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Homework Help: Another implicit differentiation Problem

  1. Nov 4, 2005 #1
    The question I'm having trouble with is as follows:
    Given that siny = 2sinx show that:
    a) (dy/dx)^2 = 1+3sec^2(y), by differentiating this equation with respect to x show that
    b)d^2y/dx^2 = 3sec^2ytany and hence that
    c) coty(d^2y/dx^2) - (dy/dx)^2 + 1 = 0
    Part (c) is straight forward and after a fair bit of work I got (a)...part (b) however is a *big* problem for me.
    my favourite method of trying to solve the problem (because this question doesn't come anywhere close to the 4 examples the book has shown up to now) is to firstly to find the square root of both sides of the equation to get back to dy/dx and then differentiate working on the principle that this is the square root of a quotient.

    dy/dx = sqrt(1+ 3sec^(2)y)...
    (y'') = 1/2(y')(6cosysiny/cos^4(y))(1 + 3sec^2(y))^(1/2)...
    (y'') = (y')(3sec^ytany)(1 + 3sec^2)^(1/2)
    I am stuck here with the bit that I want nested within rubbish :frown:
    can someone please show me what I should have done to reach the correct answer?
     
    Last edited: Nov 4, 2005
  2. jcsd
  3. Nov 4, 2005 #2

    Fermat

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    I found it easier to just differentiate (implicitly) twice.
     
  4. Nov 4, 2005 #3
    Thanks for the reply fermat though I'm still having problems! :confused:
    If I just differentiate twice the equation siny = 2sinx...
    differentiating once...cosy(y') = 2cosx and therefore (y') = 2cosx/cosy
    differentiating twice...(y')(-siny) + (y'')cosy = -2sinx
    by substituting for (y') and using 2sinx = siny I get...
    (y'')cosy = 2cosxsiny/cosy - siny...
    (y'') = 2cosxsiny/cos^2(y) - tany...and I am getting nowhere :frown:

    I can differentiate 1+3sec^2(y) neither once or twice implicitly because 1+3sec^2(y) is the square of (y') and so I would not get (y'') by doing this...that was my reason for taking the square root of both sides, and then differentiating :frown:

    As with many of the questions I have dealt with before this, it feels like its a massive jump from what has been explained in the book...and studying in my spare time with no one in my circle of friends, family, or colleagues to help me I have to just *infer* certain rules that have not been covered or even touched upon...more often than not I can fill in the gaps but here I'm struggling.
     
    Last edited: Nov 4, 2005
  5. Nov 4, 2005 #4

    Fermat

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    There's the error!
    When you differentiate cosy(y'), you should get,

    (-siny*.y')*y' + cosy*y''
     
  6. Nov 4, 2005 #5

    Fermat

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    I'm not too sure I follow your reasoning here, but if you differentiate (y')² you get,
    2y'*y''
    dunno if that would help though :confused:
     
  7. Nov 4, 2005 #6
    it does help :smile: (because I overlooked that!), though I still can't see a solution to my problem with it because now I get:
    2(y')(y'') = (y')(3sec^ytany)(1 + 3sec^2)^(1/2)
    2(y'') = (3sec^ytany)(1 + 3sec^2)^(1/2)
     
  8. Nov 4, 2005 #7

    Fermat

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    Well, if

    (y')² = 1 + 3sec²y

    then you should get,

    2y'.y'' = 6secy*secy.tany.y'

    Edit: forgot to add the y' bit.
     
    Last edited: Nov 4, 2005
  9. Nov 4, 2005 #8
    ah...I can see where you are coming from now :smile: but with reference to what you said about my reasoning before...isn't (y')^2 (y')(y')?...but you can express (y'') as (y')(y') also...:confused: but the two are not the same...and by differentiating (1 + sec^2y) I am finding the derivative of a squared derivative, not the second derivative...agh! either I am stupid...the book's author expects me to be at college whilst I use it...or both :frown:

    p.s. noticed the y' bit but knew what you meant without it
     
    Last edited: Nov 4, 2005
  10. Nov 4, 2005 #9

    Fermat

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    Yes.

    Nope!
    y'' is the 2nd derivative. (y')(y'), or (y')² is the square of the 1st derivative.

    [tex]\frac{d^2y}{dx^2} \mbox{ is not = } \left(\frac{dy}{dx}\right)^2[/tex]

    The derivative of a squared (1st) derivative will give you the 2nd derivative.
    (y')² when differentiated gives 2y'*y'', which includes both the 1st derivative and the 2nd derivative.

    You said,

    and I didn't underestand why you wouldn't get y'' for the reason you gave, that 1+3sec²y is the square of y'.

    Anyway, have you solved the problem now ?
     
    Last edited: Nov 4, 2005
  11. Nov 5, 2005 #10
    The problem is solved now :smile: wasn't comfortable with (y')^2 because I had neither seen it or knew what it would yield if differentiated.
    Thankyou for your help :smile:
     
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