Results 1 to 5 of 5

Math Help - derivatives...blah

  1. #1
    Banned
    Joined
    Oct 2007
    Posts
    45

    derivatives...blah

    Find the derivative of the function below.


    how would i do this one?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by mathaction View Post
    Find the derivative of the function below.


    how would i do this one?
    Hello,

    first rearrange the equation:

    g(w) = w^{3.3} \cdot 7^{-w}

    You are supposed to know that 7^{-w} = e^{-\ln(7) \cdot w}

    g(w) = w^{3.3} \cdot e^{-\ln(7) \cdot w}. Now use product rule to derivate this function:

    g'(w) = e^{-\ln(7) \cdot w} \cdot 3.3 \cdot w^{2.3} + w^{3.3} \cdot e^{-\ln(7) \cdot w} \cdot (-\ln(7)) . Factor this term: The common factor is e^{-\ln(7) \cdot w} = 7^{-w}

    g'(w) = 7^{-w} \cdot \left(  3.3 \cdot w^{2.3} + w^{3.3} \cdot  (\ln(7) \right) = 7^{-w} \cdot w^{2.3} \left(  3.3  + w \cdot  (\ln(7) \right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2007
    Posts
    45
    thanks...what about these two...



    If f(x) = (7x + 8)(6x - 5), find f '(x) and f ''(x).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by mathaction View Post
    thanks...what about these two...



    If f(x) = (7x + 8)(6x - 5), find f '(x) and f ''(x).
    hi,

    1. rearrange and use product rule:

    y(t) = \sqrt{t} \cdot (t^2+9)

    y'(t)=(t^2+9) \cdot \frac12 \cdot t^{-\frac12} + \sqrt{t} \cdot 2t = \frac{(t^2+9) + 4t^2}{2\sqrt{t}} = \frac{5t^2+9}{2\sqrt{t}}

    2. Don't expand the brackets.

    f(x) = (7x+8)(6x-5)~\overrightarrow{\text{ derivate }} f'(x) = 7(6x-5) + 6(7x+8)~\overrightarrow{\text{ derivate }} f''(x) = 7 \cdot 6+6 \cdot 7

    3. Do yourself and do us a favour: If you have a new question start a new thread because nobody will notice "from outside this thread" that you have an other problem to solve.
    Last edited by earboth; October 2nd 2007 at 06:24 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,802
    Thanks
    692
    Hello, mathaction!

    y(t) \;=\;\frac{t^{\frac{1}{2}}}{t^2+9}
    Evidently, earboth despises the Quotient Rule . . .

    We have: . y'(t)\;=\;\frac{(t^2+9)\cdot\frac{1}{2}t^{\text{-}\frac{1}{2}} \,- \,t^{\frac{1}{2}}\cdot2t}{(t^2+9)^2} \;=\;\frac{\frac{1}{2}t^{\text{-}\frac{1}{2}}(t^2+9) \,- \,2t^{\frac{3}{2}}}{(t^2+9)^2}

    Multiply top and bottom by 2t^{\frac{1}{2}}\!:\;\;y'(t)\;=\;\frac{(t^2+9) - 4t^2}{2t^{\frac{1}{2}}(t^2+9)}

    Therefore: . y'(t)\;=\;\frac{9-3t^2}{2\sqrt{t}\,(t^2+9)^2}



    If f(x) \:= \:(7x + 8)(6x - 5), .find f'(x) and f''(x)

    With such a simple product, I would multiply it out . . .

    \begin{array}{ccc}f(x) & = & 42x^2 + 13x - 40 \\<br />
f'(x) & = & 84x + 13 \\ f''(x) & = & 84 \end{array}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. property of zero blah
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 28th 2009, 06:20 AM
  2. Replies: 4
    Last Post: February 10th 2009, 09:54 PM
  3. Angles and Blah...
    Posted in the Geometry Forum
    Replies: 2
    Last Post: March 6th 2008, 12:38 PM
  4. Volume :S blah
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 3rd 2007, 04:51 AM
  5. More differentiation... blah
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 16th 2007, 12:13 PM

Search Tags


/mathhelpforum @mathhelpforum