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Math Help - derivative

  1. #1
    Faz
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    derivative

    Find the derivatives of the follwoing functions :

    a) y= In(x + squareroot of (x^2 + 1));

    b) y = x * Squareroot of x * (3In x -2)
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  2. #2
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    Quote Originally Posted by Faz View Post
    Find the derivatives of the follwoing functions :

    a) y= In(x + squareroot of (x^2 + 1));
    a) y = \ln (x+ \sqrt{x^2+1})

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

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

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

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

    Did you get it now???
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  3. #3
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    Quote Originally Posted by Faz View Post
    Find the derivatives of the follwoing functions :

    b) y = x * Squareroot of x * (3In x -2)
    Is your question like this ??

    y=x\sqrt {x} (3\ln x -2)<br />
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  4. #4
    Faz
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    yes

    yes very correct
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Faz View Post
    find derivatives of function

    Note that x\sqrt{x}=x^{\frac{3}{2}}

    Thus, apply the product rule to x^{\frac{3}{2}}\left(3\ln(x)-2\right). You should get \frac{d}{\,dx}\left[x^{\frac{3}{2}}\right]\left(3\ln(x)-2\right)+x^{\frac{3}{2}}\cdot\frac{d}{\,dx}\left[3\ln(x)-2\right]=\dots
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  6. #6
    Senior Member vincisonfire's Avatar
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    \frac{dy}{dx}=\sqrt{x}(3ln(x)-2)+\frac{x(3ln(x)-2)}{2\sqrt{x}}+\frac{3x\sqrt{x}}{x}
    Remember \frac{d(uv)}{dx}=u\frac{d(v)}{dx}+ v\frac{d(u)}{dx}
    Last edited by mr fantastic; December 7th 2008 at 04:42 AM. Reason: No edit - I just wanted to flag this post as a (moved) reply to questions that had been re-posted after replies were given
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  7. #7
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    Quote Originally Posted by Faz View Post
    how do i solve this ?

    y=x*squareroot of X(3lnx-2)
    Hi man,

    why is there a big X?

    y = x * \sqrt{x*(3ln(x)-2)}

    u:= x

    u '=1

    v:= \sqrt{x*(3ln(x)-2)} = (x*(3ln(x)-2))^0.5

    a:= x
    b:= 3ln(x)-2
    a' = 1
    b' = 3/x

    => v' = 0.5*[x*3/x + 1*(3ln(x)-2)]^{-0.5}

    = 0.5 * [3 + 3ln(x) -2]^{-0.5} = 0.5* [3*ln(x)+1]^{-0.5}

    => y' = u*v' + v' * u (Product rule)

    Can you solve the rest? (I already told you u, u', v, v'). I guess you can

    Regards
    Rapha
    Last edited by mr fantastic; December 7th 2008 at 04:49 AM. Reason: No edit - I just wanted to flag this post as a (moved) reply to questions that had been re-posted after replies were given
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Faz View Post
    y=x*squareroot of x * (3lnx-2)
    The question is:

    Find the derivative of:

    y=x \sqrt{x}(3 \ln(x)-2)

    This involves a certain amount of guess work about where to put the brackets.

    Rewrite this as:

    y=x^{3/2}(3 \ln(x)-2)

    Then we need the product rule:

     <br />
\frac{dy}{dx}=\left[ \frac{d}{dx}x^{3/2}\right](3 \ln(x)-2)+x^{3/2}\left[\frac{d}{dx}(3 \ln(x)-2)\right]<br />

    CB
    Last edited by mr fantastic; December 7th 2008 at 04:49 AM. Reason: No edit - I just wanted to flag this post as a (moved) reply to questions that had been re-posted after replies were given
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