Question

Let $y$ be an implicit function of $x$ defined by $x^{2x}-2x^{x}coty-1=0.$ 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

$x^{2x}-2x^{x}coty-1=0$
$\Rightarrow 2coty=x^x-x^{-x}\Rightarrow 2coty=u-\frac{1}{u}\text{where}u=x^x$
Differentiating both sides with respect to $x$, we get
$\Rightarrow -2cose{c}^{2}y\frac{dy}{dx}=(1+\frac{1}{u^2})\frac{du}{dx}$
where $u=x^x\Rightarrow logu=xlogx$
$\Rightarrow \frac{1}{u}\frac{du}{dx}=1+logx\Rightarrow \frac{du}{dx}=x^x(1+logx)$
$\therefore$ We get $-2cose{c}^{2}y\frac{dy}{dx}=(1+x^{-2x}).x^x(1+logx)$
$\Rightarrow \frac{dy}{dx}=\frac{(x^x+x^{-x})(1+\log x)}{-2(1+co{t}^{2}y)}\cdots \left(i\right)$
Now when $x=1,x^{2x}-2x^{x}coty-1=0,gives1-2coty-1=0\Rightarrow coty=0$
$\therefore$ From equation (i), at $x=1$ and $coty=0$, we get $y^{\prime }\left(1\right)=\frac{(1+1)(1+0)}{-2(1+0)}=-1$