For simplification, when b is even (like in this case where b = 2x), I invite you to use b'=b/2 then x=(-b' +- sqrt((b')^2 - a c))/a. Thank you for this amazing video ;-)
Let y = sinh inverse x Clearly - infinity < y < infinity Then sinhy = x Now coshy = + sqrt (1 + (sinhy)^2) (since coshy is > 1 or = 1 therefore coshy is always > 0 for all y belongs to R ) => coshy = sqrt (1 + x^2) Now coshy + sinhy = e^y => sqrt (1 + x^2) + x = e^y => e^y = x + sqrt (1 + x^2) => y = ln (x + sqrt (1 + x^2) ) Hence we conclude that y = sinh inverse x = ln (x + sqrt (1 + x^2) )
You r great sir. I have a question that how can be draw the graph of ln(x+sqrt(x^2+16)) Is this function is similar to inverse sine hyperbolic or not??? Please answer🙏🙏
Cannongabang It's a widespread method in algebra actually. If you find a well-defined function g such as f(g(x))=x=g(f(x)), g is automatically f's inverse and f is one-to-one
How is it confusing? It's always been taught like that and you basically get introduced to it before you even know that sin^2(x) is a thing. Standard notation that has been used for decades if not centuries.
This comes from the nature of the exponential function always being positive. Since exp(y) is what we're solving for, the negative solution is not applicable to our problem, and er can ignore it for our purpose.
It's because of the way the equation is structured. e^y is the unknown variable, and it's inside of a 3 term polynomial set equal to zero. In the first term, it's 1*our unknown variable where our unknown variable is squared, and in the second term, it's -2x*e^y, without the square. The third term is simply a constant. It's essentially structured like the standard form of quadratic equation that would allow it to be solved with the quadratic formula, it doesn't matter whether or not the unknown variable we're solving is x, u, theta, t, or even e^y.
Gaurav Mishra sqrt(x^2) is the same magnitude as x growing at the same rate as x's... However for every x you give it, you'll get a number ever so slightly larger than x thanks to that + 1 in there, and. Since they grow at the same rate, x will never be larger than sqrt(x^2 +1)
Gaurav Mishra hehe, I know right? The internet is 'obviously' supposed to be a place for mindless bickering and trolling... But we can use it for math and helping people!
Gaurav Mishra x=√(x²) so if anything under the square root is larger than that the whole expression is, intuitively larger A more rigorous proof may be this:. Assuming the principal root, √(x²)=|x| √(x²+1) = √(x²(1+1/x²)) = √(x²)×√(1+1/x²) = |x|√(1+1/x²) So, we know the square root of anything greater than one is greater than one since √1=1 and √x is an increasing function, and multiplying a number by something greater than one produces a product larger than the number: mathematically, |x|√(1+1/x²)>|x| Now, if x is nonnegative, it's trivial that x+√(x²+1) is positive. So we want this for the problem. If x is negative, then the statement I proved before is helpful because it shows that x+√(x²+1) must be positive; if x is non-positive then x+|x| = 0. Now let's assume the negative root. If x is non-positive, it's trivial that x+√(x²+1) (or rather x-√(x²+1) with the principal root) is negative. And if x is nonnegative we use the statement I proved along with the fact that with this x that x-|x| = 0 to show that the expression is again negative. Have a nice day
Hello, How can i transform u=exp(x/A)*C1*A - exp(-x/A)*c2*A (wich verify u''= u (second derivate) = 1/A^2 * u to u = sinh((x-c1)/A) (wich verify as the same way : u''= u (second derivate) = 1/A^2 * u?
You, sir, are a brilliant mathematician and an outstanding teacher!
Stephen Lanford thank you!
Coolest Mathematician ever!!!.
I love you🙇🏽
For simplification, when b is even (like in this case where b = 2x), I invite you to use b'=b/2 then x=(-b' +- sqrt((b')^2 - a c))/a. Thank you for this amazing video ;-)
Iam stady from IRAQ... 😘 اتابعك من 🇮🇶
Wow! That was so clearly presented that even with my so-so math ability, it was easy to understand. Thank you!
