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Homework Help: Tricky Trig Differentiation

  1. May 30, 2008 #1
    1. The problem statement, all variables and given/known data

    x = 5sin(2y +6) find dy in terms of x

    2. Relevant equations

    3. The attempt at a solution

    dx/dy = 8 cos (2y+6)

    dy/dx = 1 / (8cos(2y+6)

    but according to the mark scheme the final answer is

    + or - 1 / (2 sqrt[16/x^2])

    i dont see how on earth they got there :O

    Can anyone? Thanks :)
  2. jcsd
  3. May 30, 2008 #2


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    The result they give is more directly found by inverting the original function so that it becomes an inverse-sine function: that would account for the 1/sqrt form of their answer.

    As for your method, it may be better to use implicit differentiation to solve for dy/dx, rather than finding dx/dy and using the reciprocal. BTW, I don't see how you get an '8' there -- wouldn't the Chain Rule give you 10?
  4. May 30, 2008 #3
  5. May 30, 2008 #4


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    Having worked this through now, I believe you have a typo here:
    This approach:
    does work and gives the same result as implicit differentiation, but only because your function f(y) is continuous in y everywhere. Such a method using reciprocal derivatives should be used with some caution...

    How they get their answer is by a substitution. You know that sin(2y+6) = (x/4). How would you express cos(2y+6) in terms of x? (And I believe you also have a typo in your copy of their answer that you presented...)
  6. May 30, 2008 #5
    yes yes your right

    x = 4sin(2y +6) find dy in terms of x

    + or - 1 / (2 sqrt[16 - x^2])

    substitution... ohh that sounds like a better method. how do i do that here?

    y = (sin^-1(x/4) - 6)/2

    is that right? What do i subistute in?
  7. May 30, 2008 #6


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    I don't know what stage you are at in rules of differentiation. Have you done derivatives of inverse trig functions? The relevant rule is

    d/dx [sin^-1 (u)] = [ 1 / sqrt( 1 - u^2 ) ] ยท (du/dx) .

    For our function, u = x/4 .

    Alternatively, you can use your result

    dy/dx = 1 / [8 cos(2y+6)] ,

    together with

    sin(2y + 6) = (x/4) and

    sin u = sqrt( 1 - u^2 ) ,

    to get to the given answer.
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