1. ## derivative of arctan^3

I have a question with finding the derivative of the function below:

$\displaystyle y= arctan^3(\frac{x}{ln(x)})$.

This is what I get when I take the derivative of the function:

$\displaystyle \frac{d}{dx} 3(sec^2y)(tan^2y) = \frac{ln(x)-1}{(lnx)^2}$

I don't know how to solve for $\displaystyle (sec^2y)(tan^2y)$ in terms of x. I cannot figure out how. I know it is dealing with $\displaystyle sin^2y+cos^2y=1$ and dividing by either $\displaystyle sin^2y$ or $\displaystyle cos^2y$ depending on what you want. But I don't see how you can get a polynomial out of that.

2. ## Re: derivative of arctan^3

$\displaystyle y = \left( \tan ^{-1} \left(\frac{x}{\ln (x)}\right)\right)^3$

$\displaystyle y^{\frac{1}{3}} = \tan ^{-1} \left(\frac{x}{\ln (x)}\right)$

$\displaystyle \tan (y^{\frac{1}{3}}) = \frac{x}{\ln (x)}$

now derive

you did something wrong since you considered

$\displaystyle \tan ^3 (y) = \frac{x}{\ln (x)}$

3. ## Re: derivative of arctan^3

Thanks, I looked it over today and I got the right answer. Thanks bud.