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Q.
Let $x, y \in R$ satisfying the equation $\cot ^{-1} x+\cot ^{-1} y+\cot ^{-1}(x y)=\frac{11 \pi}{12}$, then the value of $\frac{d y}{d x}$ at $x =1$ is
Continuity and Differentiability
Solution:
Given, $\cot ^{-1} x +\cot ^{-1} y +\cot ^{-1}( xy )=\frac{11 \pi}{12}$ .......(1)
Put $ x=1$, we get
$\cot ^{-1} 1+\cot ^{-1} y +\cot ^{-1} y =\frac{11 \pi}{12} \Rightarrow 2 \cot ^{-1} y =\left(\frac{11 \pi}{12}-\frac{\pi}{4}\right)=\frac{8 \pi}{12}=\frac{2 \pi}{3} \text {. }$
$\therefore \cot ^{-1} y =\frac{\pi}{3} \Rightarrow y =\frac{1}{\sqrt{3}}$
So, $ P\left(x=1, y=\frac{1}{\sqrt{3}}\right)$.
Now, differentiating both sides of equation (1) with respect to $x$, we get
$\frac{-1}{1+x^2}-\frac{1}{1+y^2} \cdot \frac{d y}{d x}-\frac{1}{1+x^2 y^2} \cdot\left(x \cdot \frac{d y}{d x}+y\right)=0$
Put $x=1, y=\frac{1}{\sqrt{3}}$, we get
$ \frac{-1}{2}-\left(\frac{1}{1+\frac{1}{3}}\right) \frac{ dy }{ dx }-\left(\frac{1}{1+\frac{1}{3}}\right)\left(1 \cdot \frac{ dy }{ dx }+\frac{1}{\sqrt{3}}\right)=0 $
$\Rightarrow \frac{-1}{2}-\frac{3}{4}\left(\frac{1}{\sqrt{3}}\right)=2 \times \frac{3}{4} \cdot\left(\frac{ dy }{ dx }\right)$
$\Rightarrow -\left(\frac{2 \sqrt{3}+3}{4 \sqrt{3}}\right)=\frac{3}{2}\left(\frac{ dy }{ dx }\right)$
$\Rightarrow \frac{ dy }{ dx }=\frac{-2}{3}\left(\frac{2 \sqrt{3}+3}{4 \sqrt{3}}\right)=\frac{-1}{3}-\frac{1}{2 \sqrt{3}}=-\left(\frac{1}{3}+\frac{1}{2 \sqrt{3}}\right) $