If $$f\left( x \right) = {\left( {\frac{{{{\sin }^m}x}}{{{{\sin }^n}x}}} \right)^{m + n}}.{\left( {\frac{{{{\sin }^n}x}}{{{{\sin }^p}x}}} \right)^{n + p}}.{\left( {\frac{{{{\sin }^p}x}}{{{{\sin }^m}x}}} \right)^{p + m}},$$          then $$f'\left( x \right)$$  is equal to :

Solution :
$$\eqalign{ & {\text{We have,}} \cr & f\left( x \right) = {\left( {{{\sin }^{m - n}}x} \right)^{m + n}}.{\left( {{{\sin }^{n - p}}x} \right)^{n + p}}.{\left( {{{\sin }^{p - m}}x} \right)^{p + m}} \cr & = {\sin ^{{m^2} - {n^2}}}x.{\sin ^{{n^2} - {p^2}}}x.{\sin ^{{p^2} - {m^2}}}x \cr & = {\left( {\sin \,x} \right)^{{m^2} - {n^2} + {n^2} - {p^2} + {p^2} - {m^2}}} \cr & = {\left( {\sin \,x} \right)^0} \cr & = 1 \cr & \therefore \,\,f'\left( x \right) = 0 \cr} $$