Question

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

• A. $1$
• B. $\log 2$
• C. $-\log 2$
• D. $-1$

Answer 1

The correct option is D $-1$

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

at $x=1$, we have

$1-2\cot y-1=0$

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

Differentiating (i) w.r.t. x, we have

$2x^{2x}(1+\ln x)-2[x^x(-{\csc }^{2}y)\frac{dy}{dx}+\cot y{x}^x(1+\ln x)]=0$

At $P(1,\frac{\pi }{2})$, we have

$2(1+\ln 1)-2[1(-1){\left(\frac{dy}{dx}\right)}_P+0]=0$

$\Rightarrow$ $2+2{\left(\frac{dy}{dx}\right)}_P=0$

$\therefore$ ${\left(\frac{dy}{dx}\right)}_P=-1$