Let f (x) = x4 (5 sin x – 3 cos x) ... (i)
Differentiating (i) with respect to x, we get

d

dx

[f (x)] = (x4)′ [5 sin x − 3 cos x] + x4 [5 sin x − 3 cos x]′
= 4x3 [5 sin x – 3 cos x] + x64 [5 cos x + 3 sin x]
= 20 x3 sin x – 12x3 cos x + 5x4 cos x + 3x4 sinx
= x3 sin x (20 + 3x) + x3 cos x (5x – 12)