Limit Continuity And Differentiability Ques 138

Answer:

Correct Answer: 138.(b)

Solution:

138. We have, $x \log _e\left(\log _e x\right)-x^{2}+y^{2}=4$, which can be written as

$$ y^{2}=4+x^{2}-x \log _e\left(\log _e x\right) $$

Now, differentiating Eq. (i) w.r.t. $x$, we get

$$ 2 y \frac{d y}{d x}=2 x-x \frac{1}{\log _e x} \cdot \frac{1}{x}-1 \cdot \log _e\left(\log _e x\right) $$

[by using the product rule of differentiation]

$$ \Rightarrow \quad \frac{d y}{d x}=\frac{2 x-\frac{1}{\log _e x}-\log _e\left(\log _e x\right)}{2 y} $$

Now, at $x=e, y^{2}=4+e^{2}-e \log _e\left(\log _e e\right)$

$$ \begin{alignedat} & =4+e^{2}-e \log _e(1)=4+e^{2}-0 \\ & = e^{2} + 4 \ & \Rightarrow \quad y=\sqrt{e^{2}+4} \\ & \frac{d y}{d x}=\frac{2 e-1-0}{2 \sqrt{e^{2}+4}}=\frac{2 e-1}{2 \sqrt{e^{2}+4}} \quad \text { [using Eq. (ii)] } \end{aligned} $$