Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - Derivative of Natural Log

  1. #1
    Junior Member
    Joined
    Mar 2007
    Posts
    56

    Derivative of Natural Log

    Can someone please tell me what I'm doing wrong here, and show me how to do it correctly?

    I need to differentiate g(x) = LN a-x/a+x

    So,

    g'(x) = 1/a-x/a+x *d/dx (a-x/a+x)
    g'(x) = a+x/a-x *[(a+x)*d/dx(a-x) - (a-x)*d/dx(a+x)/(a+x)^2]
    g'(x) = a+x/a-x *[(a+x)*(1-1) - (a-x)(1+1)/(a+x)^2]
    g'(x) = a+x/a-x *[(a+x)*(0) - (a-x)(2)/(a+x)^2]
    g'(x) = a+x/a-x *[0 - 2(a-x)/(a+x)^2]
    g'(x) = a+x/a-x * -2a+2x/a^2+2ax+x^2
    g'(x) = a+x*(-2a+2x)/a-x*(a^2+2ax+x^2)
    g'(x) = -2a^2+2ax-2ax+2x^2/a^3+2a^2x+ax^2-a^2x-2ax^2-x^3)
    g'(x) = -2a^2+2x^2/(a^3+a^2x-ax^2-x^3)


    the answer is supposed to be g'(x) = -2a/a^2-x^2, but that's not what I get when I simplify . . .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    Here is the answer. I know it is right. The answer you have at the bottom is incorrect.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2007
    Posts
    56
    My book says that the answer is g'(x) = -2a/a^2-x^2, but I'll look over your work.


    The problem is g(x) = the natural log of a-x/a+x, not the natural log of a-x/ a+x, so I think you may ahve done it wrong . . .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1
    Quote Originally Posted by zachb View Post
    Can someone please tell me what I'm doing wrong here, and show me how to do it correctly?

    I need to differentiate g(x) = LN a-x/a+x

    So,

    g'(x) = 1/a-x/a+x *d/dx (a-x/a+x)
    g'(x) = a+x/a-x *[(a+x)*d/dx(a-x) - (a-x)*d/dx(a+x)/(a+x)^2]
    g'(x) = a+x/a-x *[(a+x)*(1-1) - (a-x)(1+1)/(a+x)^2]
    g'(x) = a+x/a-x *[(a+x)*(0) - (a-x)(2)/(a+x)^2]
    g'(x) = a+x/a-x *[0 - 2(a-x)/(a+x)^2]
    g'(x) = a+x/a-x * -2a+2x/a^2+2ax+x^2
    g'(x) = a+x*(-2a+2x)/a-x*(a^2+2ax+x^2)
    g'(x) = -2a^2+2ax-2ax+2x^2/a^3+2a^2x+ax^2-a^2x-2ax^2-x^3)
    g'(x) = -2a^2+2x^2/(a^3+a^2x-ax^2-x^3)


    the answer is supposed to be g'(x) = -2a/a^2-x^2, but that's not what I get when I simplify . . .
    With your notation this is what you told me. I just went by what you said. Try using parenthesis. This is what you told me:

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member frenzy's Avatar
    Joined
    Mar 2007
    From
    Planet Express
    Posts
    58
    ln[(a-x)/(a+x)]=ln(a-x)-ln(a+x)

    take the derivative

    -1/(a-x)-1/(a+x)=-2a/(a^2-x^2)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1
    Quote Originally Posted by frenzy View Post
    ln[(a-x)/(a+x)]=ln(a-x)-ln(a+x)

    take the derivative

    -1/(a-x)-1/(a+x)=-2a/(a^2-x^2)

    When you take the derivative of ln(a-x)-ln(a+x) you do not get 2a/(a^2-x^2)

    you get:




    The way that first problem was posted was terrible absolutely terrible notation.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member frenzy's Avatar
    Joined
    Mar 2007
    From
    Planet Express
    Posts
    58
    Quote Originally Posted by qbkr21 View Post
    When you take the derivative of ln(a-x)-ln(a+x) you do not get 2a/(a^2-x^2)

    you get:




    The way that first problem was posted was terrible absolutely terrible notation.
    You might want to retry that derivative again.

    What I wrote is correct.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    I am not going to argue with you but the derivative of ln(f(x))= (1/(f(x))* f'(x)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Mar 2007
    Posts
    56
    Frenzy, can you please show your work so that I can see how you got that answer? I already know what the answer should be, what I'm interested in is how you got there.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member frenzy's Avatar
    Joined
    Mar 2007
    From
    Planet Express
    Posts
    58
    Quote Originally Posted by qbkr21 View Post
    I am not going to argue with you but the derivative of ln(f(x))= (1/(f(x))* f'(x)
    This is correct...BUT, how you took the derivative is NOT correct.
    Again...look at what you have written. It is wrong.

