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🙏🙏
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?
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.
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.
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
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.
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
Yes, but it requires a detour to complex numbers to do it. For arcsine, to find an explicit formula for this one, not using predefined inverse trig, we can recognize the following identity: i*sin(y) = sinh(i*y) Solve for sin(y): sin(y) = -i*sinh(i*y) Equate this to x, and solve for y: x = sin(y) = -i*sinh(i*y) Rewrite sinh(i*y) with its definition: x = -i*[e^(i*y) - e^(-i*y)]/2 Multiply every term by 2*i*e^(i*y): 2*i*x*e^(i*y) = -i*[e^(2*i*y) - 1] Let E = e^(i*y): 2*i*x*E = E^2 - 1 Solve for E, with the quadratic formula: E = i*x +/- sqrt(1 - x^2) Recall that E = e^y e^y = i*x +/- sqrt(1 - x^2) Take the complex log: y = log(i*x +/- sqrt(1 - x^2)) This is a way to express y=arcsin(x), using arithmetic, powers, and logs. It ultimately will require arctangent to make use of this formula, since complex log needs to work with the magnitude and angle of the input. To get the angle, given cartesian form, you need a 4-quadrant angle resolver arctangent function.
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 ;-)
Wow! That was so clearly presented that even with my so-so math ability, it was easy to understand. Thank you!
Iam stady from IRAQ... 😘 اتابعك من 🇮🇶
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) )
U saved my life
:))
Wow!:" The 'Yan Can Cook of Calculus!" Awesome Thank you.
You are simply the best, no questions.
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
this was so helpful and clear thank you! :)
Finally,i understand this topic..thank you so much sir!
Love you bro 🌟, excellent teacher you are!
Thank you for an outstanding video/lecture on the Inverse Sinh(x).
Thank you! Just what I was looking for!
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
You, Sir, are incredible. You have my out most gratitude
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🙏🙏
Outstanding explanation sirjiii
Please what do u mean by negative 2x in the rot is equal to 4x square
Thank you veryy much. It is really helping me to face final exam
You're a MEGA trip!! X-D
Thanks!!
Hove you done conversions from sine to sinh & tangent to tanh?
are you have the vidoo on your playlist for expression for the invers trigonometric function??
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?
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!!
make it easy and i got it very well Thanks !
Perfect video. Thanks a lot 👍🏽
Thanks you so much sir for such an informative video 😇❤️
You sir, an outstanding teacher!
Great work sir ! thanks a lot of you sir !
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.
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 ?
Thanks sir..... you're a brilliant young man #genius
very clear, helped a lot, thank you
Thank you so much dear
Tq so much sir i am from India 🇮🇳
Hey, I’m just curious. Did you study math? And if you did, what addict of math did you focus on?
l love ur vidz
how do we know that sinh x though is 1 to 1?
great video thank you!
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?
THANK YOU SO MUCH!
Is it really a quadratic equation because of - 1??
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
Brilliant 👏🏻👏🏻👏🏻👏🏻👏🏻
That was amazing !!! Thank you so much
You're welcome!
What of expressing in log form
damn! you are a genius
Thank you sir ❤
Witch means sinh(whatever)=(e^whatever-e^-whatever)/2, Isn't it?
Thank you for the lesson!
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.
excellent......you pronunciation is like brucilee ......i can now kungfu maths exams
you sir are the GOAT
Thank you so much
Desde Colombia, que genialidad
Great video!
Thank you sir!
nice job friend I wish you happy life
that is very impressive..thank u sir
Brilliant expression! So cool!
: )
wow i loved it!!
Nice vid bruh
thank you :0
love it!
Danke
Merci
Thanks
شكرا جزيلا
Thank u so much
thank u sir for sharing, i can understand well ;)
thanks alot😍
Thank you
argument of hyperbolic sine
this guy's fantastic
thanks!
Nice work
thank youu :D
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
Thank you sir
BEST 👍🏻
Thank you sir.
thanks a lot man
Great Video but x-√(x^2-1) is not always negativ
Thanks bro +
thank you
Thank u math god
Thankyou sir
crack ! excellent explication !
I tried to do it myself, but instead of writing e^y=x+sqrt(x^2+1) I accidentally took 1 for x
Simply amazing
牛逼
Thanks sir
Can this be done for arcsin(x), arccos(x) etc?
Yes, but it requires a detour to complex numbers to do it.
For arcsine, to find an explicit formula for this one, not using predefined inverse trig, we can recognize the following identity:
i*sin(y) = sinh(i*y)
Solve for sin(y):
sin(y) = -i*sinh(i*y)
Equate this to x, and solve for y:
x = sin(y) = -i*sinh(i*y)
Rewrite sinh(i*y) with its definition:
x = -i*[e^(i*y) - e^(-i*y)]/2
Multiply every term by 2*i*e^(i*y):
2*i*x*e^(i*y) = -i*[e^(2*i*y) - 1]
Let E = e^(i*y):
2*i*x*E = E^2 - 1
Solve for E, with the quadratic formula:
E = i*x +/- sqrt(1 - x^2)
Recall that E = e^y
e^y = i*x +/- sqrt(1 - x^2)
Take the complex log:
y = log(i*x +/- sqrt(1 - x^2))
This is a way to express y=arcsin(x), using arithmetic, powers, and logs. It ultimately will require arctangent to make use of this formula, since complex log needs to work with the magnitude and angle of the input. To get the angle, given cartesian form, you need a 4-quadrant angle resolver arctangent function.
goat
very nice
great video
fenderbender28 thank you
Thanks
and god created infinitesimal calculus, and saw it was hard. and so god said, let there be UA-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 ...
Lovely
thank you
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
Cool
u're cool too!
good!!
❤️❤️❤️