CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
log2
loader
C
log2
loader
D
1
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image