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

  1. #1
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    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.
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  2. #2
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    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.
    g = \ln{(x\sqrt{x^2 - 5})}.

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


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


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

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

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

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

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


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

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

     = \frac{2x^2 - 5}{x(x^2 - 5)}.
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  3. #3
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    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.
    H(z) = \ln{\left(\sqrt{\frac{1 - z^2}{1 + z^2}}\right)}

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

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


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


    \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)}.


    \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}

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

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

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


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

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

     = -\frac{2z}{(1 - z^2)(1 + z^2)}.
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    Thanks a lot! Where I went wrong was the differentiation of \sqrt{x^2 - 5}. Makes much sense now.
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