Limit Continuity And Differentiability Ques 62
- If $x \log _e\left(\log _e x\right)-x^2+y^2=4(y>0)$, then $\frac{d y}{d x}$ at $x=e$ is equal to
(2019 Main, 11 Jan I)
(a) $\frac{e}{\sqrt{4+e^2}}$
(b) $\frac{(2 e-1)}{2 \sqrt{4+e^2}}$
(c) $\frac{(1+2 e)}{\sqrt{4+e^2}}$
(d) $\frac{(1+2 e)}{2 \sqrt{4+e^2}}$
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Answer:
Correct Answer: 62.(b)
Solution: (b) 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) $ $\quad$ ……..(i)
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 product rule of derivative]
$ \Rightarrow \quad\left(\frac{d y}{d x}\right)=\frac{2 x-\frac{1}{\log _e x}-\log _e\left(\log _e x\right)}{2 y} $ $\quad$ ……..(ii)
Now, at $x=e, y^2=4+e^2-e \log _e\left(\log _e e\right) \quad $ [using Eq. (i)]
$ \begin{aligned} & =4+e^2-e \log _e(1)=4+e^2-0 \\ & =e^2+4 \\ & \begin{array}{l} y=\sqrt{e^2+4} \quad[\because y>0] \end{array} \\ & \therefore \text { At } x=e \text { and } y=\sqrt{e^2+4} \text {, } \\ & \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} $
BETA
