Question

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

A
1
B
log2
C
log2
D
1

Solution

## The correct option is D $$-1$$$$x^{2x}-2x^{x}\cot y-1=0$$ ........ (i)at $$x=1$$, we have$$1-2\cot y-1=0$$$$\Rightarrow$$   $$\cot y=0$$   $$\therefore$$   $$y=\dfrac{\pi}{2}$$Differentiating (i) w.r.t. x, we have$$\displaystyle 2x^{2x}\left ( 1+\ln x \right )-2\left [ x^{x}\left ( -\csc^{2} y \right )\frac{dy}{dx}+\cot yx^{x}\left ( 1+\ln x \right ) \right ]=0$$At $$P(1, \frac{\pi}{2})$$, we have$$\displaystyle 2\left ( 1+\ln 1 \right )-2\left [ 1\left ( -1 \right )\left ( \frac{dy}{dx} \right )_{P}+0 \right ]=0$$$$\Rightarrow$$   $$\displaystyle 2+2\left ( \frac{dy}{dx} \right )_{P}=0$$   $$\therefore$$   $$\displaystyle \left ( \frac{dy}{dx} \right )_{P}=-1$$Mathematics

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