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Math Help - If y= log {x + (1 = x^2)^1/2} what is dy/dx?

  1. #1
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    If y= log {x + (1 + x^2)^1/2} what is dy/dx?

    I hope someone can help me with this problem:

    If y= log {x + (1 + x^2)^1/2} what is dy/dx?

    apprantly the answer is dy/dx = (1 + x^2)^-1/2 , but I cannot see how to do it. Is it a function of a function? My lecturer showed us this on the board - but I still can't understand - so I would appreciate it if someone could help - thank you.
    Last edited by goldilocks; September 28th 2007 at 11:58 PM. Reason: change of a = to a + - typo
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  2. #2
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    Quote Originally Posted by goldilocks View Post
    I hope someone can help me with this problem:

    If y= log {x + (1 + x^2)^1/2} what is dy/dx?

    apprantly the answer is dy/dx = (1 + x^2)^-1/2 , but I cannot see how to do it. Is it a function of a function? My lecturer showed us this on the board - but I still can't understand - so I would appreciate it if someone could help - thank you.
    Hello,

    from the answer you gave I assume that you mean:

    f(x) = \ln(x+\sqrt{1+x^2}) . Calculate the derivative using chain rule:

    f'(x) = \frac{1}{x+\sqrt{1+x^2}} \cdot \left({1+\frac12 \cdot (1+x^2)^{-\frac12} \cdot 2x}\right) = \frac{1+\frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}

    Now rationalize(?) the denominator by multiplying this fraction by \frac{x-\sqrt{1+x^2}}{x-\sqrt{1+x^2}}. You'll get:

    f'(x)=\frac{1+\frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^  2}} \cdot \frac{x-\sqrt{1+x^2}}{x-\sqrt{1+x^2}} = \frac{x+\frac{x^2}{\sqrt{1+x^2}}-\sqrt{1+x^2} -x}{x^2-1-x^2} = \frac{\frac{x^2-1-x^2}{\sqrt{1+x^2}}}{-1} = \frac{1}{\sqrt{1+x^2}}
    Last edited by earboth; September 29th 2007 at 04:24 AM.
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  3. #3
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    Hello, goldilocks!

    earboth did a lovely job . . . I used a slightly different simplication.


    Differentiate: . y \:= \:\ln\left[x + (1 + x^2)^{\frac{1}{2}}\right]

    Chain Rule: . \frac{dy}{dx}\;=\;\frac{1}{x + (1+x^2)^{\frac{1}{2}}}\cdot\left[1 + \frac{1}{2}(1+x^2)^{-\frac{1}{2}}\cdot2x\right] \;=\;\frac{1 + \frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}


    Multiply top and bottom by \sqrt{1+x^2}

    . . \frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}\cdot\frac{1 + \frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}} \;=\;\frac{\overbrace{\sqrt{1+x^2} + x}^{\text{cancel}}}{\sqrt{1+x^2}\underbrace{\left(  x+\sqrt{1+x^2}\right)}_{\text{cancel}}} \;=\;\frac{1}{\sqrt{1+x^2}}

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  4. #4
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    Great - thank you both, that made it a lot clearer to see it like that. Thank you very much :-)
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  5. #5
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    Quote Originally Posted by goldilocks View Post
    If y= log {x + (1 + x^2)^1/2} what is dy/dx?
    Have you ever hear about hyperbolic functions?, 'cause this is the \mathrm{arcsinh}\,x function.

    So y=\mathrm{arcsinh}\,x\implies\sinh y=x

    Now let's differentiate

    y'\cosh y=1\implies y'=\frac1{\cosh y}

    Since \cosh^2y-\sinh^2y=1, yields

    y'=\frac1{\sqrt{1+x^2}}
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