Results 1 to 9 of 9

Math Help - Check some derivatives please.

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    93

    Check some derivatives please.

    Hi:
    Could somebody check this for me please.
    thanks you

    Attached Thumbnails Attached Thumbnails Check some derivatives please.-q1.gif  
    Last edited by mr fantastic; April 5th 2010 at 03:51 AM. Reason: Changed post title
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Hello stealthmaths
    Quote Originally Posted by stealthmaths View Post
    Hi:
    Could somebody check this for me please.
    thanks you

    They look fine to me, except that in (i) you have found \frac{d^2y}{dx^2}.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    Posts
    93
    ahh yes. I put the first derivative instead of the second.
    Thank you Grandad

    ahh wait. I still need to find the second derivative of (ii)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Feb 2010
    Posts
    93
    Hi:
    Could somebody please check the last part of this question please?
    part iib)

    Attached Thumbnails Attached Thumbnails Check some derivatives please.-q1.gif  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Hello stealthmaths
    Quote Originally Posted by stealthmaths View Post
    Hi:
    Could somebody please check the last part of this question please?
    part iib)
    You have a sign wrong. See the attached - should be a minus sign at the point I've indicated.

    I should set it out like this. First notice that it's easier to divide by 3 at the start, and then factorise where you can:


    \tfrac{1}{3}y= \frac{x+1}{x^2-9}


    \Rightarrow \frac13\frac{dy}{dx}= \frac{(x^2-9)(1)-(x+1)(2x)}{(x^2-9)^2}
    =\frac{-x^2-2x-9}{(x^2-9)^2}
    \Rightarrow \frac13\frac{d^2y}{dx^2}= \frac{(x^2-9)^2(-2x-2)-(-x^2-2x-9)2(x^2-9)(2x)}{(x^2-9)^4}
    =2\frac{(x^2-9)^2(-x-1)+(x^2+2x+9)(x^2-9)(2x)}{(x^2-9)^4}

    =2\frac{(x^2-9)(-x-1)+(x^2+2x+9)(2x)}{(x^2-9)^3}


    =2\frac{-x^3-x^2+9x+9+2x^3+4x^2+18x}{(x^2-9)^3}

    \Rightarrow \frac{d^2y}{dx^2}=\frac{6(x^3+3x^2+27x+9)}{(x^2-9)^3}

    But I get things wrong sometimes, too, so check my working!

    Grandad
    Attached Thumbnails Attached Thumbnails Check some derivatives please.-temp.jpg  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Feb 2010
    Posts
    93
    Grandad:
    I'm confused. You divide by 3 and just take the 1/3 out of the problem (on the top line you gave me)?
    Last edited by stealthmaths; April 9th 2010 at 01:11 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Feb 2010
    Posts
    93
    Somehow I doubt you rarely make a mistake Grandad(not unintentionally anyway)
    I have gone through the work and I think it looks fine - except I'm not sure about the last line where you have a 6 - should this not be a 2?
    I have marked it in red:
    Thank you
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Hello stealthmaths

    I'm sure you'll kick yourself. I began by dividing both sides of the equation by 3 to keep the numbers easier - hence the \tfrac13y, the \frac13\frac{dy}{dx} and the \frac13\frac{d^2y}{dx^2}.

    So at the end, I've simply multiplied by 3 to get back to what we want, namely \frac{d^2y}{dx^2}.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Feb 2010
    Posts
    93
    ahh!
    I think that is a kick, slap and punch Grandad.
    You make me stronger

    Thank you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivatives: check my answers please
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 10th 2010, 01:02 PM
  2. Can someone double check these derivatives?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 6th 2010, 01:09 PM
  3. Derivatives CHECK ANSWERS
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 28th 2009, 01:59 AM
  4. Easy Derivatives - Can You Check...?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 2nd 2008, 08:20 PM
  5. Check my Derivatives =)
    Posted in the Calculus Forum
    Replies: 7
    Last Post: March 6th 2007, 09:09 PM

Search Tags


/mathhelpforum @mathhelpforum