Wow!:" The 'Yan Can Cook of Calculus!" Awesome Thank you.
You are simply the best, no questions.
this was so helpful and clear thank you! :)
At 3:33 you could add +x^2-x^2 so that you would get (e^y-x)^2-x^2-1=0 and then solve for y which in my opinion feels more natural
You're a MEGA trip!! X-D
Thanks!!
Hove you done conversions from sine to sinh & tangent to tanh?
Thank you! Just what I was looking for!
Let y = sinh inverse x
Clearly - infinity < y < infinity
Then sinhy = x
Now coshy = + sqrt (1 + (sinhy)^2)
(since coshy is > 1 or = 1 therefore coshy is always > 0
for all y belongs to R )
=> coshy = sqrt (1 + x^2)
Now coshy + sinhy = e^y
=> sqrt (1 + x^2) + x = e^y
=> e^y = x + sqrt (1 + x^2)
=> y = ln (x + sqrt (1 + x^2) )
Hence we conclude that
y = sinh inverse x = ln (x + sqrt (1 + x^2) )
how do we know that sinh x though is 1 to 1?
great video thank you!
Love you bro 🌟, excellent teacher you are!
You, Sir, are incredible. You have my out most gratitude
牛逼
Outstanding explanation sirjiii
Finally,i understand this topic..thank you so much sir!
That was amazing !!! Thank you so much
You're welcome!
very clear, helped a lot, thank you
Thank you for an outstanding video/lecture on the Inverse Sinh(x).
Perfect video. Thanks a lot 👍🏽
That was really helpful it just didn't come to my mind that it is a quadratic equation and I just couldn't figure it out
Great work sir ! thanks a lot of you sir !
U saved my life
:))
You sir, an outstanding teacher!
make it easy and i got it very well Thanks !
THANK YOU SO MUCH!
Brilliant 👏🏻👏🏻👏🏻👏🏻👏🏻
Hey, I’m just curious. Did you study math? And if you did, what addict of math did you focus on?
You r great sir.
I have a question that how can be draw the graph of ln(x+sqrt(x^2+16))
Is this function is similar to inverse sine hyperbolic or not???
Please answer🙏🙏
Aside from the fact that you assume the existance of the inverse before finding it, it is always cool to derive the inverse :)
Cannongabang It's a widespread method in algebra actually. If you find a well-defined function g such as f(g(x))=x=g(f(x)), g is automatically f's inverse and f is one-to-one
Thank you veryy much. It is really helping me to face final exam
Thank you sir ❤
Thank you so much dear
Thanks sir..... you're a brilliant young man #genius
Thank you for the lesson!
Thank you sir!
are you have the vidoo on your playlist for expression for the invers trigonometric function??
why do we write inverse trigs as ^-1? i think they can be interpreted as sin^-1 = csc, etc
we can just use arcsin, etc
It's just a standard notation.
for people who don't know
like sin^2 x is (sin x)^2 but sin^-1 x is *not* 1/sin x
How is it confusing? It's always been taught like that and you basically get introduced to it before you even know that sin^2(x) is a thing. Standard notation that has been used for decades if not centuries.
because it just isn't consistent with anything else?
Great video!
Brilliant expression! So cool!
: )
thank you :0
Great solution 😃 it helps me to do my homework... Thanks 🙏
it is my homework too
does all of the teacher in the world give the same homework
is it impossible ?
Why not consider the minus part despite being negative? I just want a reason why the minus part is not included anymore in finding y
This comes from the nature of the exponential function always being positive. Since exp(y) is what we're solving for, the negative solution is not applicable to our problem, and er can ignore it for our purpose.
Desde Colombia, que genialidad
love it!
