To find: Differentiation of (x² + 3x + 1) sin x

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

Let us take u = x² + 3x + 1 and v = sin x

Putting the above obtained values in the formula :-

(uv)′ = u′v + uv′

[(x² + 3x + 1) sin x]’ = (2x + 3) × sinx + (x² + 3x + 1) × cosx

= sinx (2x + 3) + cosx (x² + 3x + 1)

Ans) (2x + 3) sinx + (x² + 3x + 1) cosx