Question

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Differentiate the following function with respect to x

$\sqrt{ \frac{1 - cosmx}{1 + cosmx} }$

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Answer 1

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

Given function is

$\rm :\longmapsto\: \sqrt{\dfrac{1 - cosmx}{1 + cosmx} }$

Let assume that

$\rm :\longmapsto\: y = \sqrt{\dfrac{1 - cosmx}{1 + cosmx} }$

We know,

$\boxed{ \tt{ \: 1 - cos2x = {2sin}^{2}x \: }}$

and

$\boxed{ \tt{ \: 1 + cos2x = {2cos}^{2}x \: }}$

So, using these, we get

$\rm :\longmapsto\:y = \sqrt{\dfrac{ {2sin}^{2}mx}{ {2cos}^{2}mx }}$

$\rm :\longmapsto\:y = \sqrt{\dfrac{ {sin}^{2}mx}{ {cos}^{2}mx }}$

$\rm :\longmapsto\:y = \sqrt{ {tan}^{2} mx}$

$\rm :\longmapsto\:y = tan \: mx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}tan \: mx$

We know,

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

So, using this we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {sec}^{2}mx \: \dfrac{d}{dx}mx$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {sec}^{2}mx \: \times m\dfrac{d}{dx}x$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}x = 1 \: \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = m \: {sec}^{2}mx \: \times 1$

$\bf\implies \:\boxed{ \tt{ \: \dfrac{dy}{dx} = m \: {sec}^{2}mx \: \: }}$

More to know :-

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

Answer 2

Answer:

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

Given function is

$\rm :\longmapsto\: \sqrt{\dfrac{1 - cosmx}{1 + cosmx} }$

Let assume that

$\rm :\longmapsto\: y = \sqrt{\dfrac{1 - cosmx}{1 + cosmx} }$

We know,

$\boxed{ \tt{ \: 1 - cos2x = {2sin}^{2}x \: }}$

and

$\boxed{ \tt{ \: 1 + cos2x = {2cos}^{2}x \: }}$

So, using these, we get

$\rm :\longmapsto\:y = \sqrt{\dfrac{ {2sin}^{2}mx}{ {2cos}^{2}mx }}$

$\rm :\longmapsto\:y = \sqrt{\dfrac{ {sin}^{2}mx}{ {cos}^{2}mx }}$

$\rm :\longmapsto\:y = \sqrt{ {tan}^{2} mx}$

$\rm :\longmapsto\:y = tan \: mx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}tan \: mx$

We know,

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

So, using this we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {sec}^{2}mx \: \dfrac{d}{dx}mx$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {sec}^{2}mx \: \times m\dfrac{d}{dx}x$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}x = 1 \: \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = m \: {sec}^{2}mx \: \times 1$

$\bf\implies \:\boxed{ \tt{ \: \dfrac{dy}{dx} = m \: {sec}^{2}mx \: \: }}$

More to know :-

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