Results 1 to 12 of 12

Math Help - differenciation

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    152

    differenciation

    1)

    <br />
(x + \frac {1}{x})' = \frac {1}{2}x^{-0.5} - \frac {x^{0.5}}{x}= \frac {1}{2\sqrt{x}}  - \frac {\sqrt{x}}{x}  = \frac{\sqrt{x}}{2x}<br />

    The solution is supposed to be  \frac{\sqrt{x}(x-1)}{2x^2}

    2)
    <br />
(ln \sqrt{\frac{a+x}{a-x}})'  = \frac{a-x}{2(a+x)}<br />

    The solution is supposed to be \frac{a}{a^2-x^2}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    #1) Think of it as x+x^{-1} Use the power rule like normal.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2008
    Posts
    152
    if i use that i get \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by metlx View Post
    if i use that i get \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}}
    You shouldn't. d/dx x = 1 and d/dx x^{-1}=-1(x)^{-2}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    Quote Originally Posted by metlx View Post
    if i use that i get \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}}
    HOW do you get that? Please show your work. The power rule: (x^n)'= nx^{n-1} won't give fractional powers if you don't start with them!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Dec 2008
    Posts
    152
    aww. i just realized I forgot the square root.

    1) is  (\sqrt{x} + \frac{1}{\sqrt{x}})'
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by metlx View Post
    aww. i just realized I forgot the square root.

    1) is  (\sqrt{x} + \frac{1}{\sqrt{x}})'
    Please read the replies you're getting. There should be NO SQUARE ROOTS unless you simplify after taking the derivative.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Dec 2008
    Posts
    152
    no i forgot to add the square root into the initial derivative..

    I wrote (x + \frac {1}{x})' instead of (\sqrt{x} + \frac{1}{\sqrt{x}})'
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Dec 2008
    Posts
    152
    by the way, 2) is no longer needed. Managed to solve it
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Then your first part if right. \frac{d}{dx} \frac{1}{\sqrt{x}}= \frac{d}{dx} x^{-\frac{1}{2}}=-\frac{1}{2}x^{-\frac{3}{2}}
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Dec 2008
    Posts
    152
    I'm guessing that the solution \frac{\sqrt{x}(x-1)}{2x^2} that is in the book is wrong then?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by metlx View Post
    I'm guessing that the solution \frac{\sqrt{x}(x-1)}{2x^2} that is in the book is wrong then?
    \frac{1}{2\sqrt{x}}-\frac{1}{2}x^{-\frac{3}{2}}

    I am busy right now so I can't do this for you, but try to get these two terms into one expression. Then see if the book's answer makes more sense.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Differenciation
    Posted in the Calculus Forum
    Replies: 10
    Last Post: April 9th 2010, 09:16 PM
  2. Differenciation qn
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 6th 2008, 08:23 AM
  3. differenciation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 5th 2008, 12:23 PM
  4. Differenciation Help needed.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 20th 2007, 03:27 PM
  5. integration and differenciation
    Posted in the Calculus Forum
    Replies: 6
    Last Post: January 12th 2007, 08:01 AM

Search Tags


/mathhelpforum @mathhelpforum