I am currently studying for my AP Calculus AB midterms and this helped tremendously. I have been scared of inverse trig for some time now, and your videos are helping me work to overcome this fear. Thank you.
My teacher tried explaining this, differentiating arctan and improper integrals in 3 hours and I got nowhere, I’ve watched 15 mins worth of your videos and I get it, thank you so much
There's a general formula for finding the derivative of a function from it's inverse function. You start by defining g(x)=f-1(x) (the inverse of f(x)). You can then write that f(g(x))=x since that is the very definition of an inverse function. If we then differentiate both sides with respect to x, we get d/dx[f(g(x))]=d/dx[x]. On the left we use the chain rule and get f'(g(x))*d/dx[g(x)], and on the right we get 1. Solve for d/dx[g(x)] and you get d/dx[g(x)]=1/f'(g(x)). You can even find the integral of a function from its inverse function but that i'm not going to tell. If someone finds/wants the answer please write me a comment because i would love to know :) Hint: It involves f(g(x))*g'(x)
Amazing method to find the derviative of inverses. My teacher was trying to explain it with all these whacky formulas and stuff but it just didn't hit. This easily cleared up everything and finally makes so much sense!
Thank You shoo much..now i understand what is inverse trigo derrivative..my lect always said to us to memorizing this thing..but...meh..what is the function if we mem. without knowing where it come from..thanks..feeling happy rite nw..
This was pretty straightforward (even more so if you take the graph of the sine function; the sine is an odd function so the sign doesn't change, and for every change of y, the change of x is the period of the cosine wave. Now substitute to get the derivative wrt sine and there we have the answer.
Two months into Calc 1 and I am finally able to take derivatives of inverse trig functions. It takes a lot of hard work. Looking at different problems. Using the Pythagorean identity, sin^2x+cos^2x=1.
ALTERNATIVE METHOD. Using the derivative of inverse property, arcsin'(x)=1/sin'(arcsin(x))=1/cos(arcsin(x)). I didn't simplify it further but on DESMOS it produces the same graph as this answer. I would imagine with some trig properties it reduces to this answer
y=arcsin(x) has a range restriction. y is only Q1 or Q4 angles. In both of these quadrants, cos(y) is positive, so you can disregard the negative branch of your square root. Great question. You can see this considered more closely in a similar proof for y=arcsec(x).
Because arcsine is not the same as (sin(x))^-1, that would be cosecant. sin^-1(x) is simply shorthand notation for arcsine, there is no actual exponent
Slavik Egorov nope. This technique is called the implicit differentiation, where we make y as a function of x. So imagine that it's actually like saying f(x) = y= sin^-1(x), which makes it like saying sin(f(x)) = x
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
When I'm done with school, I'm gonna start math from the beginning (not really the beginning, but from where I'm comfortable) just because you make so much sense of it all. I love the way you teach.
Update: I have learnt a great deal. I've recently completed ODEs, particularly wronskian theorem and uniqueness theorem. Ive gotten over my fear of trigonometry and statistics. It's been great ever since i started working in insurance, especially since I now have some money to play the stock market. Keep at it, guys. KA doesn't have everything, but it will, at the very least, position you to better understand published texts without tuition.
varun009 what happened
we need update bro
did you do all of it? hopefully the pandemic gave you time...
@@luke_will_taylor yep. Read the update.
I used to hate maths in highschool, but it was just because I didn't know how to study maths. I did like it very much, but I wasn't disciplined enough to really learn it, so I thought I hated it. So I went through the entirety of KA's fundamentals and now I'm a Computer Engineering student.
that Khan guy is a pretty smart dude
I am currently studying for my AP Calculus AB midterms and this helped tremendously. I have been scared of inverse trig for some time now, and your videos are helping me work to overcome this fear. Thank you.
My teacher tried explaining this, differentiating arctan and improper integrals in 3 hours and I got nowhere, I’ve watched 15 mins worth of your videos and I get it, thank you so much
There's a general formula for finding the derivative of a function from it's inverse function. You start by defining g(x)=f-1(x) (the inverse of f(x)). You can then write that f(g(x))=x since that is the very definition of an inverse function. If we then differentiate both sides with respect to x, we get d/dx[f(g(x))]=d/dx[x]. On the left we use the chain rule and get f'(g(x))*d/dx[g(x)], and on the right we get 1. Solve for d/dx[g(x)] and you get d/dx[g(x)]=1/f'(g(x)). You can even find the integral of a function from its inverse function but that i'm not going to tell. If someone finds/wants the answer please write me a comment because i would love to know :)
Hint: It involves f(g(x))*g'(x)
Your voice calms me ,
tells me that every thing is ok , you gonna do good in tomorrow’s exam and your whole math journey in your life 🙏🏻♥️
The Drum roll is inherent for Sal's work. He makes it's all so clear and enlightening :D
khan, you nailed this
Amazing method to find the derviative of inverses. My teacher was trying to explain it with all these whacky formulas and stuff but it just didn't hit. This easily cleared up everything and finally makes so much sense!
the way my textbook explains this is ABYSMAL! Thank you so much!
