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Homework Help: Help with 1 quick implicit differentiation problem

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    what is the implicit differentiation ?

    2. Relevant equations

    2sin(x)cos(y)=1


    3. The attempt at a solution

    d/dx[2sin(x)cos(y)]= d/dx[1]

    2cos(x)*-sin(y)*dy/dx=0

    I haev a bad feeling i did this wrong...
     
  2. jcsd
  3. Nov 9, 2008 #2

    Mark44

    Staff: Mentor

    Re: Help with 1 quick implicit differentiation problem ! URGENT

    To differentiate 2sin(x)cos(y), you have to use the product rule. You also need to use the chain rule when you differentiate a function of y (where y is assumed to be implicitly a function of x). l'm pretty sure you realized this, since you ended up with a factor of dy/dx, but you forgot the product rule.

    After you have differentiated both sides, solve algebraically for dy/dx and you're done.
     
  4. Nov 9, 2008 #3
    Re: Help with 1 quick implicit differentiation problem ! URGENT

    I also graphed this using a program:

    23k819f.jpg

    Are these appropriate ranges to find the implicit differentiation? I'm using this picture so it'll be easier for me to find out what the implicit differentiation of this is.

    x and y ranges -6.28 to 6.28 ... are these appropriate figures or can I use a different value that shows it even better ? Thanks in advance.

    Edit: Thanks Matt44, give me a sec !
    Edit#2:
    using the product rule as you suggested (finally makes sense...)
    d/dx[2sin(x)cos(y)]=
    u= 2sin(x) u'=2cos(x) v= cos(y) v'= -sin(y)*dy/dx
    === 2sin(x)*-sin(y)*dy/dx + cos(y) * 2cos(x)
    === Is the answer dy/dx= -cos(y)*2cos(x)/ -2sin(x)*sin(y) ?

    Doesnt look too right ... ><
     
    Last edited: Nov 9, 2008
  5. Nov 9, 2008 #4

    HallsofIvy

    User Avatar
    Science Advisor

    Re: Help with 1 quick implicit differentiation problem ! URGENT

    Cancelling "-2" in numerator and denominator gives dy/dx= cos(x)cos(y)/sin(x)sin(y) and that is exactly right.
     
  6. Nov 9, 2008 #5
    thanks hallofivy ! do you think -6.28 and 6.28 are good ranges for x and y ? or should it become something smaller ?

    Matt44 and HallsOfIvy . u two are always there to answer my problems haha. thx a lot :) !
     
  7. Nov 9, 2008 #6

    HallsofIvy

    User Avatar
    Science Advisor

    What do you mean by "good ranges"? For a graph? Yes, those will do. The larger you make range the more "hills and valleys" you will see.
     
  8. Nov 9, 2008 #7
    sorry for the confusion. yup, thats what i meant . thank you.
     
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