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Inverse hyperbolic functions

  1. Mar 7, 2008 #1
    1. The problem statement, all variables and given/known data
    a) Find y' if y = x^2arcsinh(2x)
    b) Find y' if y = xarctanh(x) + ln[(1-x^2)^1/2]
    c) Find y' if y = arccoth[sqrt(x^2+4)]

    2. Relevant equations
    d/dx arcsinh(x) = 1/sqrt(1+x^2)
    d/dx arctan(x) = 1/(1-x^2)
    d/dx arccoth(x) = 1/(1-x^2)

    3. The attempt at a solution
    I am still having trouble mastering the Chain Rule. Trying these problems has made me second guess myself even more, so I have tried each one two different ways, both of which are apparently wrong. I'll write out what I have. I'm not necessarily expecting to get an answer, but I would appriceate it if someone could point out what I'm doing wrong.

    a) y' = 2xarcsinh(2x) * [1/sqrt(1+2x^2)] * 2
    y' = 2xarcsinh(2x) * [2/sqrt(1+2x^2)]
    y' = (4xarcsinh(4x))/sqrt(1+2x^2)

    OR

    y' = 2x * [1/sqrt(1+2x^2)] * 2
    y' = 4x/sqrt(1+2x^2)

    b) y' = (1)arctanh(x) * [1/(1-x^2)] + [1/sqrt(1-x^2)] * [1/2(1-x^2)^-1/2] * (2x)
    y' = [arctanh(x)/(1-x^2)] + [(x/sqrt(1-x^2))/sqrt(1-x^2)]

    Then, multiply the top and bottom by sqrt(1-x^2) to get

    y' = [arctanh(x)/(1-x^2)] + [x/(1-x^2)]
    y' = [arctanh(x)+x]/(1-x^2)

    OR, if I take y' of xarctanh(x) to be (1) * [1/(1-x^2)] = (1-x^2), I get

    y' = (1+x)/(1-x^2)

    c) y' = [1/1-sqrt(x^2+4)^2] * [1/2(x^2+4)^-1/2] * 2x
    y' = [(x/sqrt(x^2+4))/1-x^2+4]

    Then, multiply the top and bottom by sqrt(x^2+4) to get

    y' = [x/(1-x^2+4)sqrt(x^2+4)]
    y' = [x/(5-x^2)sqrt(x^2+4)]
    ____________________________________________

    Sorry for this huge list. These problems are similar to each other, so most likely if I can figure out one, I will be able to get the rest. I've never excelled in math, so sorry if these are dumb questions..

    If anyone has problems following what I did please ask me.

    Thanks for your help.

    Jimmy
     
    Last edited: Mar 7, 2008
  2. jcsd
  3. Mar 7, 2008 #2
    I've just tried part b again and got y' = [2x/(1-x^2], which was, of course, wrong.
     
  4. Mar 7, 2008 #3
    for part a, you have to use both the product rule and the chain rule.
    I think the answer will be
    y = 2x arcsinh(2x) + 2x^2/sqrt(1+(2x)^2)

    I dont remember the derivatives of the hyperbolics so i used the ones in your 'relevant eq.' section.

    First, i took the deriv of x^2 and left the arcsinh term, completing the first term in the product rule. Then, for the second part of the product rule you leave the term you differentiated to get the first term and differentiate the second. The derivative of arcsinh(x) is 1 / sqrt(1+x^2), as you say. So instead of using x, use 2x (the argument in your case). Then after this you must use the chain rule. To do this you take the derivative of your argument of arcsinh, 2x. The derivative of 2x is 2. So you multiply that out to get the answer I gave above ( i didnt check my answer, so let me know if i gave it wrong)
     
  5. Mar 7, 2008 #4
    Engage,

    Thanks for your help, but I think your answer is also wrong. I follow your work, and I agree with you. However, the program I am using still says it's incorrect. I worked it out your way and simplified, and it was still wrong.

    I'll keep working at this.
     
  6. Mar 7, 2008 #5
    weird, i just checked it on maple and it said it was right. what program are you using?
     
  7. Mar 7, 2008 #6
    I'm answering the questions on WebAssign.net.

    I tried it again, twice, and it's still not accepting it :(
     
  8. Mar 7, 2008 #7
    Weird... Are you sure its arcsinh, not sinh or arcsin? I checked again on maple and it says that what I gave you is the right answer. This is the exact code that maple gave me back:

    2*x*arcsinh(2*x)+2*x^2/sqrt(1+4*x^2)
     
  9. Mar 7, 2008 #8
    Oh! Look at your first repsonse, and look at the Maple code: the denominators are different. I tried it with 4x^2 instead of 2x^2 and it accepted it.


    Thanks!
     
    Last edited: Mar 7, 2008
  10. Mar 7, 2008 #9
    i did (2x)^2 in my first response, which is the same as 4x^2; i was trying to put emphasis that you replace the x with (2x) when you use tabulated integrals. But i hope that helped
     
  11. Mar 7, 2008 #10
    I figured out how to do these problems. Thanks to EngageEngage for help.
     
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