Question

If $x{\log }_e\left({\log }_{e}x\right)-x^2+y^2=4(y>0)$, then $\frac{dy}{dx}$ at $x=e$ is equal to :

• A. $\frac{(2e-1)}{2\sqrt{4+e^2}}$
• B. $\frac{e}{\sqrt{4+e^2}}$
• C. $\frac{(1+2e)}{2\sqrt{4+e^2}}$
• D. $\frac{(1+2e)}{\sqrt{4+e^2}}$

Answer 1

The correct option is A $\frac{(2e-1)}{2\sqrt{4+e^2}}$

$y^2=4+x^2-x{\log }_e\left({\log }_{e}x\right)\dots \left(1\right)$
For $x=e$
$y^2=4+e^2-e\times 0$
$\Rightarrow y=\sqrt{4+e^2}$

Differentiate equation $\left(1\right)$ w.r.t. $x$, we get
$2y\frac{dy}{dx}=2x-{\log }_e\left({\log }_{e}x\right)-x\times \frac{1}{{\log }_{e}x}\times \frac{1}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{2x-{\log }_e\left({\log }_{e}x\right)-\frac{1}{{\log }_{e}x}}{2y}$

$\Rightarrow \frac{dy}{dx}{|}_{x=e}=\frac{2e-1}{2\sqrt{4+e^2}}$