A differentiation and integraion question

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Homework Help Overview

The discussion revolves around differentiation and integration, specifically focusing on the function f(x) = ln(x/√(a-x²)) and the integral ∫1/x(25-x²) dx. Participants are exploring the derivative of the logarithmic function and methods for integrating the given expression.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to differentiate the function and expresses confusion regarding the application of the logarithmic differentiation rule. Some participants suggest finding the derivative of the logarithm and multiplying by the derivative of the inner function. For the integral, there is mention of using partial fractions, but uncertainty remains about the approach.

Discussion Status

Participants are actively engaging with the problem, offering suggestions for differentiation and integration techniques. There is a recognition of potential discrepancies in the original problem statement and the provided answers, prompting verification of the details. The discussion is ongoing with various interpretations being explored.

Contextual Notes

Some participants express a lack of familiarity with LaTeX, which is suggested for clarity in mathematical expressions. There is also an indication that the original poster may be struggling with the differentiation process and the integration technique, highlighting areas where further clarification might be needed.

kevester
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Homework Statement


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



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 I am a bit lost..
 
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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]
 
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for part b yes that's 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...
 
kevester said:
for part b yes that's 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...

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:

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