Question

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

• A. $-1$
• B. $1$
• C. $log2$
• D. $-log2$

Answer 1

Answer: (A)

$x^{2x}-2x^{x}coty-1=0$
Now at $x=1$,
$1-2coty-1=0$
$\Rightarrow coty=0$
$\Rightarrow y=\frac{\pi }{2}$
Now differentiating (i) w.r.t. $x$, we get
$2x^{2x}(1+logx)-2[x^x(-cose{c}^{2}y)\frac{dy}{dx}+coty{x}^x(1+logx)]=0$
Now at $(1,\pi /2)$,
$2(1+log1)-2[1(-1)(\frac{dy}{dx}{)}_{(1,\pi /2)}+0]=0$
$\Rightarrow 2+2(\frac{dy}{dx}{)}_{(1,\pi /2)}=0$
$\Rightarrow (\frac{dy}{dx}{)}_{(1,\pi /2)}=-1$