Added: 4 years ago
From: donylee
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  • The nickname of hyperbolic tangeant is darn I believe.

    sinh=shin

    cosh=korsh

    tanh=darn

    coth=korf

    sech=seech

    csch=ko-seech

  • Yes, the derivative of coshx is sinhx, not -sinhx. If you take the derivative of the definition of cosh, which is (e^x + e^-x)/2, you get (e^x - e^-x)/2. Which is the definition of positive sinh.

    And the tanhx is pronounced the "tanth" of x.

  • yeh... isnt when u differentiate

    sinx = cosx

    cosx =-sinx

    tanx = sec^2x

    or issit different for sinhx and coshx?

  • usefull lucter...thank you indeed

  • r u sure that d/dxcosh=sinh? not -sinh?

  • Thank you so much man. This is very helpful.

  • isnt it pronounced "shine"

  • lol, its pronouced shine in britain I believe, hes from Sginapore, great lecture though

  • Yup, could be different from different countries. Anyways, I got the pronunciation from George F. Simmons amazing calculus book 'Calculus with Analytic Geometry' where he mentioned that hyperbolic sine is pronounced as 'cinch' which rhymes with 'pinch'.

    It looks to have that 'ch' sound from the 'c' in sinc.

  • Can you make videos of the hyperboloid.

    and can you thel me what is parabol and hyperbol focus formilay?

  • isnt it jus

    sinx

    cosx

    tanx?

  • These are 'hyperbolic' functions and not the standard trigonometry functions.

    We write them as 'sinh x' with a 'h' after 'sin'. They rarely bear connections with the standard trig functions. Their derivations too are hyperbolic functions. The functions themselves are defined in terms of natural log number e and are useful in certain integrations.

    See my video 'The Catenary' and you'll know what I mean.

  • sinx

    cosx

    tanx are functions made from a basic circunference... (also but rarely called circular functions)

    sinhx

    coshx

    tanhx are functions made from a hyperbole ... (also known as hiperbolic functions)

  • tanch

  • tanh = "tanch or than"

  • tangent = tan ;pp

  • Exvelente. Puedes colocar este limite:

    lim  (x^n-1)/(x-1)

    x->1

  • thank you

  • nice work! i wish if you talked in french :)

  • Hey BlueScience,

    I guess you from France. They learn hyperbolic functions there? Hehe.

    Thanks for watching.

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