Question

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

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

Answer 1

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

When $x=e$,then $0-e^2+y^2=4,y=\sqrt{e^2+4}$

Differentiating with respect to x, We get:

$x.\frac{1}{\ell nx}.\frac{1}{x}+\ell n\left(\ell nx\right)-2x+2y.\frac{dy}{dx}=0$ at $x=e$ we get
$1-2e+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{2e-1}{2y}$
$\Rightarrow \frac{dy}{dx}=\frac{2e-1}{2\sqrt{4+e^2}}$ as $y\left(e\right)=\sqrt{4+e^2}$