Question

find dy/dx if y=x ⁴sin3x​

Answer 1

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

$\rm :\longmapsto\:\boxed{ \tt{ \: y = {x}^{4}sin3x}}$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{dy}{dx} \: \: }}$

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

Given that,

$\rm :\longmapsto\:y = {x}^{4}sin3x$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}{x}^{4}sin3x$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}uv = u\dfrac{d}{dx}v + v\dfrac{d}{dx}u \: \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {x}^{4}\dfrac{d}{dx}sin3x + sin3x\dfrac{d}{dx} {x}^{4}$

We know,

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

and

$\boxed{ \tt{ \: \dfrac{d}{dx}sinx = cosx \: \: }}$

So, on substituting these values, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {x}^{4}cos3x\dfrac{d}{dx}3x + sin3x( {4x}^{3})$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 3 {x}^{4}cos3x + 4 {x}^{3}sin3x$

$\bf\implies \:\boxed{ \tt{ \: \dfrac{dy}{dx} = {x}^{3}(3xcos3x + 4sin3x) \: \: }}$

Additional Information :-

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