Question

3) root(1-cosx) / root(1+cosx) find dy/dx ..? explain step by step please​

Answer 1

Step-by-step explanation:

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Answer 2

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

Given function is

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

Let assume that

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

We know that

$\boxed{ \tt{ \: 1 - cos2x = {2sin}^{2}x \: }} \\ \\ \bf \: and \\ \\ \boxed{ \tt{ \: 1 + cos2x = {2cos}^{2}x \: }}$

So, using these Identities, we get

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

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

$\rm :\longmapsto\:y = tan\dfrac{x}{2}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} tan\dfrac{x}{2}$

We know that

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

So, using this, we get

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2} {sec}^{2}\dfrac{x}{2} \: \dfrac{d}{dx} \: x$

We know,

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

So, using this, we get

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

$\\ \bf\implies \:\:\dfrac{dy}{dx} = \dfrac{1}{2} {sec}^{2}\dfrac{x}{2}$

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More to know :-

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