Question

Differentiative (2x³+3)(x²+4) with respect x​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:( {2x}^{3} + 3)( {x}^{2} + 4)$

Let assume that

$\rm :\longmapsto\:y = ( {2x}^{3} + 3)( {x}^{2} + 4)$

$\rm :\longmapsto\:y = {2x}^{5} + {8x}^{3} + {3x}^{2} + 12$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx} [ {2x}^{5} + {8x}^{3} + {3x}^{2} + 12]$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{d}{dx}{2x}^{5} + \dfrac{d}{dx} {8x}^{3} + \dfrac{d}{dx} {3x}^{2} + \dfrac{d}{dx} 12$

$\rm :\longmapsto\:\dfrac{dy}{dx} =2 \dfrac{d}{dx}{x}^{5} + 8\dfrac{d}{dx} {x}^{3} + 3\dfrac{d}{dx} {x}^{2} + \dfrac{d}{dx} 12$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 2( {5x}^{5 - 1}) + 8( {3x}^{3 - 1}) + 3( {2x}^{2 - 1} ) + 0$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 10{x}^{4} + 24{x}^{2} + 6{x}^{1}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 10{x}^{4} + 24{x}^{2} + 6x$

Hence,

$\rm \implies\:\boxed{\tt{ \dfrac{dy}{dx} = 10{x}^{4} + 24{x}^{2} + 6x \: }}$

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Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$