Let y be an implicit function of x defined by $x^{2 x}-2 x^x \cot y-1=0$. Then, y'(1) equals

-1

We have,

$x^{2 x}-2 x^x \cot y-1=0$         .....(i)

When x = 1, we get

$1-2 \cot y-1=0 \Rightarrow \cot y=0 \Rightarrow y=\frac{\pi}{2}$

Differentiating (i) w.r. to x, we get

$2 x^{2 x}(1+\log x)-2 x^x(1+\log x) \cot y+2 x^x ~cosec^2 y \frac{d y}{d x}=0$

Putting $x=1, y=\frac{\pi}{2}$, we get

$2-2 \times 0+2 \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-1$