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A differentiation and integraion question

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data
    a) if f(x)= ln(x/√(a-x^2)) show that f'(x) = a^2/x(a^2-x^2)


    ∫1/x(25-x^2) dx



    3. The attempt at a solution
    for a) i tried differentiating the top (ans. = 1) then the bottom.. obviously the bottom's where hte prob is at lol.. i kno d/dx ln[f(x)] --> 1/(f(x) χ f '(x) but i still lost..

    for b) i tried integrating it but idk im a bit lost..
     
  2. jcsd
  3. May 13, 2012 #2

    sharks

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    Gold Member

    For part (a):
    [tex]f(x)= \ln \left(\frac {x}{\sqrt{a-x^2}}\right)[/tex]
    First, find the derivative of ln. Then, multiply by the derivative of the fraction.

    For part (b), is this what you meant? Try to use LaTeX for clarity.
    [tex]\int \frac {1}{x(25-x^2)} \,.dx[/tex]
     
    Last edited: May 13, 2012
  4. May 13, 2012 #3
    for part b yes thats what i mean. for part a.. i dnt get u? as i said above the differential of a ln(f(x) = 1/f(x) * f ' (x) is that what u did?

    idk what LaTeX is...
     
  5. May 13, 2012 #4

    sharks

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    Gold Member

    Yes, that's exactly what i suggested for part (a).

    LaTeX is just an easy programming language for displaying equations, matrices, vectors and formulas clearly, as you can see in this post.

    For part (a):[tex]f'(x)= \frac{1}{\left(\frac {x}{\sqrt{a-x^2}}\right)}.\frac{d\left(\frac {x}{\sqrt{a-x^2}}\right)}{dx}[/tex]
    Now, to find the derivative of: [tex]\frac {x}{\sqrt{a-x^2}}[/tex]
    Use the substitution, [itex]u=a-x^2[/itex].
    But the answer doesn't check out with what you have provided in post #1. So, verify if the problem and/or answer are correct in post #1.

    For part (b):
    Express in partial fractions:
    [tex]\int \frac {1}{x(25-x^2)} \,.dx=\int \frac{1}{25x}\,.dx+\int \frac{x}{625-25x^2}\,.dx[/tex]
    [tex]\int \frac {1}{x(25-x^2)} \,.dx=\int \frac{1}{25x}\,.dx+\int \frac{x}{625-25x^2}\,.dx[/tex]
    [tex]\int \frac{1}{25x}\,.dx=\frac{1}{25}\ln x[/tex]
    To find the following integral:
    [tex]\int \frac{x}{625-25x^2}\,.dx[/tex]
    Use the formula:
    [tex]\int \frac{f'(x)}{\sqrt{f(x)}}\,dx=2\sqrt{f(x)}+C[/tex]
    You just rearrange the real constant coefficient of the numerator to match that of f'(x).
     
    Last edited: May 13, 2012
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