Question

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

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

Answer 1

Answer: (D)

$x^{2x}-2x^x\cot y-1=0$
$\Rightarrow 2\cot y=x^x-x^{-x}$
$\Rightarrow 2\cot y=u-\frac{1}{u}$ where $u=x^x$
Differentiating both sides with respect to x, we get
$-2cose{c}^{2}y\frac{dy}{dx}=\left(1+\frac{1}{u^2}\right)\frac{du}{dx}$
where $u=x^x\Rightarrow \log u=x\log x$
$\Rightarrow \frac{1}{u}\frac{du}{dx}=1+\log x$
$\Rightarrow \frac{du}{dx}=x^x\left(1+\log x\right)$
$\therefore$ We get
$-2cose{c}^{2}y\frac{dy}{dx}=\left(1+x^{-2x}\right).x^x\left(1+\log x\right)$
$\Rightarrow \frac{dy}{dx}=\frac{\left(x^x+x^{-x}\right)\left(1+\log x\right)}{-2\left(1+{\cot }^{2}y\right)}$ ...(i)
Now when $x=1,x^{2x}-2x^x\cot y-1=0$, gives $1-2\cot y-1=0$
$\Rightarrow \cot y=0$
$\therefore$ From equation (i), at x = 1 and
$\cot y=0$ , we get
$y^{\prime }\left(1\right)=\frac{\left(1+1\right)\left(1+0\right)}{-2\left(1+0\right)}=-1$