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Math Help - Derivative

  1. #1
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    Derivative

    Given that  x = sec\frac{y}{2}

     0 \leq y < \pi

    show that  \frac{dy}{dx} = \frac{2}{x \sqrt{x^{2}-1}}
     \frac{dx}{dy} = \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}

     \frac{dy}{dx} = \frac{1}{ \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}}<br />

     sec y = 2x so replace  \frac{secy}{2} with 2x

    and using the indentity  1 + tan^{2}y = sec^{2}y

     sec^{2}y - 1 = tan^{2}y

     (\frac{tany}{2})^{2} = \frac{sec^{2}y -1}{4}

     2x = secy

     (2x)^{2} = sec^{2}y

     4x^{2} = sec^{2}y

    Can I rearrange and replace it like this?

    So I'll get, ;

     \frac{1}{ \frac{1}{2}\times \frac{2x}{2} \times \frac{sec^{2}y-1}{4}}

    cancel the two's and replace sec^2y

     \frac{1}{\frac{1}{2} \times x \times  \frac{4x^{2}-1}{4}}

    so after simplifying I get,  \frac{1}{\frac{x}{2} \times \frac{4x^{2}-1}{4}}

    Can someone please help me from here, as I have tired to rearrange it but don't get the desired expression.

    And I am not even sure what I have done so far is correct, which is probably why I am not getting the right expression?

    Please check my working,

    Thank you!
    Last edited by Tweety; January 17th 2010 at 04:44 AM.
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  2. #2
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    Quote Originally Posted by Tweety View Post
     \frac{dx}{dy} = \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}

     \frac{dy}{dx} = \frac{1}{ \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}}<br />

     sec y = 2x so replace  \frac{secy}{2} with 2x

    and using the indentity  1 + tan^{2}y = sec^{2}y

     sec^{2}y - 1 = tan^{2}y

     (\frac{tany}{2})^{2} = \frac{sec^{2}y -1}{4}

     2x = secy

     (2x)^{2} = sec^{2}y

     4x^{2} = sec^{2}y

    Can I rearrange and replace it like this?

    So I'll get, ;

     \frac{1}{ \frac{1}{2}\times \frac{2x}{2} \times \frac{sec^{2}y-1}{4}}

    cancel the two's and replace sec^2y

     \frac{1}{\frac{1}{2} \times x \times  \frac{4x^{2}-1}{4}}

    so after simplifying I get,  \frac{1}{\frac{x}{2} \times \frac{4x^{2}-1}{4}}

    Can someone please help me from here, as I have tired to rearrange it but don't get the desired expression.

    And I am not even sure what I have done so far is correct, which is probably why I am not getting the right expression?

    Please check my working,

    Thank you!
    My attempt:

    x=\sec\left(\frac12 y\right)~\implies~\cos\left(\frac12 y\right)=\frac1x

    and therefore

    y = 2\arccos\left(\frac1x\right)

    \frac{dy}{dx}=2\left(-\dfrac{1}{\sqrt{1-\left(\frac1x \right)^2}}  \right) \cdot \left(-\frac1{x^2}\right) = 2\left(\dfrac{1}{\frac1x \cdot \sqrt{x^2-1}}  \right) \cdot \left(\frac1{x^2}\right)

    which simplifies to the given result.
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  3. #3
    Super Member General's Avatar
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    Quote Originally Posted by Tweety View Post
     \frac{dx}{dy} = \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}

     \frac{dy}{dx} = \frac{1}{ \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}}<br />

     sec y = 2x so replace  \frac{secy}{2} with 2x

    and using the indentity  1 + tan^{2}y = sec^{2}y

     sec^{2}y - 1 = tan^{2}y

     (\frac{tany}{2})^{2} = \frac{sec^{2}y -1}{4}

     2x = secy

     (2x)^{2} = sec^{2}y

     4x^{2} = sec^{2}y

    Can I rearrange and replace it like this?

    So I'll get, ;

     \frac{1}{ \frac{1}{2}\times \frac{2x}{2} \times \frac{sec^{2}y-1}{4}}

    cancel the two's and replace sec^2y

     \frac{1}{\frac{1}{2} \times x \times \frac{4x^{2}-1}{4}}

    so after simplifying I get,  \frac{1}{\frac{x}{2} \times \frac{4x^{2}-1}{4}}

    Can someone please help me from here, as I have tired to rearrange it but don't get the desired expression.

    And I am not even sure what I have done so far is correct, which is probably why I am not getting the right expression?

    Please check my working,

    Thank you!
    The identity is:
    sec^2(\theta) - 1 = tan^2(\theta)
    For any \theta.
    substitute \theta=\frac{y}{2}
    sec^2(\frac{y}{2}) - 1 = tan^2(\frac{y}{2})
    since sec(\frac{y}{2}) = x, then the last one can be written as:
     x^2 - 1 = tan^2(\frac{y}{2})
    Hence:
    tan(\frac{y}{2})=\sqrt{x^2-1}
    Not substitute this in \frac{dy}{dx} = \frac{1}{ \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}}

    ------------


    Quote Originally Posted by Tweety View Post
     \color{red} 2x = secy
    This is wrong.
    You probably multiply this equation x=sec(\frac{y}{2}) by 2, and you canceled the 2`s from the right side.
    Which is wrong, one of this 2`s is inside the angle of a trigonometric function.
    You can not cancel it with another 2 which outside the angle.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Tweety View Post
     \frac{dx}{dy} = \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}

     \frac{dy}{dx} = \frac{1}{ \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}}<br />

     sec y = 2x so replace  \frac{secy}{2} with 2x


    Thank you!

    this  sec y = 2x so replace  \frac{secy}{2} with 2x

    is not true since
    you have this

    \sec \frac{y}{2} = x

    not this

    \frac{\sec y}{2} = x

    first step is true

    \frac{dy}{dx} = \frac{2}{\sec \frac{y}{2} \tan \frac{y}{2}}

    want to find tan y/2 with respect to x
    you have

    \sec \frac{y}{2} = x

    \sec ^2 \frac{y}{2} = x^2

    \tan ^2 \frac{y}{2} + 1 = x^2 \Rightarrow \tan ^2 \frac{y}{2} = x^2 -1 \Rightarrow \tan \frac{y}{2} = \sqrt{x^2-1}

    sub sec y/2 and tan y/2

    \frac{dy}{dx} = \frac{2}{x\sqrt{x^2-1}}
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  5. #5
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    Thanks alot, I understand where I went wrong.

    Just to clarify, writting  sec\frac{y}{2} is not the same as writting  \frac{secy}{2} ?
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  6. #6
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    Quote Originally Posted by Tweety View Post
    Thanks alot, I understand where I went wrong.

    Just to clarify, writting  sec\frac{y}{2} is not the same as writting  \frac{secy}{2} ?
    In general no. To write unambiguously use brackets. Compare my answer.
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