Question

Find (dy)/(dx) of y=cosec(tan sqrt(x)).​

Answer 1

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

Given function is

$\rm :\longmapsto\:y = cosec(tan \sqrt{x})$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} cosec(tan \sqrt{x})$

We know,

$\red{ \boxed{ \sf{ \:\dfrac{d}{dx}cosecx = - \: cosecx \: cotx \: \: }}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: cosec(tan \sqrt{x}) \: cot(tan \sqrt{x})\dfrac{d}{dx}tan \sqrt{x}$

We know,

$\red{ \boxed{ \sf{ \:\dfrac{d}{dx}tanx = {sec}^{2}x}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: cosec(tan \sqrt{x}) \: cot(tan \sqrt{x}) {sec}^{2} \sqrt{x} \dfrac{d}{dx} \sqrt{x}$

We know,

$\red{ \boxed{ \sf{ \:\dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: cosec(tan \sqrt{x}) \: cot(tan \sqrt{x}) {sec}^{2} \sqrt{x} \times \dfrac{1}{2 \sqrt{x} }$

Therefore,

$\red{ \boxed{ \sf{ \:\dfrac{dy}{dx} = - \: \frac{cosec(tan \sqrt{x}) \: (cot(tan \sqrt{x}) \: {sec}^{2} \sqrt{x} }{2 \sqrt{x} }}}}$

Additional Information :-

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