A differentiation and integraion question

[kevester]

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..

 

[DryRun]

For part (a):
f(x)= \ln \left(\frac {x}{\sqrt{a-x^2}}\right)
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.
\int \frac {1}{x(25-x^2)} \,.dx

 

[kevester]

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...

 

[DryRun]

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):f'(x)= \frac{1}{\left(\frac {x}{\sqrt{a-x^2}}\right)}.\frac{d\left(\frac {x}{\sqrt{a-x^2}}\right)}{dx}
Now, to find the derivative of: \frac {x}{\sqrt{a-x^2}}
Use the substitution, u=a-x^2.
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:
\int \frac {1}{x(25-x^2)} \,.dx=\int \frac{1}{25x}\,.dx+\int \frac{x}{625-25x^2}\,.dx
\int \frac {1}{x(25-x^2)} \,.dx=\int \frac{1}{25x}\,.dx+\int \frac{x}{625-25x^2}\,.dx
\int \frac{1}{25x}\,.dx=\frac{1}{25}\ln x
To find the following integral:
\int \frac{x}{625-25x^2}\,.dx
Use the formula:
\int \frac{f'(x)}{\sqrt{f(x)}}\,dx=2\sqrt{f(x)}+C
You just rearrange the real constant coefficient of the numerator to match that of f'(x).