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
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} $