Hey there, I've got a question thats been bugging me for ages...I have no idea how to start it!
Differentiate tan(arcsin (1-x^2)^1/2)
Any help will be appreciated, thanks!
An alternative is to note that
$\displaystyle \displaystyle \begin{align*} \tan{\theta} &\equiv \frac{\sin{\theta}}{\cos{\theta}} \\ &\equiv \pm \frac{\sin{\theta}}{\sqrt{1 - \sin^2{\theta}}} \end{align*}$
Therefore
$\displaystyle \displaystyle \begin{align*} \tan{\left[\arcsin{\left(\sqrt{1 - x^2}\right)}\right]} &\equiv \pm \frac{\sin{\left[\arcsin{\left(\sqrt{1 - x^2}\right)}\right]}}{\sqrt{1 - \sin^2{\left[\arcsin{\left(\sqrt{1 - x^2}\right)}\right]}}} \\ &\equiv \pm \frac{\sqrt{1 - x^2}}{\sqrt{1 - \left(\sqrt{1 - x^2}\right)^2}} \\ &\equiv \pm \frac{\sqrt{1 - x^2}}{\sqrt{1 - \left(1 - x^2\right)}} \\ &\equiv \pm \frac{\sqrt{1 - x^2}}{\sqrt{x^2}} \\ &\equiv \pm \sqrt{\frac{1 - x^2}{x^2}} \\ &\equiv \pm \left(x^{-2} - 1 \right)^{\frac{1}{2}}\end{align*}$
This should be much easier to differentiate...