Let y be an implicit function of x defined by $x^{2 x}-2 x^x \cot y-1=0$. Then, y'(1) equals
Answer & explanation
Correct answer: option 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
$2-2 \times 0+2 \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-1$