Differentiate:

To find: Differentiation of $\left(x^{2}+3 x+1\right) \sin x$

Formula used: (i) (uv)' $=u^{\prime} v+u v^{\prime}$ (Leibnitz or product rule)

(ii)

$\frac{d x^{n}}{d x}=n x^{n-1}$

(iii)

$\frac{d \sin x}{d x}=\cos x$

Let us take $u=x^{2}+3 x+1$ and $v=\sin x$

$u^{\prime}=\frac{d u}{d x}=\frac{d\left(x^{2}+3 x+1\right)}{d x}=2 x+3$

$v^{\prime}=\frac{d v}{d x}=\frac{d \sin x}{d x}=\cos x$

Putting the above obtained values in the formula :-

$(u v)^{\prime}=u^{\prime} v+u v^{\prime}$

$\left[\left(x^{2}+3 x+1\right) \sin x\right]^{\prime}=(2 x+3) \times \sin x+\left(x^{2}+3 x+1\right) \times \cos x$

$=\sin x(2 x+3)+\cos x\left(x^{2}+3 x+1\right)$

Ans) $(2 x+3) \sin x+\left(x^{2}+3 x+1\right) \cos x$