Results 1 to 4 of 4

Thread: stuck on a calculus problem from online homework(logarithmic differentiation)

  1. #1
    Newbie
    Joined
    Jun 2010
    Posts
    4

    stuck on a calculus problem from online homework(logarithmic differentiation)

    I just registered and am excited to join the community. My problem is: stuck on a calculus problem from online homework(logarithmic differentiation)-age.cgi.gif. I feel like I am very close but have overlooked some things. I am aware that derivative of ln is 1/x but I can't seem to get the final answer right. Also, one that has been bugging me for a few days is stuck on a calculus problem from online homework(logarithmic differentiation)-symimage.cgi.gif. For this one I have gotten stuff like 1/xsqrt(x^2-5)*sqrt(x^2-5)+1/(2(sqrt(X^2-5))) *2x^2. Any help would be much appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by nvr20 View Post
    I just registered and am excited to join the community. My problem is: Click image for larger version. 

Name:	age.cgi.gif 
Views:	21 
Size:	1.1 KB 
ID:	18047. I feel like I am very close but have overlooked some things. I am aware that derivative of ln is 1/x but I can't seem to get the final answer right. Also, one that has been bugging me for a few days is Click image for larger version. 

Name:	symimage.cgi.gif 
Views:	22 
Size:	832 Bytes 
ID:	18046. For this one I have gotten stuff like 1/xsqrt(x^2-5)*sqrt(x^2-5)+1/(2(sqrt(X^2-5))) *2x^2. Any help would be much appreciated.
    $\displaystyle g = \ln{(x\sqrt{x^2 - 5})}$.

    Let $\displaystyle u = x\sqrt{x^2 - 5}$ so that $\displaystyle g = \ln{u}$.


    $\displaystyle \frac{dg}{du} = \frac{1}{u} = \frac{1}{x\sqrt{x^2 - 5}}$.


    $\displaystyle \frac{du}{dx} = x\,\frac{d}{dx}(\sqrt{x^2 - 5}) + \sqrt{x^2 - 5}\,\frac{d}{dx}(x)$

    $\displaystyle = x\cdot \frac{x}{\sqrt{x^2 - 5}} + \sqrt{x^2 - 5}\cdot 1$

    $\displaystyle = \frac{x^2}{\sqrt{x^2 - 5}} + \sqrt{x^2 - 5}$

    $\displaystyle = \frac{x^2\sqrt{x^2 - 5}}{x^2 - 5} + \frac{(x^2 - 5)\sqrt{x^2 - 5}}{x^2 - 5}$

    $\displaystyle = \frac{(2x^2 - 5)\sqrt{x^2 - 5}}{x^2 - 5}$.


    Therefore $\displaystyle \frac{dg}{dx} = \frac{dg}{du}\,\frac{du}{dx}$

    $\displaystyle = \left(\frac{1}{x\sqrt{x^2 - 5}}\right)\left(\frac{(2x^2 - 5)\sqrt{x^2 - 5}}{x^2 - 5}\right)$

    $\displaystyle = \frac{2x^2 - 5}{x(x^2 - 5)}$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by nvr20 View Post
    I just registered and am excited to join the community. My problem is: Click image for larger version. 

Name:	age.cgi.gif 
Views:	21 
Size:	1.1 KB 
ID:	18047. I feel like I am very close but have overlooked some things. I am aware that derivative of ln is 1/x but I can't seem to get the final answer right. Also, one that has been bugging me for a few days is Click image for larger version. 

Name:	symimage.cgi.gif 
Views:	22 
Size:	832 Bytes 
ID:	18046. For this one I have gotten stuff like 1/xsqrt(x^2-5)*sqrt(x^2-5)+1/(2(sqrt(X^2-5))) *2x^2. Any help would be much appreciated.
    $\displaystyle H(z) = \ln{\left(\sqrt{\frac{1 - z^2}{1 + z^2}}\right)}$

    $\displaystyle = \ln{\left(\frac{1 - z^2}{1 + z^2}\right)^{\frac{1}{2}}}$

    $\displaystyle = \frac{1}{2}\ln{\left(\frac{1 - z^2}{1 + z^2}\right)}$.


    Now let $\displaystyle u = \frac{1 - z^2}{1 + z^2}$ so that $\displaystyle H = \frac{1}{2}\ln{u}$.


    $\displaystyle \frac{dH}{du} = \frac{1}{2u} = \frac{1}{2\left(\frac{1 - z^2}{1 + z^2}\right)} = \frac{1 + z^2}{2(1 - z^2)}$.


    $\displaystyle \frac{du}{dx} = \frac{(1 + z^2)\frac{d}{dx}(1 - z^2) - (1 - z^2)\frac{d}{dx}(1 + z^2)}{(1 + z^2)^2}$

    $\displaystyle = \frac{-2z(1 + z^2) - 2z(1 - z^2)}{(1 + z^2)^2}$

    $\displaystyle = \frac{-2z - 2z^3 - 2z + 2z^3}{(1 + z^2)^2}$

    $\displaystyle = \frac{-4z}{(1 + z^2)^2}$.


    So $\displaystyle \frac{dH}{dx} = \frac{dH}{du}\,\frac{du}{dx}$

    $\displaystyle = \left( \frac{1 + z^2}{2(1 - z^2)}\right)\left(\frac{-4z}{(1 + z^2)^2}\right)$

    $\displaystyle = -\frac{2z}{(1 - z^2)(1 + z^2)}$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2010
    Posts
    4
    Thanks a lot! Where I went wrong was the differentiation of \sqrt{x^2 - 5}. Makes much sense now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Mar 30th 2011, 09:50 PM
  2. Replies: 3
    Last Post: May 5th 2010, 05:19 PM
  3. Calculus II Homework Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Apr 3rd 2010, 02:49 PM
  4. Calculus Logarithmic Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 28th 2009, 05:42 PM
  5. Another homework problem: Calculus
    Posted in the Calculus Forum
    Replies: 11
    Last Post: Jun 4th 2006, 09:46 AM

Search Tags


/mathhelpforum @mathhelpforum