Differentiate $\sin^{-1}\left(\frac {\sin x + \cos x}{\sqrt{2}}\right)$ with respect to $x$ 
Differentiate $$\sin^{-1}\left(\frac {\sin x + \cos x}{\sqrt{2}}\right)$$ with respect to $x$.

I started like this: Consider $$\frac {\sin x + \cos x}{\sqrt{2}}$$, substitute $\cos x$ as $\sin \left(\frac {\pi}{2} - x\right)$, and proceed with the simplification. Finally I am getting it as $\cos \left(x - \frac {\pi}{4}\right)$. After this I could not proceed. Any help would be appreciated. Thanks in advance!
 A: Hint The angle sum rule for $\sin$ is
$$\sin(x + y) = \sin x \cos y + \cos x \sin y.$$

Additional hint In particular, if we take $y = \frac{\pi}{4}$ and rearrange, we get $$\sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right).$$

A: It's better to rewrite
$$
\frac{1}{\sqrt{2}}(\sin x+\cos x)=\sin(x+\pi/4)
$$
and then use the chain rule:
$$
f(x)=\arcsin\sin\Bigl(x+\frac{\pi}{4}\Bigr)
$$
so
$$
f'(x)=\frac{1}{\sqrt{1-\sin^2(x+\pi/4)}}\cos\Bigl(x+\frac{\pi}{4}\Bigr)
=\dots
$$
(Beware of the square root!)

 $f'(x)=\dfrac{\cos(x+\pi/4)}{|\cos(x+\pi/4)|}$ so the derivative is $1$ where $\cos(x+\pi/4)>0$ and $-1$ where $\cos(x+\pi/4)<0$; the function is not differentiable where $\cos(x+\pi/4)=0$.

A: An alternative approach is to use Implicit Differentiation:
\begin{equation}
 y = \arcsin\left(\frac{\sin(x) + \cos(x)}{\sqrt{2}} \right) \rightarrow \sin(y) = \frac{\sin(x) + \cos(x)}{\sqrt{2}}
\end{equation}
Now differentiate with respect to '$x$':
\begin{align}
\frac{d}{dx}\left[\sin(y) \right] &= \frac{d}{dx}\left[\frac{\sin(x) + \cos(x)}{\sqrt{2}} \right] \\
\cos(y)\frac{dy}{dx} &= \frac{\cos(x) - \sin(x)}{\sqrt{2}} \\
\frac{dy}{dx} &= \frac{\cos(x) - \sin(x)}{\sqrt{2}\cos(y)} 
\end{align}
Thus:
\begin{equation}
 \frac{dy}{dx} = \frac{d}{dx}\left[\arcsin\left(\frac{\sin(x) + \cos(x)}{\sqrt{2}} \right) \right] = \frac{\cos(x) - \sin(x)}{\sqrt{2}\cos\left(\arcsin\left(\frac{\sin(x) + \cos(x)}{\sqrt{2}}\right) \right)}
\end{equation}
Here this method is unnecessarily complicated in comparison to those already presented. It is however good to know if an identity is either unknown. 
A: $$\frac{d(\arcsin x)}{dx}=\frac{1}{\sqrt{1-x^2}}$$ It is as simple as that. This is followed by the substitution $v(x)=\frac{\sin x+\cos x}{\sqrt{2}}$. $$ \begin{align} \frac{d}{dx}\arcsin(v(x)) & =\frac{1}{\sqrt{1-v^2(x)}}v'(x) \\
&= \frac{1}{\sqrt{1-\Biggl(\frac{\sin x+\cos x}{\sqrt{2}}\Biggl)^2}}\Biggl(\frac{\cos x-\sin x}{\sqrt{2}}\Biggl) \\ 
&= \frac{\sqrt{2}}{\sqrt{(\cos x-\sin x)^2}}\frac{\cos x-\sin x}{\sqrt{2}} \\
&= \frac{\cos x-\sin x}{|\cos x-\sin x|} \\
&= \frac{\cos(x+\frac{\pi}{4})}{|\cos(x+\frac{\pi}{4})|} \\  \end{align}$$
