1. ## differentiation

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!

2. ## Re: differentiation

Do you know the chain rule?

3. ## Re: differentiation

yes, but I dont see how I can use that here...

4. ## Re: differentiation

Let $u=\arcsin{(1-x^2)}$, and then, if necessary, let $v=1-x^2$ and use the chain rule twice.

5. ## Re: differentiation

So I would get something like f'(x)sec^2(f(x)) with f(x) being arcsin(1-x^2)^1/2 ?

6. ## Re: differentiation

Originally Posted by maths13
So I would get something like f'(x)sec^2(f(x)) with f(x) being arcsin(1-x^2)^1/2 ?
Yeah.

7. ## Re: differentiation

Another approach might be to use:

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

Then use the chain rule from there...

8. ## Re: differentiation

Originally Posted by maths13
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 \begin{align*} \tan{\theta} &\equiv \frac{\sin{\theta}}{\cos{\theta}} \\ &\equiv \pm \frac{\sin{\theta}}{\sqrt{1 - \sin^2{\theta}}} \end{align*}

Therefore

\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...