To find $f'(x)$ when $f(x) = \sqrt{\sin^{-1}x+\sqrt{\cos^{-1}x+\sqrt{\sin^{-1}x+\sqrt{\cos^{-1}x+\cdots\infty}}}}$

Question

If $$ f(x) = \sqrt{\sin^{-1}x + \sqrt{\cos^{-1}x + \sqrt{\sin^{-1}x + \sqrt{\cos^{-1}x + \cdots\infty}}}}, $$ then how to find $\frac{d}{dx}(f(x))$?

I am trying this question by taking $f(x)=y$.

Now $$ y = \sqrt{\sin^{-1}x + \sqrt{\cos^{-1}x + \sqrt{\sin^{-1}x + \sqrt{\cos^{-1}x + \cdots\infty}}}} $$

So, from this we can clearly see that $y=\sqrt{\sin^{-1}x+\sqrt{\cos^{-1}x+y}}$.

Now we can write that $y^{2}=\sin^{-1}x+\sqrt{\cos^{-1}x+y}$.

I am facing problem to find the derivative of $f(x)$. Please help me out.

Answer 1

Differentiate the last equation you wrote in to order to get : $$ 2yy' = \frac{1}{\sqrt{1-x^2}} + \frac{y'-\frac{1}{\sqrt{1-x^2}}}{2\sqrt{y+\arccos x}} = \frac{1}{\sqrt{1-x^2}}\left(1-\frac{1}{2\sqrt{y + \arccos x}}\right) + \frac{y'}{2\sqrt{y+\arccos x}} $$ Next, isolate $y'$ : $$ y' = \frac{1}{\sqrt{1-x^2}} \frac{1-\frac{1}{2\sqrt{y+\arccos x}}}{2y-\frac{1}{2\sqrt{y+\arccos x}}} = \frac{1}{\sqrt{1-x^2}} \frac{1-2\sqrt{y+\arccos x}}{1-4y\sqrt{y+\arccos x}} $$ You can still work on this expression by replacing $\sqrt{y+\arccos x}$ by $y^2 - \arcsin x$ for instance, but basically you are done.