Let f (x) = $x^4$ (5 sin x – 3 cos x) ... (i) Differentiating (i) with respect to x, we get $\frac{d}{dx}$ [f (x)] = $(x^4)'$ [5 sin x − 3 cos x] + $x^4$ [5 sin x − 3 cos x]′ = $4x^3$ [5 sin x – 3 cos x] + $x64$ [5 cos x + 3 sin x] = 20 $x^3$ sin x – $12x^3$ cos x + $5x^4$ cos x + $3x^4$ sinx = $x^3$ sin x (20 + 3x) + $x^3$ cos x (5x – 12)