I want that shirt. Wait, no, I want the person in front of me to have that shirt.

@blackpenredpen

5 жыл бұрын

kingbeauregard : )

@OtherTheDave

5 жыл бұрын

I wonder what would happen if a student came to an exam wearing that? Make them sit in the back of the class?

Derivative of arccosh x is -arcsinh x *dabs*

@AlgyCuber

5 жыл бұрын

hmmmmm

@Aruthicon

5 жыл бұрын

Obligatory rant: arsinh and arcosh, not arcsinh and arccosh. (It's okay for normal trigonometric functions, though.) Inverse hyperbolic functions give you area, not arc length.

@DutchMathematician

5 жыл бұрын

+Futfan As Tommy Thach correctly pointed out: the inverse functions of sinh and cosh are denoted by arsinh and arcosh respectively, where 'ar' stands for 'area', not 'arc'(length). Besides that, the derivative of arcosh is NOT arsinh. In the video it is shown that arcosh'(x) = 1/√(x²-1), which is not equal to arsinh(x) (which is ln (x+√(x²+1))). (no idea what 'dabs' means ...)

are students allowed to wear that shirt during tests? xD

this video just saved me from depression

Great explanation thank you!!

I liked the second one... don't know, I have always liked more implicit (don't know how to write it) derivatives... Great video!

Can you do a video of this showing the general equation for the derivative cosh^-1x?

@blackpenredpen

5 жыл бұрын

NVA Pisces ?

Thank you！

U make it easy

This is why I love you bro!

Does anyone have the link for buying that t-shirt

Thanks a lot

Amazing!

Very helpful 👌👌

Thankyou sooooooooo much

Good job

Love the shirt.

@blackpenredpen

5 жыл бұрын

julius omega thanks!!!

Hello sir from india how we will solve underroot r4-a4 please answer

I had an idea for a video on a "concatenate" operator, a function that just puts 2 things together. For example, concat(2, 2) = 22, concat(47, 299) = 47299, etc. I'm not sure how to derive a function for this, and it may be interesting.

@k0nahrik

5 жыл бұрын

Jow that function is not continuous in the first argument, and in the second it behaves like the argument itself, so that would be 1 as the derivative

@JoJoJet100

5 жыл бұрын

@@k0nahrik I meant "derive" as in find a function for it, not differentiate

@DutchMathematician

5 жыл бұрын

+Jow What about y+x*10^(1+E(log10(y))) (E(x) being the so-called Entier or floor function - see e.g. en.wikipedia.org/wiki/Floor_and_ceiling_functions for details; log10 being the log base 10 function)

@JoJoJet100

5 жыл бұрын

@@DutchMathematician That looks like it should work, cool!

Awesome

What if you had cosh^-n (x)?

Totally appreciated 3:03 AM

Do You Wear this shirt when you are invigilator in any exam?

Now when you say "chain rule" it sounds wrong

I want you to be my Calc 2 teacher

how are you able to rewrite cosh^-1 to ln(x+sqrt(x^2-1)? I didn't now that is a thing

@DutchMathematician

5 жыл бұрын

+Low Noice Actually, it is not that difficult, using the definition of cosh(x). Suppose y=cosh(x). Finding an expression for arcosh(x) (the inverse of cosh(x)) means finding a function expressing x in terms of y (this holds for every inverse of a function y=f(x)). First, we have to restrict the domain of cosh(x), for an inverse of cosh to even exist. Since cosh(x)=cosh(-x), the inverse of cosh does not exist on the whole of R. However, if we restrict the domain of cosh to [0,+∞), it is easy to see that cosh DOES has an inverse. The range of cosh on the domain [0,+∞) is easily seen to be [1,+∞). Hence, the inverse of cosh (being arcosh or cosh^-1) is therefore a function with domain [1,+∞) and range [0,+∞). Now the computation of arcosh ... So, y=cosh(x)=1/2*(e^x+e^(-x)). Multiplying both sides with 2*e^x gives 2*y*e^x=(e^x)^2+1. Substituting u=e^x then gives the equation 2*y*u=u^2+1. Rearranging this equation gives u^2-2*y*u+1=0. Note that u=e^x, hence u>0. Applying the a,b,c-formula to this equation gives u=[-(-2*y)±√(-2*y)^2-4*(1)*(1)]/(2*1)=y±√(y^2-1) Since u>0, it is easy to see that only the '+' sign holds (whether y>0, y

Is it pronounced sinch or shine

Show that cosh^- 1 (1/x) = sec(h ^ - 1) * x

The third way is to use that cosh(x) = cos(ix), thus cosh^-1(x) = -i cos^-1(x). Now, d/dx cosh^-1(x) = d/dx (-i cos^-1(x)) = -i d/dx cos^-1(x) But by trigonometry (draw a triangle diagram and use trig identities), d/dx cos^-1(x) = -1/sqrt(1 - x^2). Then d/dx cosh^-1(x) = i/sqrt(1 - x^2), and noting that when |x| >= 1 we have sqrt(1 - x^2) = i sqrt(x^2 - 1), we get d/dx cosh^-1(x) = 1/sqrt(x^2 - 1), |x| >= 1. Done. Complex Analysis FTW.

@blackpenredpen

5 жыл бұрын

Wow, very very nice!!!

求文化衫淘宝店地址（滑稽

I haven't testet it out, but I have a problem for you: (f^-1(x))' = (f(x)') ^-1

@blackpenredpen

5 жыл бұрын

Leon Bannöhr peyam did it already, I think.

@DutchMathematician

5 жыл бұрын

Actually, your answer is almost right, but not quite ... The correct equation is ... (f^-1(x))' = 1 / f' (f^-1(x))

How old are you ?

find the definite integral of this derivative and youll have a little surprise :D

The second method is way cooler, because i don't need to remember anything

its blackpenredpenbluepen now

Your sinh looks like smh(x) XD

one legal way to cheat in exam lol

I got 1/sinh(arccosh(x)) amirite Edit: No Edit again: Yes

@AlgyCuber

5 жыл бұрын

you’re actually right

@marc6365

5 жыл бұрын

Yup

@taisinclair9033

4 жыл бұрын

Liked for the edits