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!

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- Dec 24th 2011, 10:26 AMmaths13differentiation
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! - Dec 24th 2011, 10:31 AMSironRe: differentiation
Do you know the chain rule?

- Dec 24th 2011, 10:37 AMmaths13Re: differentiation
yes, but I dont see how I can use that here...

- Dec 24th 2011, 10:40 AMQuackyRe: differentiation
Let $\displaystyle u=\arcsin{(1-x^2)}$, and then, if necessary, let $\displaystyle v=1-x^2$ and use the chain rule twice.

- Dec 24th 2011, 10:45 AMmaths13Re: differentiation
So I would get something like f'(x)sec^2(f(x)) with f(x) being arcsin(1-x^2)^1/2 ?

- Dec 24th 2011, 10:50 AMQuackyRe: differentiation
- Dec 24th 2011, 10:50 AMMarkFLRe: differentiation
Another approach might be to use:

$\displaystyle \tan\left(\arcsin\left(\sqrt{1-x^2}\right)\right)=\sqrt{x^{-2}-1}$

Then use the chain rule from there... - Dec 24th 2011, 03:19 PMProve ItRe: differentiation
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...