# Thread: Logarithm Derivative (easy...except for me)

1. ## Logarithm Derivative (easy...except for me)

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

y=ln $x^2+1$

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

2. Remember: $\frac{d}{dx}ln(u)=\frac{1}{u}\cdot\frac{d}{dx}u$

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

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

You should be able to do the second problem now.

3. Do I split the numerator and the denominator?

So... $2x/(x+1) + 2/x$

4. What you wrote is somewhat unclear. If the problem is:

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

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

And you can just simplify from there.

5. Hmm my calculus professor has the answer as

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

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

$ln x^2+1- ln x^2$

6. Maybe the problem was $\frac{d}{dx}\left[\ln\left(\frac{x^2+1}{x^2}\right)\right]$ instead of $\frac{d}{dx}\left[\frac{\ln(x^2+1)}{x^2}\right]$.

Because $\frac{d}{dx}\left[\ln\left(\frac{x^2+1}{x^2}\right)\right] = \frac{d}{dx}[\ln(x^2+1)] - \frac{d}{dx}[\ln(x^2)] = \frac{2x}{x^2+1}-\frac{2}{x}$ like you said.

7. Originally Posted by redsoxfan325
Maybe the problem was $\frac{d}{dx}\left[\ln\left(\frac{x^2+1}{x^2}\right)\right]$ instead of $\frac{d}{dx}\left[\frac{\ln(x^2+1)}{x^2}\right]$.

Because $\frac{d}{dx}\left[\ln\left(\frac{x^2+1}{x^2}\right)\right] = \frac{d}{dx}[\ln(x^2+1)] - \frac{d}{dx}[\ln(x^2)] = \frac{2x}{x^2+1}-\frac{2}{x}$ like you said.

You are correct. When I was first doing the problem I was using the quotient rule and I was getting the wrong answer

So I put two and two together (pun intended) and I figured I had to split the numerator/denominator

+rep to everyone for helping