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Math Help - derivative ques

  1. #1
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    derivative ques

    <br />
y = x\sqrt{x^2+1}<br />


    <br />
f (x) = x, <br />
g (x) = \sqrt{x^2 + 1}<br />


    <br />
f '(x) = 1, <br />
g '(x) = \frac{x}{\sqrt{x^2 + 1}}<br />

    using the product rule:

    <br /> <br />
= (\sqrt{x^2 + 1}) + x(\frac{x}{\sqrt{x^2 + 1}})<br /> <br />

    this correcT? thank you
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  2. #2
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    Quote Originally Posted by jvignacio View Post
    <br /> <br />
y'(x) = \sqrt{x^2 + 1} + x\left(\frac{x}{\sqrt{x^2 + 1}}\right)<br /> <br />

    this correcT? thank you
    Yes it is.

    Laurent.
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  3. #3
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    Quote Originally Posted by jvignacio View Post
    <br />
y = x\sqrt{x^2+1}<br />


    <br />
f (x) = x, <br />
g (x) = \sqrt{x^2 + 1}<br />


    <br />
f '(x) = 1, <br />
g '(x) = \frac{x}{\sqrt{x^2 + 1}}<br />

    using the product rule:

    <br /> <br />
= (\sqrt{x^2 + 1}) + x(\frac{x}{\sqrt{x^2 + 1}})<br /> <br />

    this correcT? thank you
    Yes, it is correct, but it can be further solved. See here,

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

     = \frac{x^2 + 1 + x^2}{\sqrt{x^2 + 1}}

     = \frac{2x^2 + 1}{\sqrt{x^2 + 1}}
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  4. #4
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    Quote Originally Posted by Shyam View Post
    Yes, it is correct, but it can be further solved. See here,

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

     = \frac{x^2 + 1 + x^2}{\sqrt{x^2 + 1}}

     = \frac{2x^2 + 1}{\sqrt{x^2 + 1}}
    thanks u!!!
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  5. #5
    Super Member 11rdc11's Avatar
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    You may not have learned this method yet but I think it is easier than remembering the product and quotient rules sometimes, logarithmic differentiation. If you would like me to show you more examples I be happy too.



    ln y = ln (x\sqrt{x^2+1})

    ln y = lnx + ln\sqrt{x^2+1}

    ln y = lnx +\frac{lnx^2+1}{2}

    now take the derivative

    \frac{y'}{y} = \frac{1}{x} + \bigg(\frac{1}{2}\bigg)\bigg<br />
(\frac{2x}{x^2+1}\bigg)

    \frac{y'}{y} = \frac{x^2+1+x^2}{x^3+x} = \frac{2x^2 + 1}{x^3 + x}

    y' = \bigg(y\bigg)\bigg(\frac{2x^2 + 1}{x^3 + x}\bigg)

    y' = \bigg(\frac{x\sqrt{x^2+1}}{1}\bigg)\bigg(\frac{2x^  2 + 1}{x^3 + x}\bigg) = \frac{(\sqrt{x^2+1})(2x^2+1)}{x^2+1} = \frac{2x^2+1}{\sqrt{x^2+1}}

    So

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