Differentiate

To find: Differentiation of $\left(\frac{5 x^{2}+6 x+7}{2 x^{2}+3 x+4}\right)$

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 $u=\left(5 x^{2}+6 x+7\right)$ and $v=\left(2 x^{2}+3 x+4\right)$

$u^{\prime}=\frac{d u}{d x}=\frac{d\left(5 x^{2}+6 x+7\right)}{d x}=10 x+6$

$v^{\prime}=\frac{d v}{d x}=\frac{d\left(2 x^{2}+3 x+4\right)}{d x}=4 x+3$

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)

$\left(\frac{5 x^{2}+6 x+7}{2 x^{2}+3 x+4}\right)^{\prime}=\frac{(10 x+6) \times\left(2 x^{2}+3 x+4\right)-\left(5 x^{2}+6 x+7\right) \times(4 x+3)}{\left(2 x^{2}+3 x+4\right)^{2}}$

$=\frac{20 x^{3}+30 x^{2}+40 x+12 x^{2}+18 x+24-20 x^{3}-15 x^{2}-24 x^{2}-18 x-28 x-21}{\left(2 x^{2}+3 x+4\right)^{2}}$

$=\frac{3 x^{2}+12 x+3}{\left(2 x^{2}+3 x+4\right)^{2}}$

$=\frac{3\left(x^{2}+4 x+1\right)}{\left(2 x^{2}+3 x+4\right)^{2}}$

Ans $)=\frac{3\left(x^{2}+4 x+1\right)}{\left(2 x^{2}+3 x+4\right)^{2}}$