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Thread: Another Hyperbolic function

  1. #1
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    Another Hyperbolic function

    Given that, $\displaystyle sinh \ y = x $, show that $\displaystyle y = ln [x+(1+x^2)^\frac{1}{2}] $
    By differentiating this result, show that

    $\displaystyle (1+x^2)(\frac {dy}{dx})^2 = 1$


    Iíve done first part, but how can I differentiating this result?
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  2. #2
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    Hello,

    Well, i think you got to differentiate $\displaystyle \sinh(\ln(x+(1+x^2)^{1/2}))$, which is sinh(y)=x. This derivate should be 1, because the derivate of x is 1.

    For the derivate of $\displaystyle \sinh(\ln(x+(1+x^2)^{1/2}))$, use the fog derivate.
    When you differentiate sinh, you got cosh. And cosh is $\displaystyle \sqrt{1+\sinh^2}$
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    Quote Originally Posted by Moo View Post
    Hello,

    Well, i think you got to differentiate $\displaystyle \sinh(\ln(x+(1+x^2)^{1/2}))$, which is sinh(y)=x. This derivate should be 1, because the derivate of x is 1.

    For the derivate of $\displaystyle \sinh(\ln(x+(1+x^2)^{1/2}))$, use the fog derivate.
    When you differentiate sinh, you got cosh. And cosh is $\displaystyle \sqrt{1+\sinh^2}$
    I think sinh y = x is for first portion. Can you tell me the way how can I direct differentiate this? I tried by using ln x, dy/dx (xy).... etc. But can't solved yet.
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  4. #4
    Moo
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    Well,

    We got sinh(y)=x.

    So if we differentiate, we have :

    $\displaystyle \frac{dy}{dx} \cosh(y) = \frac{d}{dx}x = 1$ (1)

    (1)≤ will give : $\displaystyle (\frac{dy}{dx})^2 \cosh^2(y) = 1$

    $\displaystyle \cosh^2(y)=1+\sinh^2(y)= 1+x^2$

    And you obtain what you want :-)
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    Quote Originally Posted by Moo View Post
    Well,

    We got sinh(y)=x.

    So if we differentiate, we have :

    $\displaystyle \frac{dy}{dx} \cosh(y) = \frac{d}{dx}x = 1$ (1)

    (1)≤ will give : $\displaystyle (\frac{dy}{dx})^2 \cosh^2(y) = 1$

    $\displaystyle \cosh^2(y)=1+\sinh^2(y)= 1+x^2$

    And you obtain what you want :-)
    Look in my question, there are two parts,

    first, Given that, sinh y = x, show that $\displaystyle y = ln [x+(1+x^2)^\frac{1}{2}] $

    I solved that. No problem.


    Second, By differentiating this result, show that

    $\displaystyle (1+x^2)(\frac {dy}{dx})^2 = 1$


    So why I start from sinh y = x?
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  6. #6
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    Good luck then , i've tried to differentiate the expression of y, and it's far too complicated when the solution is here... I can't even know if i'll be able to get the result they're waiting from you
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  7. #7
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    Quote Originally Posted by geton View Post
    Given that, $\displaystyle sinh \ y = x $, show that $\displaystyle y = ln [x+(1+x^2)^\frac{1}{2}] $
    By differentiating this result, show that

    $\displaystyle (1+x^2)(\frac {dy}{dx})^2 = 1$


    I’ve done first part, but how can I differentiating this result?
    $\displaystyle y = \ln [x+(1+x^2)^{1/2}] $.

    Let $\displaystyle u = x+(1+x^2)^{1/2} = x + \sqrt{1 + x^2}$ so that $\displaystyle y = \ln u$.

    Chain rule: $\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{du}{dx}$.

    To get $\displaystyle \frac{du}{dx}$ you need to differentiate $\displaystyle (1+x^2)^{1/2}$. So .... pretend for a moment that you have $\displaystyle y = (1+x^2)^{1/2}$:

    Let $\displaystyle w = 1 + x^2$ so that $\displaystyle y = w^{1/2}$.

    Chain rule: $\displaystyle \frac{dy}{dx} = \frac{dy}{dw} \cdot \frac{dw}{dx} = \frac{1}{2} \, w^{-1/2} \cdot 2x = \frac{x}{\sqrt{w}} = \frac{x}{\sqrt{1 + x^2}} $.

    So $\displaystyle \frac{du}{dx} = 1 + \frac{x}{\sqrt{1 + x^2}} = \frac{\sqrt{1 + x^2} +x}{\sqrt{1 + x^2}} $.

    Therefore:

    $\displaystyle \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{x+ \sqrt{1+ x^2}} \cdot \frac{\sqrt{1 + x^2} +x}{\sqrt{1 + x^2}}$

    $\displaystyle = \frac{\sqrt{1 + x^2} +x}{(x+ \sqrt{1+ x^2})\sqrt{1 + x^2}} = \frac{1}{\sqrt{1 + x^2}}$.

    It should be all blue sky from here.
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