Question

If $x \log_e(\log_e x) - x^2 + y^2 = 4(y > 0)$, 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

Answer: (C)

Differentiating with respect to $x$,
$x. \frac{1}{\ell nx} . \frac{1}{x} +\ell n\left(\ell nx\right)-2x+2y . \frac{dy}{dx} =0$
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}}} y=\left(e\right) = \sqrt{4+e^{2}} $