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. $1$
• C. $log \,2$
• D. $-log \,2$

Answer 1

Answer: (A)

$x^{2x} - 2x^x \, cot\,y - 1 = 0$
Now at $ x = 1$,
$ 1 - 2 \,cot\,y - 1 = 0$
$\Rightarrow cot\,y = 0$
$\Rightarrow y = \frac{\pi}{2}$
Now differentiating (i) w.r.t. $x$, we get
$2x^{2x} ( 1 + log\,x) - 2[ x^x ( - cosec^2 \,y) \frac{dy}{dx} + cot\,y\,x^x ( 1 + log\,x)] = 0$
Now at $(1, \pi/2)$,
$2( 1 + log\,1) - 2 [ 1 (-1) (\frac{dy}{dx})_{(1, \pi/2)} + 0] = 0$
$\Rightarrow 2 + 2 (\frac{dy}{dx})_{(1, \pi/2)} = 0$
$\Rightarrow (\frac{dy}{dx})_{(1, \pi/2)} = -1$