Question

If \(x\log_e(\log_e x) - x^2 + y^2 = 4\ (y \gt 0)\), then \(\frac{dy}{dx}\) at \(x = e\) is equal to :

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

Hint:

Differentiate implicitly (product rule on \(x\ln\ln x\)), then find y at \(x=e\) from the given relation.

Answer 1

The Correct Option is C

At \(x = e\): \(\ln(\ln e) = \ln 1 = 0\), so \(0 - e^2 + y^2 = 4 \Rightarrow y = \sqrt{4 + e^2}\) (as \(y\gt0\)).
Differentiate: \(\ln(\ln x) + x\cdot\frac{1}{x\ln x} - 2x + 2y\frac{dy}{dx} = 0 \Rightarrow \ln(\ln x) + \frac{1}{\ln x} - 2x + 2y\,y' = 0\).
At \(x=e\): \(0 + 1 - 2e + 2\sqrt{4+e^2}\,y' = 0 \Rightarrow y' = \frac{2e - 1}{2\sqrt{4+e^2}}\).
Answer: (3) \(\frac{(2e-1)}{2\sqrt{4+e^2}}\)