$\frac{\mathrm{d}}{\mathrm{dx}}\left({\sin }^{-1} \left(\frac{3\cos \mathrm{x}+4\sin \mathrm{x}}{5}\right)\right)=$

The given differentiation is 

$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{dx}}\left({\sin }^{-1} \left(\frac{3\cos \mathrm{x}+4\sin \mathrm{x}}{5}\right)\right)& =& \frac{\mathrm{d}}{\mathrm{dx}}\left({\sin }^{-1} \left(\frac{3}{5}\cos x+\frac{4}{5}\sin x\right)\right)\\ & =& \frac{d}{dx}\left({\sin }^{-1}\left(\sin A\cos x+\cos A\sin x\right)\right)\end{array}$

Where $SinA=\frac{3}{5},\cos A=\frac{4}{5}$ and $\tan A=\frac{3}{4}$

Hence, 

$\begin{array}{rcl}\frac{d}{dx}\left({\sin }^{-1}\left(\sin \left(A+x\right)\right)\right)& =& \frac{d}{dx}\left(A+x\right)\\ & =& \frac{d}{dx}\left({\tan }^{-1}\left(\frac{3}{4}\right)+x\right)\\ & =& 0+1\\ & =& 1\end{array}$