# Derivatives of Natural Logs question

• Jul 25th 2007, 02:47 PM
starswept
Derivatives of Natural Logs question
Hello, I'm having difficulty with the following question:

Solve the equation f'(x)=0.

f(x)= (x^2 + 1)^-1 ln (x^2 + 1)

I've worked out the derivative to be ln(x+1) divided by (x^2 + 1)^2. However, I'm unsure if this is correct, and if it is, I don't know how to solve for the x values once I set the derivative equal to zero- the natural log in the numerator throws me off. Thank you for any help!
• Jul 25th 2007, 02:55 PM
galactus
$\displaystyle f(x)=\frac{ln(x^{2}+1)}{x^{2}+1}$

$\displaystyle f'(x)=\frac{-2x(ln(x^{2}+1)-1)}{(x^{2}+1)^{2}}$

Now, what makes the numerator = 0?.

$\displaystyle -2x(ln(x^{2}+1)-1)=0$

0 is obviously one solution.

$\displaystyle ln(x^{2}+1)-1=0$

$\displaystyle ln(x^{2}+1)=1$

$\displaystyle x^{2}+1=e$

$\displaystyle x=\pm\sqrt{e-1}$
• Jul 25th 2007, 03:03 PM
starswept
Quote:

Originally Posted by galactus
$\displaystyle f(x)=\frac{ln(x^{2}+1)}{x^{2}+1}$

$\displaystyle f'(x)=\frac{-2x(ln(x^{2}+1)-1)}{(x^{2}+1)^{2}}$

Now, what makes the numerator = 0?.

$\displaystyle -2x(ln(x^{2}+1)-1)=0$

0 is obviously one solution.

$\displaystyle ln(x^{2}+1)-1=0$

$\displaystyle ln(x^{2}+1)=1$

$\displaystyle x^{2}+1=e$

$\displaystyle x=\pm\sqrt{e-1}$

Thanks very much- but how did you get the derivative? I can't seem to work out the derivative correctly.
• Jul 25th 2007, 03:13 PM
galactus
Actually, I ran it through my calculator, but we can do it the 'old-fashioned' way.

We can use the quotient rule:

$\displaystyle \frac{(x^{2}+1)(\frac{2x}{x^{2}+1})-ln(x^{2}+1)(2x)}{(x^{2}+1)^{2}}$

$\displaystyle \frac{2x-2xln(x^{2}+1)}{(x^{2}+1)^{2}}$

Factor:

$\displaystyle \frac{2x(1-ln(x^{2}+1))}{(x^{2}+1)^{2}}$

Which is the same thing as I posted. My calculator just displayed it that way.