Question

differentiate 4^x cosec x (can anyone help with step by step solution)​

Answer 1

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

Given function is

$\rm :\longmapsto\: {4}^{x} \: cosecx \:$

Let assume that

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

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

$\rm :\longmapsto\: \dfrac{d}{dx}y \: =\dfrac{d}{dx} [ \: {4}^{x} \: cosecx \: ]$

We know,

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

So, using this, we get

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

We know

$\boxed{ \tt{ \: \dfrac{d}{dx} {a}^{x} \: = \: {a}^{x} \: loga \: }} \\ \\ and \\ \\ \boxed{ \tt{ \: \dfrac{d}{dx}cosecx = - cosecx \: cotx}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {4}^{x}[ - cosecx \: cotx] \: + \: cosecx \: {4}^{x} \: log4$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {4}^{x}cosecx \:[ - cotx\: + \: log4 \: ]$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to know :-

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