Ok we have 2 problems here....one is just a slightly harder than the other(Headbang)

y=ln $\displaystyle x^2+1$

y=ln $\displaystyle (x2+1)/x^2$

Please post steps!!! Thanks :)

Printable View

- May 2nd 2009, 02:35 PMChrisLogarithm Derivative (easy...except for me)
Ok we have 2 problems here....one is just a slightly harder than the other(Headbang)

y=ln $\displaystyle x^2+1$

y=ln $\displaystyle (x2+1)/x^2$

Please post steps!!! Thanks :) - May 2nd 2009, 02:42 PMPinkk
Remember: $\displaystyle \frac{d}{dx}ln(u)=\frac{1}{u}\cdot\frac{d}{dx}u$

So for the first problem, $\displaystyle u=x^{2}+1$,

$\displaystyle \frac{d}{dx}ln(x^{2}+1)=\frac{1}{x^{2}+1}\cdot\fra c{d}{dx}(x^{2}+1)=\frac{2x}{x^{2}+1}$

You should be able to do the second problem now. - May 2nd 2009, 03:11 PMChris
Do I split the numerator and the denominator?

So... $\displaystyle 2x/(x+1) + 2/x $ - May 2nd 2009, 03:16 PMPinkk
What you wrote is somewhat unclear. If the problem is:

$\displaystyle \frac{ln(x^{2}+1)}{x^{2}}$ You'll need to invoke log differentiation AND the quotient rule:

$\displaystyle \frac{dy}{dx}=\frac{x^{2}\frac{2x}{x^{2}+1}-2x\,ln(x^{2}+1)}{x^{4}}$

And you can just simplify from there. - May 2nd 2009, 05:47 PMChris
Hmm my calculus professor has the answer as

$\displaystyle 2x/(x^2+1)-2/x $

The only way I could figure out how to get that answer was to make the problem

$\displaystyle ln x^2+1- ln x^2 $ - May 2nd 2009, 06:01 PMredsoxfan325
Maybe the problem was $\displaystyle \frac{d}{dx}\left[\ln\left(\frac{x^2+1}{x^2}\right)\right]$ instead of $\displaystyle \frac{d}{dx}\left[\frac{\ln(x^2+1)}{x^2}\right]$.

Because $\displaystyle \frac{d}{dx}\left[\ln\left(\frac{x^2+1}{x^2}\right)\right] = \frac{d}{dx}[\ln(x^2+1)] - \frac{d}{dx}[\ln(x^2)] = \frac{2x}{x^2+1}-\frac{2}{x}$ like you said. - May 2nd 2009, 06:36 PMChris