Question

Differentiate 4x^3 + 4x with respect to x^2 +2

Answer 1

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

Let assume that

$\rm \: u = {4x}^{3} + 4x$

and

$\rm \: v = {x}^{2} + 2$

Now, Consider

$\rm \: u = {4x}^{3} + 4x$

Differentiate both sides w. r. t. x we get

$\rm \: \dfrac{d}{dx}u = \dfrac{d}{dx}({4x}^{3} + 4x)$

$\rm \: \dfrac{du}{dx} = 4\dfrac{d}{dx}{x}^{3} + 4\dfrac{d}{dx}x$

We know,

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

So, using this result, we get

$\rm \: \dfrac{du}{dx} = 4( {3x}^{2}) + 4 \times 1$

$\bf\implies \:\dfrac{du}{dx} = 12 {x}^{2} + 4$

Now, Consider

$\rm \: v = {x}^{2} + 2$

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

$\rm \: \dfrac{d}{dx}v = \dfrac{d}{dx}({x}^{2} + 2)$

$\rm \: \dfrac{dv}{dx} = \dfrac{d}{dx}{x}^{2} + \dfrac{d}{dx}2$

$\rm \: \dfrac{dv}{dx} = {2x}^{2 - 1} + 0$

$\bf\implies \:\dfrac{dv}{dx} = 2x$

Now, Consider

$\rm \: \dfrac{du}{dv}$

$\rm \: = \: \dfrac{du}{dx} \div \dfrac{dv}{dx}$

$\rm \: = \: \dfrac{12 {x}^{2} + 4}{2x}$

$\rm \: = \: 6x + \dfrac{2}{x}$

Hence,

$\rm \: \bf\implies \:\dfrac{du}{dv}= \: 6x + \dfrac{2}{x}$

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

Additional Information :-

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