Hey there! Sign in to join this conversationNew here? Join for free

Why are they equal?? Watch

Announcements
    • Very Important Poster
    • Welcome Squad
    • Thread Starter
    Online

    20
    ReputationRep:
    In Class yesterday I was doing a bit of differentiating and I realized that

    \frac{d}{dx}arctan(sinh(x))=sech  (x)=\frac{d}{dx}arcsin(tanh(x))

    I later that day used graphing software to plot them and saw that they lied on top of each other suggesting that they are equivalent. but I can't see how one would prove it. so far I have written the 2 out in exponential form:

    \frac{e^{iy}-e^{-iy}}{2i}=\frac{e^x-e^{-x}}{e^x+e^{-x}}
    \frac{e^{iy}-e^{-iy}}{i\left(e^{iy}+e^{-iy}\right)}=\frac{e^x-e^{-x}}{2}

    but can't see how one goes to the other. HELP.

    P.S. this is not work, this is just to satisfy my curiosity but there is no maths for fun forum I believe.
    Offline

    10
    ReputationRep:
    (Original post by Aph)
    In Class yesterday I was doing a bit of differentiating and I realized that

    \frac{d}{dx}arctan(sinh(x))=sech  (x)=\frac{d}{dx}arcsin(tanh(x))

    I later that day used graphing software to plot them and saw that they lied on top of each other suggesting that they are equivalent. but I can't see how one would prove it. so far I have written the 2 out in exponential form:

    \frac{e^{iy}-e^{-iy}}{2i}=\frac{e^x-e^{-x}}{e^x+e^{-x}}
    \frac{e^{iy}-e^{-iy}}{i\left(e^{iy}+e^{-iy}\right)}=\frac{e^x-e^{-x}}{2}

    but can't see how one goes to the other. HELP.

    P.S. this is not work, this is just to satisfy my curiosity but there is no maths for fun forum I believe.
    Integrate? Edit: as in return right to the start and integrate.
    • TSR Support Team
    • Very Important Poster
    • Reporter Team
    • Welcome Squad
    Offline

    19
    ReputationRep:
    It's because arcsin(tanh(x)) \equiv arctan(sinh(x)) that their derivatives are also equal.
    Offline

    13
    ReputationRep:
    (Original post by Aph)
    In Class yesterday I was doing a bit of differentiating and I realized that

    \frac{d}{dx}arctan(sinh(x))=sech  (x)=\frac{d}{dx}arcsin(tanh(x))

    I later that day used graphing software to plot them and saw that they lied on top of each other suggesting that they are equivalent. but I can't see how one would prove it. so far I have written the 2 out in exponential form:

    \frac{e^{iy}-e^{-iy}}{2i}=\frac{e^x-e^{-x}}{e^x+e^{-x}}
    \frac{e^{iy}-e^{-iy}}{i\left(e^{iy}+e^{-iy}\right)}=\frac{e^x-e^{-x}}{2}

    but can't see how one goes to the other. HELP.

    P.S. this is not work, this is just to satisfy my curiosity but there is no maths for fun forum I believe.
    If y = \sin^{-1}(\tanh x) then \sin y = \tanh x. Therefore,

    \displaystyle \tan^2 y =\frac{\sin^2 y}{1-\sin^2 y} = \frac{\tanh^2 x}{1-\tanh^2 x} = \frac{\sinh^2 x}{\cosh^2 x - \sinh^2 x} = \sinh^2 x

    Therefore \tan y  = \pm \sinh x

    Now ponder domains and choose the correct branch of the square root, and you're home and dry.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Break up or unrequited love?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.