    Quote Originally Posted by zachb View Post
    Frenzy, can you please show your work so that I can see how you got that answer? I already know what the answer should be, what I'm interested in is how you got there.
    d/dx ln(a-x)= [1/(a-x)]*d/dx[(a-x)]=1/(a-x)]*(-1)=-1/(a-x)

    d/dx -ln(a+x)= -[1/(a+x)]*d/dx[(a+x)]=-1/(a+x)]*(+1)=-1/(a+x)



    -1/(a-x)-1/(a+x)=(-1(a+x)-1(a-x))/[(a-x)*(a+x)]


    =(-a-x-a+x))/[a^2+ax-ax-x^2]=(-a-a)/(a^2-x^2)=-2a/(a^2-x^2)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    If you go to my second post where I show the problem aligned in the calculator. Then if you take the derivative of whats on the screen and put it on paper you get my exact answerr. Now maybe this whole scenario revolves around the fact that I didn't understand the way:

    g(x)= ln x-a/x+a

    There again this is terrible notation because it could me several different things:

    g(x)= ln(x-a)/x+a ???

    g(x)= ln(x)-a/x+a ???

    or g(x)= ln(x-a/x+a) ???

    I mean if my professor handed me this in class I would hand it right back to him simply because this in its entirety is "mathematically incorrect". A professor who does this obviously should be teaching grade school or not at all.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member frenzy's Avatar
    Joined
    Mar 2007
    From
    Planet Express
    Posts
    58
    Quote Originally Posted by qbkr21 View Post
    If you go to my second post where I show the problem aligned in the calculator. Then if you take the derivative of whats on the screen and put it on paper you get my exact answerr. Now maybe this whole scenario revolves around the fact that I didn't understand the way:

    g(x)= ln x-a/x+a

    There again this is terrible notation because it could me several different things:

    g(x)= ln(x-a)/x+a ???

    g(x)= ln(x)-a/x+a ???

    or g(x)= ln(x-a/x+a) ???

    I mean if my professor handed me this in class I would hand it right back to him simply because this in its entirety is "mathematically incorrect". A professor who does this obviously should be teaching grade school or not at all.
    no doubt that the problem was missing some prentices



    Going back to your 2nd post you have..

    d/dx(a-x)=1-1=0
    and
    d/dx(a+x)=1+1=2
    These are not correct.

    you should have

    d/dx(a-x)=0-1=-1
    and
    d/dx(a+x)=0+1=1

    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Mar 2007
    Posts
    56
    Quote Originally Posted by frenzy View Post
    no doubt that the problem was missing some prentices



    Going back to your 2nd post you have..

    d/dx(a-x)=1-1=0
    and
    d/dx(a+x)=1+1=2
    These are not correct.

    you should have

    d/dx(a-x)=0-1=-1
    and
    d/dx(a+x)=0+1=1
    I don't understand why d/dx(a-x)=0-1=-1
    and
    d/dx(a+x)=0+1=1

    Is that because "a" is a constant term?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member frenzy's Avatar
    Joined
    Mar 2007
    From
    Planet Express
    Posts
    58
    Quote Originally Posted by zachb View Post
    I don't understand why d/dx(a-x)=0-1=-1
    and
    d/dx(a+x)=0+1=1

    Is that because "a" is NOT a constant term?
    a is a constant therefore

    d/dx(a)=0

    The derivative of any constant is zero.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    Mar 2007
    Posts
    56
    Okay, but I don't understand how we're supposed to know that "a" IS a constant term, and x is not. Is there some kind of rule for determining which term is constant when you have two or more varaibles? I thought that constant terms were just integers. For example, in f(x)=x^2-7x+5 the constant term is 5 and when you find f'(x) it's f'(x)=2x-7+0.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Natural Derivative
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 14th 2009, 11:27 AM
  2. Derivative with natural logarithm
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 5th 2009, 03:01 PM
  3. Derivative of Natural Log
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 28th 2009, 02:45 PM
  4. first derivative with natural log
    Posted in the Calculus Forum
    Replies: 9
    Last Post: October 9th 2008, 03:43 PM
  5. Derivative of a Natural Log
    Posted in the Calculus Forum
    Replies: 11
    Last Post: April 20th 2008, 12:52 PM

Search Tags


/mathhelpforum @mathhelpforum