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Differentiate

Question:

Differentiate

$\left(\frac{2 x^{2}-4}{3 x^{2}+7}\right)$

 

Solution:

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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