Differentiate:

To find: Differentiation of $\left(x^{2}-4 x+5\right)\left(x^{3}-2\right)$

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

(ii)

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

Let us take $u=\left(x^{2}-4 x+5\right)$ and $v=\left(x^{3}-2\right)$

$u^{\prime}=\frac{d u}{d x}=\frac{d\left(x^{2}-4 x+5\right)}{d x}=2 x-4$

$v^{\prime}=\frac{d v}{d x}=\frac{d\left(x^{3}-2\right)}{d x}=3 x^{2}$

Putting the above obtained values in the formula:-

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

$\left.\left(x^{2}-4 x+5\right)\left(x^{3}-2\right)\right]^{\prime}=(2 x-4) \times\left(x^{3}-2\right)+\left(x^{2}-4 x+5\right) \times\left(3 x^{2}\right)$

$=2 x^{4}-4 x-4 x^{3}+8+3 x^{4}-12 x^{3}+15 x^{2}$

$=5 x^{4}-16 x^{3}+15 x^{2}-4 x+8$

Ans) $5 x^{4}-16 x^{3}+15 x^{2}-4 x+8$