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    Could anyone help me with question 4, I've been on it for an hour and got no clue how to do it.
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    (Original post by ibrahim541)
    Could anyone help me with question 4, I've been on it for an hour and got no clue how to do it.
    \sinh (x) = \frac{1}{2} \left( e^x - e^{-x} \right)

    What similarities can you see between y and \sinh (x) \ ?
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    (Original post by Khallil)
    \sinh (x) = \frac{1}{2} \left( e^x - e^{-x} \right)

    What similarities can you see between y and \sinh (x) \ ?
    Y(x) = 2sinh(x) ?
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    (Original post by ibrahim541)
    Y(x) = 2sinh(x) ?
    Exactly

    Do you know how to differentiate hyperbolic functions?
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    (Original post by Khallil)
    Exactly

    Do you know how to differentiate hyperbolic functions?
    I do and after I differentiate it I get dy/dx=2cosh(x) but it wants it in terms of y and I've got no idea how to rearrange the original formula to get that
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    There's a useful identity: \cosh^2 (x) - \sinh^2 (x) = 1
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    (Original post by ibrahim541)
    I do and after I differentiate it I get dy/dx=2cosh(x) but it wants it in terms of y and I've got no idea how to rearrange the original formula to get that
    Ah I see. Have you learnt Osbourne's rule yet?

    It states that to convert any trigonometric identity to a hyperbolic identity, you must change all cos terms into cosh and replace any product of two sin terms with -sinh^2 :

    \cos \to \cosh

    \sin \to \sinh

    \sin \times \sin \to -\sinh^2

    For example:

    \cos^2 x + \sin^2 x = 1 \longrightarrow \cosh^2 x - \sinh^2 x = 1
    \sin^2 x is the product of two \sin x terms so it is replaced by -\sinh^2 x
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    Ahh okay, that helps a lot. so I got dx/dy=1/2(y^2+1)^0.5 can anyone clarify if this is right ?
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    (Original post by ibrahim541)
    ...
    y = 2\sinh x \ \Rightarrow \ \dfrac{dy}{dx} = 2\cosh x

    \cosh^2 x = 1 + \sinh^2x \ \Rightarrow \ \cosh x = \sqrt{1+\underbrace{\sinh^2x}}


    Use the fact that \ \sinh x = \frac{y}{2} \ to simplify the expression for \cosh x
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    (Original post by Khallil)
    y = 2\sinh x \ \Rightarrow \ \dfrac{dy}{dx} = 2\cosh x

    \cosh^2 x = 1 + \sinh^2x \ \Rightarrow \ \cosh x = \sqrt{1+\sinh^2x}

    Use the fact that \ \sinh x = \frac{y}{2} \ to simplify the expression for \cosh x
    Ah so its dx/dy=1/2(1+(y/2)^0.5)^0.5, thanks you so much for your help
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    (Original post by ibrahim541)
    Ah so its dx/dy=1/2(1+(y/2)^0.5)^0.5, thanks you so much for your help
    I think you mean:

    \dfrac{dy}{dx} = 2 \sqrt{ 1 + \left( \frac{y}{2} \right)^2 }
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    (Original post by Khallil)
    I think you mean:

    \dfrac{dy}{dx} = 2 \sqrt{ 1 + \left( \frac{y}{2} \right)^2 }
    Okay got it, thanks again
 
 
 
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