If $$y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right)$$, then ${{dy} \over {dx}}$ at $x = {{5\pi } \over 6}$ is :
Solution
We have,
<br/><br/>$$
y(x)=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)
$$
<br/><br/>$$
=\cot ^{-1} \frac{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|+\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|-\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}
$$
<br/><br/>$\left[\text { as } \cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}=1\right.$
<br/><br/>and $\left.\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\right]$
<br/><br/>$$
\begin{aligned}
& =\cot ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}}\right) \forall x \in\left(\frac{\pi}{2}, \pi\right) \\
&
\end{aligned}
$$
<br/><br/>$$
=\cot ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)=\cot ^{-1}\left(\tan \frac{x}{2}\right)
$$
<br/><br/>$=\frac{\pi}{2}-\tan ^{-1}\left(\tan \frac{x}{2}\right)$
<br/><br/>$$
\begin{aligned}
& \therefore y^{\prime}(x)=\frac{\pi}{2}-\frac{x}{2} \\\\
& \Rightarrow \frac{d y}{d x}=y^{\prime}(x)=-\frac{1}{2}=y^{\prime}\left(\frac{5 \pi}{6}\right)
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Differentiation · Topic: Derivatives of Standard Functions
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