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
$-2 \, cosec^{2} \,y \,\frac{dy}{dx} = \left(1+ \frac{1}{u^{2}}\right) \frac{du}{dx}$
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
$- 2\, cosec^{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
$y' \left(1\right) = \frac{\left(1+1\right)\left(1+0\right)}{-2\left(1+0\right)}=-1$