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?

Substitution is a good way to solve these kinds of problems. In the original problem, the term $\sqrt{1-x^2}$ suggests the substitution $x=\sin\theta\ \text{or}\cos\theta$. Also, notice that $\sin\frac\pi 4=\cos\frac\pi 4=\frac{1}{\sqrt2}$. Now Substituting $x=\sin\theta$ where $\theta\in[-\frac\pi 2,\frac\pi 2]$, we get
$$\begin{align}\sin^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt 2}\right) &= \sin^{-1}\left(\frac{1}{\sqrt2}(\sin\theta+\cos\theta)\right) \\ &= \sin^{-1}\left(\frac{1}{\sqrt2}\sin\theta+\frac{1}{\sqrt2}\cos\theta\right)\\ &= \sin^{-1}\left(\cos\frac\pi 4\sin\theta+\sin\frac\pi 4\cos\theta\right) \\ &= \sin^{-1}\left(\sin\left(\frac\pi 4+\theta\right)\right). 
\end{align}$$
Now we have $\sin^{-1}\left(\sin\left(\frac\pi 4+\theta\right)\right)=\frac\pi 4+\theta=\frac\pi 4+\sin^{-1}x$ when $\theta\in[-\frac\pi 2,\frac\pi 4]$ and $\sin^{-1}\left(\sin\left(\frac\pi 4+\theta\right)\right)=\pi-\left(\frac\pi 4+\theta\right)=\frac{3\pi} 4-\theta=\frac{3\pi} 4- \sin^{-1}x\ \text{when}\ \theta\in[\frac\pi 4,\frac\pi 2]$.
You just have to find the derivatives in both cases now.
