Question

find dy/dx (2x-1)(3x+1)(4x2 -3)​

Answer 1

(2×-1) (3×+1) (4×2-3) =-30

Answer 2

Appropriate Question :-

$\rm \: Find \: \dfrac{dy}{dx} \: if \: y = (2x - 1)(3x + 1)( {4x}^{2} - 3)$

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

Given expression is

$\rm \: y = (2x - 1)(3x + 1)( {4x}^{2} - 3)$

$\rm \: y = ( {6x}^{2} - 3x + 2x - 1))( {4x}^{2} - 3)$

$\rm \: y = ( {6x}^{2} - x - 1))( {4x}^{2} - 3)$

$\rm \: y = {4x}^{2} ( {6x}^{2} - x - 1)) - 3( {6x}^{2} - x - 1)$

$\rm \: y = {24x}^{4} - {4x}^{3} - {4x}^{2} - {18x}^{2} + 3x + 3$

$\rm \: y = {24x}^{4} - {4x}^{3} - {22x}^{2} + 3x + 3$

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

$\rm \: \dfrac{d}{dx}y = \dfrac{d}{dx}\bigg( {24x}^{4} - {4x}^{3} - {22x}^{2} + 3x + 3 \bigg)$

$\rm \: \dfrac{dy}{dx} = 24\dfrac{d}{dx}{x}^{4} - 4\dfrac{d}{dx}{x}^{3} - 22\dfrac{d}{dx}{x}^{2} + 3\dfrac{d}{dx}x + \dfrac{d}{dx}3$

We know,

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

So, on applying this result, we get

$\rm \: \dfrac{dy}{dx} = 24( {4x}^{3}) - 4( {3x}^{2}) - 22(2x) + 3(1) + 0$

$\rm \: \dfrac{dy}{dx} = {96x}^{3} - {12x}^{2} - 44x + 3$

Hence,

$\rm\implies \:\boxed{\rm{  \:\rm \: \dfrac{dy}{dx} = {96x}^{3} - {12x}^{2} - 44x + 3 \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

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