[tex]\sf if\: 2y = \bigg(cot^{-1}\bigg(\dfrac{\sqrt{3}cosx+sinx}{cosx-\sqrt{3}sinx}\bigg)\bigg)^2,x\in\bigg(0,\dfrac{\pi}{2}\bigg)[/tex] Then dy/dx is equal to About the author Savannah
[tex]\begin{gathered}\Large{\sf{{\underline{Formula \: Used – }}}} \end{gathered}[/tex] [tex] \boxed{ \bf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{ n- 1} }}[/tex] [tex] \boxed{ \bf{ \: \dfrac{d}{dx}x = 1}}[/tex] [tex] \boxed{ \bf{ \: {tan}^{ – 1} x + {cot}^{ – 1}x = \dfrac{\pi}{2} }}[/tex] [tex] \boxed{ \bf{ \: {tan}^{ – 1}x + {tan}^{ – 1} y = {tan}^{ – 1} \bigg( \dfrac{x + y}{1 – xy} \bigg) }}[/tex] [tex] \boxed{ \bf{ \: {tan}^{ – 1} (tanx) = x}}[/tex] [tex]\large\underline{\sf{Solution-}}[/tex] [tex] \rm :\longmapsto\:\sf \: 2y = \bigg(cot^{-1}\bigg(\dfrac{\sqrt{3}cosx+sinx}{cosx-\sqrt{3}sinx}\bigg)\bigg)^2[/tex] We know, [tex] \boxed{ \bf{ \: {cot}^{ – 1} x = \dfrac{\pi}{2} – {tan}^{ – 1} x}}[/tex] So, Given expression can be rewritten as [tex]\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} – tan^{-1}\bigg(\dfrac{\sqrt{3}cosx+sinx}{cosx-\sqrt{3}sinx}\bigg)\bigg)^2[/tex] [tex]\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} – tan^{-1}\bigg(\dfrac{\sqrt{3} \: \dfrac{cosx}{cosx} +\dfrac{sinx}{cosx} }{1-\sqrt{3} \times \dfrac{sinx}{cosx} }\bigg)\bigg)^2[/tex] [tex]\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} – tan^{-1}\bigg(\dfrac{\sqrt{3}+tanx}{1-\sqrt{3}tanx}\bigg)\bigg)^2[/tex] Now, We know that [tex] \boxed{ \bf{ \: {tan}^{ – 1} \bigg(\dfrac{x + y}{1 – xy} \bigg) = {tan}^{ – 1}x + {tan}^{ – 1}y}} [/tex] Therefore, [tex]\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} – \bigg( tan^{-1} \sqrt{3} + {tan}^{ – 1} (tanx) \bigg)\bigg)^2[/tex] [tex]\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} – \dfrac{\pi}{3} – x\bigg)^2[/tex] [tex]\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{6} -x \bigg)^2[/tex] On differentiating both sides w. r. t. x, we get [tex]\rm :\longmapsto\: \dfrac{d}{dx} (2y) =\dfrac{d}{dx} \bigg(\dfrac{\pi}{6} -x \bigg)^2[/tex] [tex]\rm :\longmapsto\: \cancel2\dfrac{dy}{dx} = \cancel2 \bigg(\dfrac{\pi}{6} -x \bigg)\dfrac{d}{dx}\bigg(\dfrac{\pi}{6} -x \bigg)[/tex] [tex]\rm :\longmapsto\: \dfrac{dy}{dx} = \bigg(\dfrac{\pi}{6} -x \bigg)(0 – 1)[/tex] [tex]\bf\implies \:\dfrac{dy}{dx} = x – \dfrac{\pi}{6} [/tex] Reply
[tex]good \: morning \: dear \: \\ have \: a \: grt \: day \: ahead[/tex]
[tex]\begin{gathered}\Large{\sf{{\underline{Formula \: Used – }}}} \end{gathered}[/tex]
[tex] \boxed{ \bf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{ n- 1} }}[/tex]
[tex] \boxed{ \bf{ \: \dfrac{d}{dx}x = 1}}[/tex]
[tex] \boxed{ \bf{ \: {tan}^{ – 1} x + {cot}^{ – 1}x = \dfrac{\pi}{2} }}[/tex]
[tex] \boxed{ \bf{ \: {tan}^{ – 1}x + {tan}^{ – 1} y = {tan}^{ – 1} \bigg( \dfrac{x + y}{1 – xy} \bigg) }}[/tex]
[tex] \boxed{ \bf{ \: {tan}^{ – 1} (tanx) = x}}[/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
[tex] \rm :\longmapsto\:\sf \: 2y = \bigg(cot^{-1}\bigg(\dfrac{\sqrt{3}cosx+sinx}{cosx-\sqrt{3}sinx}\bigg)\bigg)^2[/tex]
We know,
[tex] \boxed{ \bf{ \: {cot}^{ – 1} x = \dfrac{\pi}{2} – {tan}^{ – 1} x}}[/tex]
So,
Given expression can be rewritten as
[tex]\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} – tan^{-1}\bigg(\dfrac{\sqrt{3}cosx+sinx}{cosx-\sqrt{3}sinx}\bigg)\bigg)^2[/tex]
[tex]\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} – tan^{-1}\bigg(\dfrac{\sqrt{3} \: \dfrac{cosx}{cosx} +\dfrac{sinx}{cosx} }{1-\sqrt{3} \times \dfrac{sinx}{cosx} }\bigg)\bigg)^2[/tex]
[tex]\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} – tan^{-1}\bigg(\dfrac{\sqrt{3}+tanx}{1-\sqrt{3}tanx}\bigg)\bigg)^2[/tex]
Now,
We know that
[tex] \boxed{ \bf{ \: {tan}^{ – 1} \bigg(\dfrac{x + y}{1 – xy} \bigg) = {tan}^{ – 1}x + {tan}^{ – 1}y}} [/tex]
Therefore,
[tex]\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} – \bigg( tan^{-1} \sqrt{3} + {tan}^{ – 1} (tanx) \bigg)\bigg)^2[/tex]
[tex]\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} – \dfrac{\pi}{3} – x\bigg)^2[/tex]
[tex]\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{6} -x \bigg)^2[/tex]
On differentiating both sides w. r. t. x, we get
[tex]\rm :\longmapsto\: \dfrac{d}{dx} (2y) =\dfrac{d}{dx} \bigg(\dfrac{\pi}{6} -x \bigg)^2[/tex]
[tex]\rm :\longmapsto\: \cancel2\dfrac{dy}{dx} = \cancel2 \bigg(\dfrac{\pi}{6} -x \bigg)\dfrac{d}{dx}\bigg(\dfrac{\pi}{6} -x \bigg)[/tex]
[tex]\rm :\longmapsto\: \dfrac{dy}{dx} = \bigg(\dfrac{\pi}{6} -x \bigg)(0 – 1)[/tex]
[tex]\bf\implies \:\dfrac{dy}{dx} = x – \dfrac{\pi}{6} [/tex]