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!

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. 
