Results 1 to 7 of 7

Math Help - More differentiation... blah

  1. #1
    Junior Member
    Joined
    Mar 2007
    Posts
    38

    More differentiation... blah

    I hate to post a bunch of questions... just cuz I'm lost... but I guess I'll post the questions and what I got and whoever can help me out with whatever they can help me out with would be nice

    I was supposed to solve these by logarithmic differentiation...

    y= (lnx)^(2x)

    i got dy/dx = (lnx)^(2x) * 2((1+lnx*ln(lnx))/lnx)

    this next one was a big mess so I don't know if anyone would want to take it on... no need to simplify though

    logarithmic differentiation again...

    y = (x^(1/2) * e^(x)^(5) * (x^(2) + 1)^(1/3))/(x+1)

    i got... a mess... I don't even think I can type it all out... blah... if this one is too much trouble I understand =)


    this one seemed simple enough just wanted to make sure I was doing things correctly

    x^(y) = y^(x)

    i got dy/dx = (-y^(2) + xylny)/(xylnx - x^(2))


    These next 2 are just regular differentiation... any help is appreciated.

    h(x) = ln(x + (x^(2) - 1)^(1/2))

    i got

    h'(x) = 1/(x(x^(2))^(1/2) + x - 1)


    last one i promise =)

    f(x) = ln [((2x^(2) + 1)/(2x-1))^(1/2)]

    i came up with

    (2x^(2) - 2x - 1)/[(2x^(2)+1) * (2x-1)]

    THANK YOU SO MUCH for any help... I really do appreciate it greatly.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, drain!

    I'll do the last two . . . and thank you for showing your work!


    h(x) .= .ln[x + (x - 1)^]

    . . . . . . . 1 + (x - 1)^{-}2x . . . . 1 + x(x - 1)^{-}
    h'(x) . = . ---------------------------- . = . -----------------------
    . . . . . . . . . x + (x - 1)^ . . . . . . . . . .x + (x - 1)^


    Multiply top and bottom by (x - 1)^

    . . . . . . . . . (x - 1)^ + x
    . . = . --------------------------------- . . . . This reduces!
    . . . . .(x - 1)^
    [x + (x - 1)^]

    . . . . . . . . . . . . 1
    . . h'(x) .= .--------------
    . . . . . . . . .(x - 1)^



    f(x) .= .ln[(2x + 1)/(2x - 1)^]

    We have: .f(x) .= .ln(2x + 1) - ln(2x - 1)

    . . . . . . . . . . . . . . 4x . . . . . . . 2 . . . . . . .6x - 4x - 1
    Then: . f'(x) . = . ---------- - ------- . = . --------------------
    . . . . . . . . . . . . .2x + 1 . . . 2x - 1 . . . .(2x + 1)(2x - 1)

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2007
    Posts
    38
    thanks for your help man! But on that last one you did, I don't know if I mistyped it, or you misread it, but the whole thing was supposed to be square rooted. And sorry for not showing my work... if you're going to take the time to help me... I guess I should take the time to help you help me

    It was supposed to be ln[sqrt{(2x^(2)+1)/(2x-1)}]

    First I got

    1/sqrt[(2x^(2) + 1)/(2x-1)] * 1/2 * [(2x^(2) + 1)/(2x-1)]^(-1/2) * [(2x-1)(4x) - (2x^(2) + 1)(2)]/(2x-1)^2] by using the quotien rule

    then

    (4x^(2) - 4x - 2)/2(2x^(2) + 1)(2x-1) which i factored out a 2 and got

    (2x^(2) - 2x - 1)/[(2x^(2) * (2x-1)]
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by drain View Post
    thanks for your help man! But on that last one you did, I don't know if I mistyped it, or you misread it, but the whole thing was supposed to be square rooted. And sorry for not showing my work... if you're going to take the time to help me... I guess I should take the time to help you help me

    It was supposed to be ln[sqrt{(2x^(2)+1)/(2x-1)}]

    First I got

    1/sqrt[(2x^(2) + 1)/(2x-1)] * 1/2 * [(2x^(2) + 1)/(2x-1)]^(-1/2) * [(2x-1)(4x) - (2x^(2) + 1)(2)]/(2x-1)^2] by using the quotien rule

    then

    (4x^(2) - 4x - 2)/2(2x^(2) + 1)(2x-1) which i factored out a 2 and got

    (2x^(2) - 2x - 1)/[(2x^(2) * (2x-1)]
    Soroban's way of simplifying the ln before differentiating is still best here. It saves you from having to use the bothersome quotient rule.

    f(x) = ln [((2x^(2) + 1)/(2x-1))^(1/2)]
    => f(x) = (1/2)*[ ln(2x^(2) + 1) - ln(2x-1)]
    => f ' (x) = (1/2) * [1/(2x^(2) + 1) * 4x - 1/(2x - 1) * 2]
    => f ' (x) = (1/2) * [4x/(2x^(2) + 1) - 2/(2x - 1)]
    => f ' (x) = 2x/(2x^(2) + 1) - 1/(2x - 1)
    => f ' (x) = (4x^2 - 2x - 2x^2 - 1)/[(2x^(2) + 1)(2x - 1)]
    => f ' (x) = (2x^(2) - 2x - 1)/[(2x^(2)+1) * (2x-1)]
    which is what you got--but with slightly harder work. Sure, it looks like i took more steps, but most were easy simplifications and could be omitted
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2007
    Posts
    38
    Yeah I didn't notice that you could factor out the exponent, until I was re-doing it Thanks to both of you!!!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello again, drain!

    Yes, I misplaced the square root . . . my fault.
    . . Luckily, Jhevon covered for me.


    x^y .= .y^x

    I got: .dy/dx .= .[-y + xyln(y]/[xyln(x) - x] . . . . Yes!
    y . = . x^ e^{x^5) [(x + 1)^{1/3}]/(x + 1)
    This one is messy . . . I won't even try to simplify it.


    Take logs: . ln(y) . = . (1/2).ln(x) + x^5 + (1/3)ln(x + 1) - ln(x + 1)

    . . . . . .y' . . . . . . . .1 . . . . . . . . . . . . . 2x . . . . 1
    Then: . -- . = . (1/2)-- + 5x^4 + (1/3)-------- - ------- . . . . and so on . . .
    . . . . . .y . . . . . . . . x . . . . . . . . . . . .x + 1---x + 1

    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2007
    Posts
    38


    Thanks man, I really appreciate you taking the time to help me out.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 26th 2010, 05:24 PM
  2. property of zero blah
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 28th 2009, 06:20 AM
  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. derivatives...blah
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 2nd 2007, 04:08 AM

Search Tags


/mathhelpforum @mathhelpforum