If 2y=^-1(√3 x+ x x-√3 x)^2, x(0,2), then dydx is equal to:

Question

If $2y=\cot^{-1}\left(\dfrac{\sqrt3\cos x+\sin x}{\cos x-\sqrt3\sin x}\right)^2$, $x\in\left(0,\dfrac\pi2\right)$, then $\dfrac{dy}{dx}$ is equal to:

Accepted Answer

$$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$$

$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3  + \tan x} \over {1 - \sqrt 3 \tan x}}} \right)} \right)^2}$$

$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\left( {{{\tan {\pi  \over 3} + \tan x} \over {1 - \tan {\pi  \over 3}\tan x}}} \right)} \right)^2}$$

$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\tan \left( {{\pi  \over 3} + x} \right)} \right)^2}$$

$$ \Rightarrow $$ 2y = $${\left( {{\pi  \over 2} - {{\tan }^{ - 1}}\tan \left( {{\pi  \over 3} + x} \right)} \right)^2}$$

As x $$ \in $$ $$\left( {0,{\pi  \over 2}} \right)$$ then

$${{{\tan }^{ - 1}}\tan \left( {{\pi  \over 3} + x} \right)}$$ = $${\left( {{\pi  \over 3} + x} \right)}$$

$$ \Rightarrow $$ 2y = $${\left( {{\pi  \over 2} - \left( {{\pi  \over 3} + x} \right)} \right)^2}$$

$$ \Rightarrow $$ 2y = $${\left( {{\pi  \over 6} - x} \right)^2}$$

$$ \therefore $$ $$2{{dy} \over {dx}} = 2\left( {{\pi  \over 6} - x} \right)\left( { - 1} \right)$$

$$ \Rightarrow $$ $${{dy} \over {dx}} = \left( {x - {\pi  \over 6}} \right)$$

Solutions · 1