Differentiate
$\left(\frac{2 x^{2}-4}{3 x^{2}+7}\right)$
To find: Differentiation of $\frac{\left(2 x^{2}-4\right)}{\left(3 x^{2}+7\right)}$
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 $u=\left(2 x^{2}-4\right)$ and $v=\left(3 x^{2}+7\right)$
$u^{\prime}=\frac{d u}{d x}=\frac{d\left(2 x^{2}-4\right)}{d x}=4 x$
$v^{\prime}=\frac{d v}{d x}=\frac{d\left(3 x^{2}+7\right)}{d x}=6 x$
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)
$\left[\frac{\left(2 x^{2}-4\right)}{\left(3 x^{2}+7\right)}\right]^{\prime}=\frac{4 x \times\left(3 x^{2}+7\right)-\left(2 x^{2}-4\right) \times 6 x}{\left(3 x^{2}+7\right)^{2}}$
$=\frac{12 x^{3}+28 x-12 x^{3}+24 x}{\left(3 x^{2}+7\right)^{2}}$
$=\frac{52 x}{\left(3 x^{2}+7\right)^{2}}$
Ans $)=\frac{52 x}{\left(3 x^{2}+7\right)^{2}}$
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