$(3 x-5)\left(4 x^{2}-3+e^{x}\right)$ 


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

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


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


$\frac{d e^{x}}{d x}=e^{x}$

Let us take $u=(3 x-5)$ and $v=\left(4 x^{2}-3+e^{x}\right)$

$u^{\prime}=\frac{d u}{d x}=\frac{d(3 x-5)}{d x}=3$

$v^{\prime}=\frac{d v}{d x}=\frac{d\left(4 x^{2}-3+e^{x}\right)}{d x}=\left(8 x+e^{x}\right)$

Putting the above obtained values in the formula :-

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

$\left[(3 x-5)\left(4 x^{2}-3+e^{x}\right)\right]^{\prime}=3 x\left(4 x^{2}-3+e^{x}\right)+(3 x-5) \times\left(8 x+e^{x}\right)$

$=12 x^{2}-9+3 e^{x}+24 x^{2}+3 x e^{x}-40 x-5 e^{x}$

$=36 x^{2}+x\left(3 e^{x}-40\right)-9-2 e^{x}$

Ans) $36 x^{2}+x\left(3 e^{x}-40\right)-9-2 e^{x}$


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