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Let $y$ be an implicit function of $x$ defined by $x^{2 x}-2 x^x \cot y-1=0$. Then $y^{\prime}(1)$ equals
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$-1$
$-1$
$x^{2 x}-2 x^x \cot y-1=0$
Now $x=1$,
$$
1-2 \cot y-1=0 \Rightarrow \cot y=0 \Rightarrow y=\frac{\pi}{2}
$$
Now differentiating eq. (1) w.r.t. ' $x$ '
$$
2 x^{2 x}(1+\log x)-2\left[x^x\left(-\operatorname{cosec}^2 y\right) \frac{d y}{d x}+\cot y x^x(1+\log x)\right]=0
$$
Now at $\left(1, \frac{\pi}{2}\right)$
$$
\begin{aligned}
& 2(1+\log 1)-2\left(1(-1)\left(\frac{d y}{d x}\right)_{\left(1, \frac{\pi}{2}\right)}+0\right)=0 \\
& \Rightarrow 2+2\left(\frac{d y}{d x}\right)_{\left(1, \frac{\pi}{2}\right)}=0 \Rightarrow\left(\frac{d y}{d x}\right)_{\left(1, \frac{\pi}{2}\right)}=-1
\end{aligned}
$$
Now $x=1$,
$$
1-2 \cot y-1=0 \Rightarrow \cot y=0 \Rightarrow y=\frac{\pi}{2}
$$
Now differentiating eq. (1) w.r.t. ' $x$ '
$$
2 x^{2 x}(1+\log x)-2\left[x^x\left(-\operatorname{cosec}^2 y\right) \frac{d y}{d x}+\cot y x^x(1+\log x)\right]=0
$$
Now at $\left(1, \frac{\pi}{2}\right)$
$$
\begin{aligned}
& 2(1+\log 1)-2\left(1(-1)\left(\frac{d y}{d x}\right)_{\left(1, \frac{\pi}{2}\right)}+0\right)=0 \\
& \Rightarrow 2+2\left(\frac{d y}{d x}\right)_{\left(1, \frac{\pi}{2}\right)}=0 \Rightarrow\left(\frac{d y}{d x}\right)_{\left(1, \frac{\pi}{2}\right)}=-1
\end{aligned}
$$
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