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Math Help - differention

  1. #1
    Member srirahulan's Avatar
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    Arrow differention

    If ,y= \ln{secx}, |x|< \frac{\pi}{2} show that \frac{d^2y}{dx^2}=e^{2y}
    Last edited by srirahulan; July 6th 2013 at 07:35 PM.
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  2. #2
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    Re: differention

    It doesn't...

    \displaystyle \begin{align*} y &= \ln{ \left[ \sec{(x)} \right] } \\ &= \ln{ \left\{ \left[ \cos{(x)} \right] ^{-1} \right\} } \\ &= -\ln{ \left[ \cos{(x)} \right] }  \end{align*}

    Now let \displaystyle \begin{align*} u = \cos{(x)} \implies y = -\ln{(u)} \end{align*}, then

    \displaystyle \begin{align*} \frac{du}{dx} &= -\sin{(x)} \\ \\ \frac{dy}{du} &= -\frac{1}{u} \\ &= -\frac{1}{\cos{(x)}} \\ \\ \frac{dy}{dx} &= -\sin{(x)} \left[ -\frac{1}{\cos{(x)}} \right] \\ &= \frac{\sin{(x)}}{\cos{(x)}} \\ \\ \frac{d^2y}{dx^2} &= \frac{\cos{(x)}\cos{(x)} - \sin{(x)} \left[ -\sin{(x)} \right] }{\left[ \cos{(x)} \right] ^2} \\ &= \frac{\cos^2{(x)} + \sin^2{(x)}}{\cos^2{(x)}} \\ &= \frac{1}{\cos^2{(x)}} \\ &= \sec^2{(x)} \end{align*}

    Now remembering that \displaystyle \begin{align*} y = \ln{ \left[ \sec{(x)} \right] } \end{align*} that means \displaystyle \begin{align*} \sec{(x)} = e^y \end{align*}, and so \displaystyle \begin{align*} \frac{d^2y}{dx^2} = e^y \end{align*}, NOT \displaystyle \begin{align*} e^{2y} \end{align*}.
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  3. #3
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    Re: differention

    Hello, srirahulan!

    \text{If }\,y\:=\:\ln(\sec x),\;|x|< \tfrac{\pi}{2},\;\text{ show that: }\:\frac{d^2y}{dx^2}\:=\:e^{2y}

    We have: . y \:=\:\ln(\sec x)\;\;[1]

    Then: . \frac{dy}{dx} \:=\:\frac{1}{\sec x}(\sec x\tan x) \:=\:\tan x

    Hence: . \frac{d^2y}{dx^2} \:=\:\sec^2\!x\;\;[2]


    From [1], we have: . y \:=\:\ln(\sec x) \quad\Rightarrow\quad \sec x \:=\:e^y

    Substitute into [2]: . \frac{d^2y}{dx^2} \:=\:(e^y)^2

    . . . . . . Therefore: . \frac{d^2y}{dx^2} \:=\:e^{2y}
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  4. #4
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    Re: differention

    It appears I am unable to edit my previous post. I for some reason forgot that the second derivative is \displaystyle \begin{align*} \sec^2{(x)} \end{align*}, not \displaystyle \begin{align*} \sec{(x)} \end{align*}.
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  5. #5
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    Re: differention

    Another way is:

    We can write y = \ln(\sec x) as e^y = \sec x.

    Now differentiate both sides:

    e^y \frac{dy}{dx}= \sec x \tan x = e^y \tan x

    Cancelling e^y on both sides, we get \frac{dy}{dx} = \tan x.

    Differentiating this equation we get,

    \frac{d^2y}{dx^2} = \sec^2 x = (e^y)^2 = e^{2y}.
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