$\left(x^{2}+2 x-3\right)\left(x^{2}+7 x+5\right)$



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

Formula used: (i) (uv)' $=u^{\prime} v+u v^{\prime}$ (Leibnitz or product rule)


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

Let us take $u=\left(x^{2}+2 x-3\right)$ and $v=\left(x^{2}+7 x+5\right)$

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

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

Putting the above obtained values in the formula :-

$(u v)^{\prime}=u^{\prime} v+u v^{\prime}$

$\left[\left(x^{2}+2 x-3\right)\left(x^{2}+7 x+5\right)\right]^{\prime}$

$=(2 x+2) \times\left(x^{2}+7 x+5\right)+\left(x^{2}+2 x-3\right) \times(2 x+7)$

$=2 x^{3}+14 x^{2}+10 x+2 x^{2}+14 x+10+2 x^{3}+7 x^{2}+4 x^{2}+14 x-6 x-21$

$=4 x^{3}+27 x^{2}+32 x-11$

Ans) $4 x^{3}+27 x^{2}+32 x-11$


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