Derivative of $\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ I'm a calculus beginner. I was asked to find the derivative of the function: $$\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}.$$
I'm able to solve it in the following way:
I first calculate the derivative of $\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ and get $\frac{1}{\sqrt 2}(1-\frac{x}{\sqrt{1-x^2}})$. Then the derivative of the given function is $\frac{1}{\sqrt{1-(\frac{x+\sqrt{1-x^2}}{\sqrt 2}})^2}\cdot \frac{1}{\sqrt 2}(1-\frac{x}{\sqrt{1-x^2}})$. Simplifying this gives the final answer $\frac{1}{\sqrt{1-x^2}}$. But the simplication process is quite lengthy and involves some bizarre calculations. Is there tricks/ways to solve these kinds of derivatives that do not involve too much calculations like above?

Let $f(x) = \sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt{2}} $. We have, for $\theta = \sin^{-1} x \in[-{\pi\over 2},{\pi\over 2}]$:
$$ f(\sin \theta) = \sin^{-1}\frac{\sin\theta+\sqrt{1-\sin^2\theta}}{\sqrt{2}} = \sin^{-1}\frac{\sin\theta+\cos\theta}{\sqrt{2}} = \sin^{-1}\big(\sin(\theta+{\pi\over 4})\big)$$
That is
$$ f(\sin\theta) = \left\{\begin{array}{ll}\theta+ {\pi\over 4} & \text{for }\theta\in [-{\pi\over 2},{\pi\over 4}] \\ \pi-(\theta+ {\pi\over 4})={3\pi\over 4}-\theta & \text{for } \theta\in [{\pi\over 4},{\pi\over 2}]\end{array} \right. $$
$$ f(x) = \left\{\begin{array}{ll}{\pi\over 4} + \sin^{-1} x& \text{for } x \in[-1,{\sqrt{2}\over 2}] \\ {3\pi\over 4} - \sin^{-1} x & \text{for } x \in[{\sqrt{2}\over 2},1]\end{array} \right. $$
Taking the derivative we have
$$ f'(x) = \left\{\begin{array}{ll}{1\over\sqrt{1-x^2}} & \text{for }x\in (-1,{\sqrt{2}\over 2}) \\ -{1\over\sqrt{1-x^2}} & \text{for } x \in({\sqrt{2}\over 2},1)\end{array} \right. $$
In the point $x={\sqrt{2}\over 2}$ the derivative doesn't exist (though left-sided and right-sided derivates do).
