Derivatives of Natural Logs question

**starswept** · July 25th 2007, 02:47 PM

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!

**galactus** · July 25th 2007, 02:55 PM

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

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

Now, what makes the numerator = 0?.

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

0 is obviously one solution.

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

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

$x^{2}+1=e$

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

**starswept** · July 25th 2007, 03:03 PM

Originally Posted by galactus

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

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

Now, what makes the numerator = 0?.

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

0 is obviously one solution.

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

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

$x^{2}+1=e$

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

Thanks very much- but how did you get the derivative? I can't seem to work out the derivative correctly.

**galactus** · July 25th 2007, 03:13 PM

Actually, I ran it through my calculator, but we can do it the 'old-fashioned' way.

We can use the quotient rule:

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

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

Factor:

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

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

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