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Thread: Derivative

  1. #1
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    Derivative

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

    $\displaystyle 0 \leq y < \pi $

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

    $\displaystyle \frac{dy}{dx} = \frac{1}{ \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}}
    $

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

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

    $\displaystyle sec^{2}y - 1 = tan^{2}y $

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

    $\displaystyle 2x = secy $

    $\displaystyle (2x)^{2} = sec^{2}y $

    $\displaystyle 4x^{2} = sec^{2}y $

    Can I rearrange and replace it like this?

    So I'll get, ;

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

    cancel the two's and replace sec^2y

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

    so after simplifying I get, $\displaystyle \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; Jan 17th 2010 at 04:44 AM.
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  2. #2
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    Quote Originally Posted by Tweety View Post
    $\displaystyle \frac{dx}{dy} = \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2} $

    $\displaystyle \frac{dy}{dx} = \frac{1}{ \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}}
    $

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

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

    $\displaystyle sec^{2}y - 1 = tan^{2}y $

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

    $\displaystyle 2x = secy $

    $\displaystyle (2x)^{2} = sec^{2}y $

    $\displaystyle 4x^{2} = sec^{2}y $

    Can I rearrange and replace it like this?

    So I'll get, ;

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

    cancel the two's and replace sec^2y

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

    so after simplifying I get, $\displaystyle \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:

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

    and therefore

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

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

    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
    $\displaystyle \frac{dx}{dy} = \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2} $

    $\displaystyle \frac{dy}{dx} = \frac{1}{ \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}}
    $

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

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

    $\displaystyle sec^{2}y - 1 = tan^{2}y $

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

    $\displaystyle 2x = secy $

    $\displaystyle (2x)^{2} = sec^{2}y $

    $\displaystyle 4x^{2} = sec^{2}y $

    Can I rearrange and replace it like this?

    So I'll get, ;

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

    cancel the two's and replace sec^2y

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

    so after simplifying I get, $\displaystyle \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:
    $\displaystyle sec^2(\theta) - 1 = tan^2(\theta)$
    For any $\displaystyle \theta$.
    substitute $\displaystyle \theta=\frac{y}{2}$
    $\displaystyle sec^2(\frac{y}{2}) - 1 = tan^2(\frac{y}{2})$
    since $\displaystyle sec(\frac{y}{2}) = x$, then the last one can be written as:
    $\displaystyle x^2 - 1 = tan^2(\frac{y}{2}) $
    Hence:
    $\displaystyle tan(\frac{y}{2})=\sqrt{x^2-1}$
    Not substitute this in $\displaystyle \frac{dy}{dx} = \frac{1}{ \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}}$

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


    Quote Originally Posted by Tweety View Post
    $\displaystyle \color{red} 2x = secy $
    This is wrong.
    You probably multiply this equation $\displaystyle x=sec(\frac{y}{2})$ 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
    $\displaystyle \frac{dx}{dy} = \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2} $

    $\displaystyle \frac{dy}{dx} = \frac{1}{ \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}}
    $

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


    Thank you!

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

    is not true since
    you have this

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

    not this

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

    first step is true

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

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

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

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

    $\displaystyle \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

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