Differentiate

To find: Differentiation of $\frac{\left(x^{2}-1\right)}{\left(x^{2}+7 x+1\right)}$

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 $u=\left(x^{2}-1\right)$ and $v=\left(x^{2}+7 x+1\right)$

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

$v^{\prime}=\frac{d v}{d x}=\frac{d\left(x^{2}+7 x+1\right)}{d x}=2 x+7$

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)

$\left[\frac{\left(x^{2}-1\right)}{\left(x^{2}+7 x+1\right)}\right]=\frac{2 x \times\left(x^{2}+7 x+1\right)-\left(x^{2}-1\right) \times(2 x+7)}{\left(x^{2}+7 x+1\right)^{2}}$

$=\frac{2 x^{3}+14 x^{2}+2 x-2 x^{3}-7 x^{2}+2 x+7}{\left(x^{2}+7 x+1\right)^{2}}$

$=\frac{7 x^{2}+4 x+7}{\left(x^{2}+7 x+1\right)^{2}}$

Ans $)=\frac{7 x^{2}+4 x+7}{\left(x^{2}+7 x+1\right)^{2}}$