JEE PYQ: Differentiation Question 19
Question 19 - 2019 (11 Jan Shift 1)
If $x\log_e(\log_e x) - x^2 + y^2 = 4$ ($y > 0$), then $\frac{dy}{dx}$ at $x = e$ is equal to:
(1) $\frac{(1+2e)}{2\sqrt{4+e^2}}$ (2) $\frac{(2e-1)}{2\sqrt{4+e^2}}$ (3) $\frac{(1+2e)}{\sqrt{4+e^2}}$ (4) $\frac{e}{\sqrt{4+e^2}}$
Show Answer
Answer: (2) $\frac{(2e-1)}{2\sqrt{4+e^2}}$
Solution
At $x = e$: $y = \sqrt{4+e^2}$. Differentiating and substituting: $\frac{dy}{dx} = \frac{2e-1}{2\sqrt{4+e^2}}$.
BETA
