Question

If $x \cdot \log_e(\log_e x) - x^2 + y^2 = 4(y > 0)$, then $\frac{dy}{dx}$ at $x = e$ is

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

Hint:

Always find the value of the dependent variable ($y$) at the given $x$ before substituting into the derivative.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to calculate the value of the derivative $\frac{dy}{dx}$ at a specific point for an implicit function.
Step 3: Detailed Explanation:
1. First, find $y$ when $x = e$:
$e \cdot \log(\log e) - e^2 + y^2 = 4$
$e \cdot \log(1) - e^2 + y^2 = 4 \implies 0 - e^2 + y^2 = 4 \implies y^2 = 4 + e^2 \implies y = \sqrt{4 + e^2}$ (since $y>0$).
2. Differentiate the given equation implicitly with respect to $x$:
$\frac{d}{dx} [x \cdot \log(\log x)] - \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 0$
$[1 \cdot \log(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x}] - 2x + 2y \frac{dy}{dx} = 0$
$\log(\log x) + \frac{1}{\log x} - 2x + 2y \frac{dy}{dx} = 0$
3. Substitute $x = e$ and $y = \sqrt{4 + e^2}$:
$\log(1) + \frac{1}{1} - 2e + 2\sqrt{4 + e^2} (\frac{dy}{dx}) = 0$
$1 - 2e + 2\sqrt{4 + e^2} (\frac{dy}{dx}) = 0$
$2\sqrt{4 + e^2} (\frac{dy}{dx}) = 2e - 1 \implies \frac{dy}{dx} = \frac{2e - 1}{2\sqrt{4 + e^2}}$.
Step 4: Final Answer:
The derivative at $x = e$ is $\frac{2e - 1}{2\sqrt{4 + e^2}}$.