Thank you so much for this! I don't know why but this makes everything so much simpler...
Thank You shoo much..now i understand what is inverse trigo derrivative..my lect always said to us to memorizing this thing..but...meh..what is the function if we mem. without knowing where it come from..thanks..feeling happy rite nw..
Thank u so much Khan academy. I fell in love with maths
This was pretty straightforward (even more so if you take the graph of the sine function; the sine is an odd function so the sign doesn't change, and for every change of y, the change of x is the period of the cosine wave. Now substitute to get the derivative wrt sine and there we have the answer.
This is just beautiful!! =) the beauty of mathematics!
Thank you Sal for still coming with interesting videos. DFTBA :D
Oh wow I never knew that. I've always just know them. Thanks for the video.
Two months into Calc 1 and I am finally able to take derivatives of inverse trig functions. It takes a lot of hard work. Looking at different problems. Using the Pythagorean identity, sin^2x+cos^2x=1.
I have a tight schedule, in India to prepare for IIT-JEE, which required me to learn the whole Derivatives topic in a week!
Great and clear cut lecture... Love it... ❤️❤️
Thank you so much for the clear explanation
You are really super ❤❤❤❤
Thanks again Sal!
ALTERNATIVE METHOD. Using the derivative of inverse property, arcsin'(x)=1/sin'(arcsin(x))=1/cos(arcsin(x)). I didn't simplify it further but on DESMOS it produces the same graph as this answer. I would imagine with some trig properties it reduces to this answer
Thank you so much
From jamaica: mi rate you!
Thank you for the video.
Thanks teacher it was very good
THANK YOU VERY MUCH SIR
Thank you very much! :)
Should there not be a plus/minus sign in front of that square root?
Well until we open the sqrt
There is no need to apply +- sign infront of it
y=arcsin(x) has a range restriction. y is only Q1 or Q4 angles. In both of these quadrants, cos(y) is positive, so you can disregard the negative branch of your square root. Great question. You can see this considered more closely in a similar proof for y=arcsec(x).
Well explained
It's a good video, but I have to make of you for encouraging us to try it out first. We're here because we've already tried and failed.
tnxx..
at 3:06, khan only took the positive square root of cos y, implicitly rejecting the negative root of cos y. why did he do that?
Why did you take the positive square root only?
thanks Genghis.... i needed this to understand how the derivative table ended up with this...
just curious, why cosy is not equal to negative square root of 1 minus sinx square?
Where did the plus and minus sign go?
Khan bro please explain your concepts in Hindi as Hindi is more understanding than English
Why not simply apply chain rule?
Alright, but shouldn't it be +- 1/(1-x^2)^1/2 ?
The inverse of sine is only defined for sine positive values ([0,π])
I can feel it
Sal, you speak maths !
Cool
Why not to use arcsin? It's always confusing for me
Good question. It's a shame that the more confusing/ambiguous notation is much more popular.
Thank you 🌹🌹
why can't I use chain rule and just answer it as, -sine^-2(x) . cos x
have you figured this out? :v
Because arcsine is not the same as (sin(x))^-1, that would be cosecant. sin^-1(x) is simply shorthand notation for arcsine, there is no actual exponent
thanks/.
♥️
can someone plz tell me what is e inverse x derivative
dy/dx sin(y)??? Is that allowed? shouldnt it be dx/dy sin(y) then you flip dx/dy to dy/dx to get 1/cos(y)
4:50 i got scared!!!
Wouldn't that be plus and minus of the final formula?
the graph of y=sin ^-1 x is always positive
But.. why is x = sin y ?
For inverse you switch the x and y
So the 'sin-1 x' is arcsin x... I thought it was (sin x)^-1
cos(x) should be +/- Sqrt(1-x^2 ); why the - just skipped here !
Mo Tahoon cuz y € [-90,90]
This comment is six months old so you probably don’t care why (y) is positive
wait isn't d/dx siny equal to zero?
Slavik Egorov nope. This technique is called the implicit differentiation, where we make y as a function of x. So imagine that it's actually like saying f(x) = y= sin^-1(x), which makes it like saying sin(f(x)) = x
principle root i mean comon... its square root ffs... comon khan stop trying to be fancy...
... or you could just draw the triangle.
I don't get it...
Mrius86 yup, and I never will. Inverse trig functions screw me up.
Go learn implicit differentiation first so that u can understand this proof of the derivation of (sinx) ^-1
Who can translate the video into Arabic ??
COULD YOU PLEASE EXPLAIN GRAPHICALLY AND INTUITIVELY (AS YOU ALMOST ALWAYS DO) WHAT IN THE HECK IS THE INVERSE OF SINE IS?????????? THANK YOU.