Differentiate

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

Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

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

Let us take $u=\left(x^{2}+3 x-1\right)$ and $v=(x+2)$

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

$v^{\prime}=\frac{d v}{d x}=\frac{d(x+2)}{d x}=1$

Putting the above obtained values in the formula:-

$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

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

$=\frac{2 x^{2}+7 x+6-x^{2}-3 x+1}{(x+2)^{2}}$

$=\frac{x^{2}+4 x+7}{(x+2)^{2}}$

Ans $)=\frac{x^{2}+4 x+7}{(x+2)^{2}}$