Official solution
f(x) = sin4x + cos4x
\[ \therefore \] f'(x) = 4sin3x cosx + 4cos3x (- sinx)
= 4sinx cosx (sin2x - cos2x)
= - 2sin2x cos2x
= - sin4x
As, f(x) is increasing function when f'(x) > 0
\[ \Rightarrow \] - sin4x > 0
\[ \Rightarrow \] sin4x < 0
\[ \therefore \] \[\pi \] < 4x < 2\[\pi \]
\[{\pi \over 4} < x < {\pi \over 2}\]
\[ \therefore \] x \[ \in \] \[\left( {{\pi \over 4},{\pi \over 2}} \right)\]