Results 1 to 3 of 3

Math Help - Calculating With Derivatives

  1. #1
    Member
    Joined
    Oct 2010
    Posts
    133

    Calculating With Derivatives

    I have done 2 of the 3 parts of the following problem but now I'm stuck. Any help would be appreciated:

    Let \mathbb{R}^+=(0,\infty). Suppose f:\mathbb{R}^+\rightarrow \mathbb{R} is differentiable at x=1 and f(xy)=f(x)+f(y) for all x,y\in \mathbb{R}^+.

    1)Prove f(1)=0
    2)Prove f(1/x)=-f(x)
    3)Prove f is differentiable on \mathbb{R}^+, and f'(x)=f'(1)/x.

    I have already proven parts 1 and 2. For part 3 I'm guessing you set it up something like this... (f(xy)-f(a))/(xy-a) and then you use the fact that f(xy)=f(x)+f(y) to rearrange to find the derivative but I'm struggling to see how. Any help would be great.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by zebra2147 View Post
    I have done 2 of the 3 parts of the following problem but now I'm stuck. Any help would be appreciated:

    Let \mathbb{R}^+=(0,\infty). Suppose f:\mathbb{R}^+\rightarrow \mathbb{R} is differentiable at x=1 and f(xy)=f(x)+f(y) for all x,y\in \mathbb{R}^+.

    1)Prove f(1)=0
    2)Prove f(1/x)=-f(x)
    3)Prove f is differentiable on \mathbb{R}^+, and f'(x)=f'(1)/x.

    I have already proven parts 1 and 2. For part 3 I'm guessing you set it up something like this... (f(xy)-f(a))/(xy-a) and then you use the fact that f(xy)=f(x)+f(y) to rearrange to find the derivative but I'm struggling to see how. Any help would be great.
    See if this gives you a kick start
    \displaystyle \lim_{x \to a}\frac{f(x)-f(a)}{x-a}=\lim_{x \to a}\frac{f\left( \frac{x}{a}\right)}{a\left[\left( \frac{x}{a}\right)-1\right]}
    Now let \displaystyle t=\frac{x}{a} then you get

    \lim_{x \to a}\frac{f\left( \frac{x}{a}\right)}{a\left[\left( \frac{x}{a}\right)-1\right]}=\lim_{t \to 1}\cdots and don't forget that f(1)=0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2010
    Posts
    470
    [duplicate]

    The main idea is that no matter where you differentiate, you can scale the parameter of the function, so that the new parameter becomes near 1 (which you know is differentiable).

    Algebraically:

    \displaystyle \frac{f(y) - f(x)}{y - x} = \frac{f(xy/x) - f(x)}{xy/x - x} = \frac{f(x) + f(y/x) - f(x)}{x(y/x - 1)} = \frac{f(y/x)}{x(y/x - 1)}
    \displaystyle = \frac{1}{x}\frac{f(y/x) - f(1)}{y/x - 1}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: November 6th 2009, 12:45 PM
  2. Replies: 4
    Last Post: February 10th 2009, 10:54 PM
  3. Trig derivatives/anti-derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 10th 2009, 02:34 PM
  4. Replies: 1
    Last Post: October 10th 2008, 08:29 AM
  5. Calculating Derivatives
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 29th 2007, 07:50 PM

Search Tags


/mathhelpforum @mathhelpforum