Question

1-COS X/1+ COS X

differentiate the following​

Answer 1

Answer:-

We have to differentiate:

$\implies \sf \: \dfrac{1 - \cos(x) }{1 + \cos(x) }$

Differentiating w.r.t x , we get;

$\implies \sf \: \dfrac{d}{dx} \bigg[ \dfrac{1 - \cos(x) }{1 + \cos(x) } \bigg ]$

using ;

$\sf \dfrac{d}{dx} \bigg(\dfrac{u}{v} \bigg) = \dfrac{v \bigg( \dfrac{du}{dx} \bigg) - u \bigg( \dfrac{dv}{dx} \bigg) }{ {v}^{2} }$

We get;

$\\ \implies \sf \: \frac{(1 + \cos(x) ) \bigg( \dfrac{d}{dx}(1 - \cos(x)) \bigg) - (1 - \cos(x)) \bigg( \dfrac{d}{dx}(1 + \cos(x)) \bigg)}{ {(1 + \cos(x)) \ }^{2} }$

using;

d/dx (u ± v...) = d/dx (u) ± d/dx (v) +....

we get,

$\\ \implies \sf \: \frac{(1 + \cos(x) ) \bigg( \dfrac{d}{dx}(1)- \dfrac{d}{dx}( \cos(x)) \bigg) - (1 - \cos(x)) \bigg( \dfrac{d}{dx}(1) + \dfrac{d}{dx} (\cos(x)) \bigg)}{ {(1 + \cos(x)) \ }^{2} }$

using, d/dx (constant) = 0 & d/dx (cos x) = - sin x we get,

$\\ \implies \sf \frac{(1 + \cos(x)) \big( -( - \sin(x) \big) - (1 - \cos(x)) \big( - \sin(x) \big) }{ {(1 + \cos(x) )}^{2} } \\ \\ \\ \implies \sf \: \frac{(1 + \cos(x))( \sin(x)) - (1 - \cos(x))( - \sin(x) ) }{ {(1 + \cos(x)) }^{2} } \\ \\ \\ \implies \sf \: \frac{ \sin(x) + \sin(x) . \cos(x) - ( - \sin(x) + \sin(x). \cos(x) ) }{ {(1 + \cos(x) )}^{2} } \\ \\ \\ \implies \sf \: \frac{ \sin(x) + \sin(x). \cos(x) + \sin(x) - \sin(x). \cos(x) }{ {(1 + \cos(x)) }^{2} } \\ \\ \\ \implies \boxed{ \red{ \sf \frac{2 \sin (x)}{ {(1 + \cos(x)) }^{2} }} }$

Answer 2

$\large\underline\blue{\bold{Given \: }}$

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

$\begin{gathered}\begin{gathered}\bf \: To \: find \:  - \begin{cases} &\sf{\dfrac{dy}{dx}}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{\green{\underline{Formula \: Used \::}}}} \end{gathered}$

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

$\rm :\implies\: \boxed{ \pink{ \bf \:\dfrac{d}{dx} {x}^{2} \: = \tt \:2x }}$

$\rm :\implies\: \boxed{ \pink{ \bf \: \dfrac{d}{dx}x\: = \tt \: 1}}$

$\rm :\implies\: \boxed{ \pink{ \bf \: 1 - cos2x\: = \tt \: 2 {sin}^{2}x }}$

$\rm :\implies\: \boxed{ \pink{ \bf \: 1 + cos2x \: = \tt \: 2 {cos}^{2}x }}$

$\large\underline\purple{\bold{Solution :- }}$

$\rm :\implies\:Let \: y \: = \: \dfrac{1 - cosx}{1 + cosx}$

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

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

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

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

$\rm :\implies\:\dfrac{dy}{dx} = 2 \: tan\dfrac{x}{2} \: \dfrac{d}{dx}(tan\dfrac{x}{2})$

$\rm :\implies\:\dfrac{dy}{dx} = 2 \: tan\dfrac{x}{2} \: {sec}^{2} \dfrac{x}{2} \: \dfrac{d}{dx}(\dfrac{x}{2} )$

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

$\rm :\implies\: \boxed{ \pink{ \bf \: \dfrac{dy}{dx} \: = \tt \: tan\dfrac{x}{2} \: {sec}^{2}\dfrac{x}{2} }}$