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Thread: derivative ques

  1. #1
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    derivative ques

    $\displaystyle
    y = x\sqrt{x^2+1}
    $


    $\displaystyle
    f (x) = x,
    g (x) = \sqrt{x^2 + 1}
    $


    $\displaystyle
    f '(x) = 1,
    g '(x) = \frac{x}{\sqrt{x^2 + 1}}
    $

    using the product rule:

    $\displaystyle

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

    $

    this correcT? thank you
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  2. #2
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    Quote Originally Posted by jvignacio View Post
    $\displaystyle

    y'(x) = \sqrt{x^2 + 1} + x\left(\frac{x}{\sqrt{x^2 + 1}}\right)

    $

    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
    $\displaystyle
    y = x\sqrt{x^2+1}
    $


    $\displaystyle
    f (x) = x,
    g (x) = \sqrt{x^2 + 1}
    $


    $\displaystyle
    f '(x) = 1,
    g '(x) = \frac{x}{\sqrt{x^2 + 1}}
    $

    using the product rule:

    $\displaystyle

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

    $

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

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

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

    $\displaystyle = \frac{2x^2 + 1}{\sqrt{x^2 + 1}}$
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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,

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

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

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



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

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

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

    now take the derivative

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

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

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

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

    So

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