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Homework Help: Need help with differentiation

  1. Nov 2, 2005 #1
    Need help with implicit differentiation

    I have only just been introduced to implicit differentiation and am cluelessly stuck on this question:
    Express d^2y/dx^2 as a function of x if siny + cosy = x
    my first attempt was just to simply differentiate each term, and ended up with -(siny+cosy)d^2y/dx^2 = 0. The answer in the back of the book is: x/(2-x^2)^(3/2)...which unlike my answer is a function of x...my problem is that I really don't know how to arrive at this answer.
    I can't think of any trig identity that can help me here and so my best guess is that I should find the three sides of a triangle with which siny and cosy make x, but I have no idea about how to go about finding these sides with the information I have been given.
    part of the denominator: sqrt(2-x^2) seems like the hypoteneuse but I am having difficulty coming up with the other two sides and to make x. I apologise if solving this should be a no-brainer but I simply don't know how to proceed. Can someone please point me in the right direction?
    Last edited: Nov 2, 2005
  2. jcsd
  3. Nov 2, 2005 #2


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    Then I suspect that the answer in the back of the book is wrong!

    sin y+ cos y= x so, by implicit differentiation, cos y y'- sin y y'= 1 or
    y'(cos y- sin y)= 1.

    Differentiate again: y" (cos y- sin y)+ y'(-sin y- cosy)= 0 which is the same as y" (cos y- sin y)= (sin y+ cos y)y'= xy' (since sin y+ cos y= x).

    But y'(cos - sin y)= 1 means that y'= 1/(cosy -sin y) so

    y"(cos y- sin y)= x/(cos y- sin y)

    y"= x/(cos y- sin y)2= x/(cos2[/sup y- 2sin ycosy+ cos2 y)= x/(1- 2sin y cos y)

    Now, we know that x= sin y+ cos y so x2= (sin y+ cos y)2= sin2+ 2sin y cos y+ cos2 y= 1+ 2 sin y cos y.
    1- 2 siny cos y= 2- (1+ 2 sin y cos y)= 2- x2.

    The correct answer is just [tex]y"= \frac{x}{2-x^2}[/tex]. There is no "3/2" power.
  4. Nov 2, 2005 #3
    That was brilliant HallsofIvy...Thankyou:smile:
  5. Nov 3, 2005 #4
    The curve y1 is y1=ax^2+ax+b.The curve y2 is y2=cx-x^2.They both have
    common tangent the line which passes through the ponit (1,0).Find the a,b,c.
    Please somebody help me.Thank you.
  6. Nov 3, 2005 #5


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    1. Do not "hijack" someone else's thread to ask a completely new question- start your own thread.

    2. Show us what you have done, what you have attempted.

    You know that any line passing through (1,0) must be of the form y= m(x-1).
    What are the conditions on m so that it is tangent to both curves (at some points on those curves)?
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