Question

Solve it
$2y = ( {cot }^{ - 1}( \frac{ \sqrt{3}cosx + sinx }{cosx - \sqrt{3}sinx }) {)}^{2}$
where x belongs to(0, π/2), find dy/dx



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Answer 1

Answer:

$x \: - \: \dfrac{\pi}{6}$

Step-by-step explanation:

To find--->

$derivative \: of \:$

$2y = ( \: {cot}^{ - 1}( \dfrac{ \sqrt{3}cosx + sinx }{cosx \: - \sqrt{3}sinx }) \: ) ^{2}$

Concept used ---->

1)

$cos( \alpha - \beta ) = cos \alpha \: cos \beta \: + sin \alpha \: sin \beta$

2)

$\sin( \alpha - \beta ) = sin \alpha \: cos \beta - cos \alpha \: sin \beta$

3)

${cot}^{ - 1} cotx \: = x$

4)

$\dfrac{d}{dx}( x \: ^{n} ) \: = nx ^{n - 1}$

Solution---->

$2y \: = (cot ^{ - 1} \: ( \dfrac{ \sqrt{3}cosx + sinx }{cosx - \sqrt{3} sinx}) \: ) ^{2}$

$= > 2y = ( \: \: {cot}^{ - 1} \: \dfrac{ \dfrac{ \sqrt{3} }{2}cosx + \dfrac{1}{2}sinx }{ \dfrac{1}{2}cosx - \dfrac{ \sqrt{3} }{2}sinx } \: \: ) ^{2}$

$= ( \: \: {cot}^{1} \: \dfrac{cosx \: cos \dfrac{\pi}{6} + sinx \: sin \dfrac{\pi}{6} }{sin \dfrac{\pi}{6} \: cosx - sinx \: cos \dfrac{\pi}{6} \: \: } \: \: ) ^{2}$

$= ( \: \: {cot}^{ - 1} \dfrac{cos( \dfrac{\pi}{6} - x)}{sin( \dfrac{\pi}{6} - x) } \: \: ) ^{2}$

$= ( {cot}^{ - 1} \: cot( \dfrac{\pi}{6} - x) \: ) ^{2}$

$= {( \: \dfrac{\pi}{6} \: - x \: ) }^{2}$

$differentiating \: with \: respect \: to \: x$

$2 \dfrac{dy}{dx} \: = \dfrac{d}{dx} {( \dfrac{\pi}{6} - x) }^{2}$

$= 2 \: {( \dfrac{\pi}{6} - x) }^{2 - 1} \dfrac{d}{dx} ( \dfrac{\pi}{6} - x)$

$\dfrac{dy}{dx} =( \dfrac{\pi}{6} - x) \: (0 - 1)$

$\dfrac{dy}{dx} = x - \dfrac{\pi}{6}$

Answer 2

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