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Thread: Calculus II Derivative Help

  1. #1

    Smile Calculus II Derivative Help

    Hello all! Well this is my first post ever, so here it goes. I am taking an online Calculus II class, which leaves me with no one to ask questions for when I get confused. I think I found the solution, but got it rather easily, which makes me want to second guess myself. Any advice or help would be very much appreciated!

    Problem: What is the derivative of ln(x+ln x).

    My answer: 1/x(x+ln x)

    Wow, so simple right? I split the problem into lnx multiplied by ln(ln x) and solved for the derivative. Being a Calculus II course, I am wondering if I was suppose to do more work.


    My answer is on the multiple choice answers (which can be a trap!), but the other answers are:


    b. x/x+ ln x
    c. 1/x+ ln x
    d. 1/x+ ln x multiplied by (1 + 1/x)

    I hope to be right and questioning for no reason, but we'll see what happens. Thank you so much for the help

  2. #2
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    Quote Originally Posted by abrammer12 View Post
    Hello all! Well this is my first post ever, so here it goes. I am taking an online Calculus II class, which leaves me with no one to ask questions for when I get confused. I think I found the solution, but got it rather easily, which makes me want to second guess myself. Any advice or help would be very much appreciated!

    Problem: What is the derivative of ln(x+ln x).

    My answer: 1/x(x+ln x)

    Wow, so simple right? I split the problem into lnx multiplied by ln(ln x) and solved for the derivative. Being a Calculus II course, I am wondering if I was suppose to do more work.


    My answer is on the multiple choice answers (which can be a trap!), but the other answers are:


    b. x/x+ ln x
    c. 1/x+ ln x
    d. 1/x+ ln x multiplied by (1 + 1/x)

    I hope to be right and questioning for no reason, but we'll see what happens. Thank you so much for the help
    (ln(f(x))' = f'(x)/f(x)
    You have the 1/f(x) [=1/(x+ln x)] correct but what about that f'(x). What is the derivative of x + ln(x)

  3. #3
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    Quote Originally Posted by abrammer12 View Post
    Hello all! Well this is my first post ever, so here it goes. I am taking an online Calculus II class, which leaves me with no one to ask questions for when I get confused. I think I found the solution, but got it rather easily, which makes me want to second guess myself. Any advice or help would be very much appreciated!

    Problem: What is the derivative of ln(x+ln x).

    My answer: 1/x(x+ln x)

    Wow, so simple right? I split the problem into lnx multiplied by ln(ln x) and solved for the derivative. Being a Calculus II course, I am wondering if I was suppose to do more work.

    1st lets suppose that ln(x+ln(x)) = ln(x)*ln(ln(x)). This clearly is a product (even you said that you are multiply these two factor). So to get your answer did you use the product rule?.
    When you take the derivative of ln of anything you always get anything'/anyhing. You have ln (x+ln(x)) so the anything is x+ln(x). The answer will be (x+ln(x))'/(x+ln(x)). So all you have to do is figure what what (x+ln(x))' is, divide it by (x+ln(x)) and clean this up.

    Now back to the ln rule! ln(A) + ln(B) = ln (AB). BUT ln(x+ln(x)) does not equal ln(x)*ln(ln(x)). That is ln(A+B) is NOT ln(A)*ln(B)

    To succeed in Calculus 2 you *must* know calculus 1, trigonometry, pre calculus, college algebra, basic algebra and arithmetic.
    Last edited by Jomo; 01-10-2015 at 02:38 PM.
    A mathematician is a blind man in a dark room looking for a black cat which isn’t there. - Charles R. Darwin

  4. #4
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    Quote Originally Posted by abrammer12 View Post
    Hello all! Well this is my first post ever, so here it goes. I am taking an online Calculus II class, which leaves me with no one to ask questions for when I get confused. I think I found the solution, but got it rather easily, which makes me want to second guess myself. Any advice or help would be very much appreciated!

    Problem: What is the derivative of ln(x+ln x).

    My answer: 1/x(x+ln x)

    What you wrote actually translates to........ [tex]\frac{1}{x} * [x + ln(x)][/tex]

    perhaps you meant:................................ [tex]\frac{1}{x * [x + ln(x)]}[/tex]

    then you should have written:.............
    ..1/[x(x+ln(x))]

    Those
    [] are very important!!

    Wow, so simple right? I split the problem into lnx multiplied by ln(ln x) and solved for the derivative. Being a Calculus II course, I am wondering if I was suppose to do more work.


    My answer is on the multiple choice answers (which can be a trap!), but the other answers are:


    b. x/x+ ln x
    c. 1/x+ ln x
    d. 1/x+ ln x multiplied by (1 + 1/x)

    I hope to be right and questioning for no reason, but we'll see what happens. Thank you so much for the help
    I hope you have figured out the correct answer from other responses.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  5. #5
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by Jomo View Post
    1st lets suppose that ln(x+ln(x)) = ln(x)*ln(ln(x)).
    Why would one wish to suppose this? The rule for logs is that ln(a*b) = ln(a) + ln(b). It is not true that ln(x + ln(x)) equals ln(x)*ln(ln(x).

    Quote Originally Posted by abrammer12 View Post
    Problem: What is the derivative of ln(x+ln x).

    My answer: 1/x(x+ln x)

    Wow, so simple right? I split the problem into lnx multiplied by ln(ln x)...
    Apologies for any confusion about the log rules. You can follow the link (above) for clarification, as needed. However, the "solution" is suggested in the first reply you received. To clarify:

    Rather than trying to break the log expression into two terms, try instead to apply the Chain Rule. You have the natural log of something on the "outside"; take the derivative of that. Then multiply that result by the derivative of the "something" (which will be the x + ln(x) sum). What do you get?
    Last edited by stapel; 01-10-2015 at 01:29 PM.

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