First slide

Derivatives

Question

$If 2 y = {({Cot}^{- 1} (\frac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x}))}^{2} x \in (0, \frac{π}{2}) then \frac{d y}{d x} is equal to$

Moderate

A

$2 x - \frac{π}{3}$
B

$\frac{π}{3} - x$
C

$\frac{π}{6} - x$
D

$x - \frac{π}{6}$

Solution

$\begin{array}{l} \cot^{- 1} (\frac{\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x}{\frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x}) = \cot^{- 1} (\frac{\sin (x + \frac{π}{3})}{\cos (x + \frac{π}{3})}) \\ = \cot^{- 1} (\tan (x + \frac{π}{3})) = \frac{π}{2} - \tan^{- 1} (\tan (x + \frac{π}{3})) \\ \frac{π}{2} - (x + \frac{π}{3}) = \frac{π}{6} - x 0 < x < \frac{π}{6} \\ \frac{π}{2} - ((x - \frac{π}{3}) - π) = \frac{7 π}{6} - x \frac{π}{6} < x < \frac{π}{2} \\ 2 y = \{\begin{cases} {(\frac{π}{6} - x)}^{2} 0 < x < \frac{π}{6} \\ {(\frac{7 π}{6} - x)}^{2} \frac{π}{6} < x < \frac{π}{2} \end{cases} \\ 2 \frac{d y}{d x} = \{\begin{array}{lc} 2 (\frac{π}{6} - x) (- 1) 0 < x < \frac{π}{6} \\ 2 (\frac{7 π}{6} - x) (- 1) \frac{π}{6} < x < \frac{π}{2} \end{array} \end{array}$

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