Ta có
$y' = \left( \sqrt{\sin^4x + 4 \cos(4x)} \right)'$
$= \left[(\sin^4x + 4 \cos(4x))^{\frac{1}{2}} \right]'$
$= \dfrac{[\sin^4x + 4\cos(4x)]'}{2\sqrt{\sin^4x + 4 \cos(4x)}}$
$= \dfrac{4\sin^3x . (\sin x)' -16\sin(4x)}{2\sqrt{\sin^4x + 4 \cos(4x)}}$
$= \dfrac{2\sin^3x \cos x - 8\sin(4x)}{\sqrt{\sin^4x + 4 \cos(4x)}}$
Vậy
$y' = \dfrac{2\sin^3x \cos x - 8\sin(4x)}{\sqrt{\sin^4x + 4 \cos(4x)}}$.
