$y=\sin px + \cos px $ $ y=\left [ (\cos px + \sin px)^2 \right ]^{\frac {1}{2}} $ $y=\left [ \cos^2 px + 2 \cos px \sin px + \sin^2 px \right ]^{\frac {1}{2}}$ $\cos px+ \sin px = \left [ 1+\sin 2 px \right ]^{\frac{1}{2}} \ \cdots \ \cdots (A)$ $SImilarly , \ \cos px -\sin px = \left [ 1-\sin 2px \right ] ^{\frac {1}{2}} \ \cdots \ \cdots (B)$ $y_1 = p \cos px - p \sin px \ \cdots \ \cdots (C)$ $y_1 =p (\cos px - \sin px)$ $y_1 =p^1 \left [1-\sin 2px \right ]^{\frac {1}{2}} \ \cdots \ \cdots (D) \ \ \ \ \ { from \ (B)}$ Differentiating (C) to get y2, $y_2= -p^2 \sin px - p^2 \cos px = p^2 \left [ (-1)^2 (\sin px + \cos px)^2 \right ]^{\frac {1}{2}} $ $y_2= p^2 \left [ 1+\sin 2px \right ]^{\frac {1}{2}} \ \cdots \ \cdots (E) \ \ \ \ \ {from \ (A)}$ Differentiating y2, $y_3 = -p^3 \cos px + p^3 \sin px = p^3 \left [ (-1)^2 )(\cos px -\sin px )^2 \right ]^{\frac {1}{2}} $ $y_3=p^3 \left [ 1-\sin 2 px \right ]^{\frac {1}{2}} \ \cdots \ \cdots (F)$ From (D), (E), (F) and continuing in similar way, $y_n=p^n \left [ 1+ (-1)^n \sin 2 px \right ]^{\frac {1}{2}}$