1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hyperbolic Differentiation

  1. Mar 30, 2014 #1
    1. The problem statement, all variables and given/known data

    y=1 / (cos h x), find dy/dx


    2. Relevant equations

    chain rule and coshx=(e^x+e^-x)/2

    3. The attempt at a solution

    IdRxUyK.jpg
     
  2. jcsd
  3. Mar 30, 2014 #2

    Curious3141

    User Avatar
    Homework Helper

    Resize your image!!! Way too big.

    The most obvious error is ##\frac{dy}{du} = \ln |u|##. Sure that's the derivative, and not the integral?

    Other than that, for Chain Rule, it's generally not helpful to substitute variables like this. Not wrong, but it can overcomplicate things.
     
  4. Mar 30, 2014 #3
    What an idiot I am. This mistake is proof that I am tired and should get some sleep. Well what other method do you suggest I use, I'd definitely like to know if it's more efficient.

    You can resize my image by clicking Ctrl and - on the keyboard. If you have a mouse with a wheel then turn the wheel towards you (downwards) while holding down Ctrl
     
  5. Mar 30, 2014 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Start with the definition of cosh(x) in terms of the exponentials and re-write 1/cosh(x). What is dy/dx when y = 1/x or x^-1?
     
  6. Mar 30, 2014 #5
    I've already solved it.
     
  7. Mar 30, 2014 #6

    Curious3141

    User Avatar
    Homework Helper

    Essentially what you did, but you don't have to spell out every step explicitly.

    You're using ##\frac{dy}{dx} = \frac{dy}{du}\frac{du}{da}\frac{da}{dx}##, but there's no need to spell out every step.

    Are you allowed to assume the derivative of the standard hyperbolic functions, like ##\frac{d}{dx}\cosh x = \sinh x##?

    If so, isn't it easier to let ##u = \cosh x## and ##y = \frac{1}{u}##? Again, you shouldn't need to spell it out like this in different variables, this is just for clarity.

    Yes, I know how to resize images - which is how I viewed yours. The problem is that it becomes impossible to view your text (and mine, when I post) in the same sizing. The font becomes too small.

    If you're going to post here regularly, you would be much better served by learning basic LaTex.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Hyperbolic Differentiation
Loading...