Please what do u mean by negative 2x in the rot is equal to 4x square
you sir are the GOAT
nice job friend I wish you happy life
Thanks you so much sir for such an informative video 😇❤️
Thank you so much
Tq so much sir i am from India 🇮🇳
Witch means sinh(whatever)=(e^whatever-e^-whatever)/2, Isn't it?
thanks alot😍
excellent......you pronunciation is like brucilee ......i can now kungfu maths exams
argument of hyperbolic sine
this guy's fantastic
thanks!
damn! you are a genius
Is it really a quadratic equation because of - 1??
I don't understand how that (e^(y)) becomes equal to the quadratic function. Is it because U= quadratic Function ?
It's because of the way the equation is structured. e^y is the unknown variable, and it's inside of a 3 term polynomial set equal to zero. In the first term, it's 1*our unknown variable where our unknown variable is squared, and in the second term, it's -2x*e^y, without the square. The third term is simply a constant. It's essentially structured like the standard form of quadratic equation that would allow it to be solved with the quadratic formula, it doesn't matter whether or not the unknown variable we're solving is x, u, theta, t, or even e^y.
that is very impressive..thank u sir
crack ! excellent explication !
Thank u so much
thank youu :D
thank u sir for sharing, i can understand well ;)
l love ur vidz
Thank you
Nice work
BEST 👍🏻
Thank you sir.
What of expressing in log form
thank you
thanks a lot man
I tried to do it myself, but instead of writing e^y=x+sqrt(x^2+1) I accidentally took 1 for x
wow i loved it!!
Thank you sir
how do you know that sqrt(x^2+1) is greater than x?
Gaurav Mishra sqrt(x^2) is the same magnitude as x growing at the same rate as x's... However for every x you give it, you'll get a number ever so slightly larger than x thanks to that + 1 in there, and. Since they grow at the same rate, x will never be larger than sqrt(x^2 +1)
Thanks
Nice to see helpful people on the Internet. Faith in humanity restored.
Gaurav Mishra hehe, I know right? The internet is 'obviously' supposed to be a place for mindless bickering and trolling... But we can use it for math and helping people!
Gaurav Mishra x=√(x²) so if anything under the square root is larger than that the whole expression is, intuitively larger
A more rigorous proof may be this:. Assuming the principal root, √(x²)=|x|
√(x²+1) = √(x²(1+1/x²)) = √(x²)×√(1+1/x²) = |x|√(1+1/x²)
So, we know the square root of anything greater than one is greater than one since √1=1 and √x is an increasing function, and multiplying a number by something greater than one produces a product larger than the number: mathematically,
|x|√(1+1/x²)>|x|
Now, if x is nonnegative, it's trivial that x+√(x²+1) is positive. So we want this for the problem. If x is negative, then the statement I proved before is helpful because it shows that x+√(x²+1) must be positive; if x is non-positive then x+|x| = 0. Now let's assume the negative root. If x is non-positive, it's trivial that x+√(x²+1) (or rather x-√(x²+1) with the principal root) is negative. And if x is nonnegative we use the statement I proved along with the fact that with this x that x-|x| = 0 to show that the expression is again negative.
Have a nice day
Simply amazing
Great Video but x-√(x^2-1) is not always negativ
Cool
u're cool too!
very nice
Thanks bro +
and god created infinitesimal calculus, and saw it was hard. and so god said, let there be TH-cam videos, and god saw it was good.
was about to write god said "let there be Asians", but i figured it will be racist ...
❤️❤️❤️
Hello, How can i transform u=exp(x/A)*C1*A - exp(-x/A)*c2*A (wich verify u''= u (second derivate) = 1/A^2 * u to u = sinh((x-c1)/A) (wich verify as the same way : u''= u (second derivate) = 1/A^2 * u?
Thankyou sir
I never write comments but I had to for this! Your method is so clear and you explained everything very well, thank you for such a great watch!!
Thank u math god
great video
fenderbender28 thank you
Danke
Merci
Thanks
شكرا جزيلا
Lovely
thank you
goat
Nice vid bruh
Thanks sir
good!!
He could've just done the y=f(x) method
And that's the method he used lol.
I will be a lot better if you do the arbitrary calculations faster thanks! Really love your work though
Tanay Saha u can fast forward
Thanks