1. ## Derivatives#2

y=(1+x^2)tan^-1(2x)
Yet again I am stuck on this one. I am studying for a test next week. I am not sure where to even start on this. Can someone work this one complete?
Thanks,
Caleb

2. Here's a hint: by the Product Rule,

$\displaystyle \frac{dy}{dx}=2x\cdot\tan^{-1}(2x)+(1+x^2)\cdot\frac{d}{dx}\tan^{-1}(2x)\cdot 2,$

and the derivative of $\displaystyle \tan^{-1}x$ is $\displaystyle \frac{1}{1+x^2}$.

3. I take it the problem is:

$\displaystyle y= [1+x^{2}] * [ \arctan 2x]$?

Use the product rule.

Hint: the derivative of $\displaystyle arctan(u)$ is $\displaystyle \frac {u'}{1+u^{2}}$

4. Ahhh,
Yes it is arctan.
Thanks Guys for all of your help. I am just not understanding derivatives at all, and you guys have been very helpful in me trying to learn them!

5. Like anything, they get easier with practice.

6. I scanned this particular problem, since I didnt know how to type it in. Any help on it?

7. Since $\displaystyle y= ln (x^{n})$ = $\displaystyle n \ln (x)$, then:

$\displaystyle \ln [\frac {3x+2}{3x-2}]^{ \frac {1}{3}}$ = $\displaystyle y= \frac {1}{3} \ln \frac {3x+2}{3x-2}$

Since you are dividing, break it up into two natural logarithms.

$\displaystyle y= \ln \frac {a}{b}$ = $\displaystyle \ln a - \ln b$

Does that help? Just use the rules of logarithms to simply so that you can find the derivative.

Hint:
if $\displaystyle y= \ln u$, then $\displaystyle y'= (\frac {1}{u}) \frac {du}{dx}$