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Math Help - differentiation 4

  1. #1
    Senior Member furor celtica's Avatar
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    differentiation 4

    given that siny= 2sinx, show that (dy/dx)^2 = 1 + 3sec^2y
    by differentiating this equation with respect to x show that
    d^2y/dx^2 = 3sec^2ytany
    and hence that coty (d^2y/dx^2) - (dy/dx)^2 + 1 = 0
    i havent gotten past the first problem:
    differentiating siny=2sinx i get dy/dx= 2cosx/cosy
    squaring this i get (dy/dx)^2 = 4cos^2/cos^2y; from here i use the first given statement and double angle identities to get a variety of results such as (4 - siny) x sec^2y but never 1+3sec^2y
    can someone please show me exactly how to get to this result?
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  2. #2
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    Quote Originally Posted by furor celtica View Post
    given that siny= 2sinx, show that (dy/dx)^2 = 1 + 3sec^2y
    by differentiating this equation with respect to x show that
    d^2y/dx^2 = 3sec^2ytany
    and hence that coty (d^2y/dx^2) - (dy/dx)^2 + 1 = 0
    i havent gotten past the first problem:
    differentiating siny=2sinx i get dy/dx= 2cosx/cosy
    squaring this i get (dy/dx)^2 = 4cos^2/cos^2y; from here i use the first given statement and double angle identities to get a variety of results such as (4 - siny) x sec^2y but never 1+3sec^2y
    can someone please show me exactly how to get to this result?
    You have a good start, you have [tex](dy/dx)^2= \frac{4 cos^2(x)}{cos^2(y)}. Now, to replace that "cos(x)" with y you need to get back to "sin(x)". Use cos^2(x)= 1- sin^2(x) to get (dy/dx)^2= \frac{4(1- sin^2(x))}{cos^2(y)}= \frac{4- 4 sin^2(x)}{cos^2(y)} = \frac{4- sin^2(y)}{cos^2(y)}= \frac{4}{cos^2(y)}- \frac{sin^2(y)}{cos^2(y)} = 4sec^2(y)- tan^2(y).

    Now use the fact that tan^2(y)= sec^2(y)- 1.
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  3. #3
    Senior Member furor celtica's Avatar
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    "Now use the fact that ."
    i wasnt aware of this identity, can you show me how it is developed, either from basic principles or another basic identity?
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  4. #4
    Senior Member furor celtica's Avatar
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    how am i supposed to solve the second problem?
    i differentiated and got
    3secytany/sqrt(3secy+1)
    does this mean that i have to prove sqrt(3secy+1) = 1?
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  5. #5
    Member Em Yeu Anh's Avatar
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    Quote Originally Posted by furor celtica View Post
    "Now use the fact that ."
    i wasnt aware of this identity, can you show me how it is developed, either from basic principles or another basic identity?
    You are aware of sin^2x + cos^2x = 1, I presume? It's a very common identity.
    Divide all terms by cos^2x:

    \frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x}

    tan^2x + 1 = sec^2x which is equivalent to tan^2x = sec^2x - 1
    Last edited by Em Yeu Anh; November 26th 2009 at 08:41 PM. Reason: LaTex errors
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  6. #6
    Senior Member furor celtica's Avatar
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    can you help me with the second problem?
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    Quote Originally Posted by furor celtica View Post
    can you help me with the second problem?
    The way you formatted your original post makes it difficult to discern exactly what the second problem is, which would explain the lack of further replies.
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  8. #8
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    Quote Originally Posted by furor celtica View Post
    given that siny= 2sinx, show that (dy/dx)^2 = 1 + 3sec^2y
    By "second problem" I presume you mean

    by differentiating this equation with respect to x show that
    d^2y/dx^2 = 3sec^2ytany
    and hence that coty (d^2y/dx^2) - (dy/dx)^2 + 1 = 0
    Perhaps the reason you are not getting responses to that is that it is wrong!

    Differentiating (dy/dx)^2= 1+ 3sec^2 y we get 2(dy/dx)(d^2y/dx^2)= 6 sec y (sec y tan y)= 6 sec^2 y tan y so that (dy/dx)(d^2y/dx^2)= 3 sec^2 y tan y. But, since dy/dx is not 1, that is NOT d^2y/dx^2= 3 sec^2 y tan y.
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