JEE PYQ: Differentiation Question 20
Question 20 - 2019 (12 Jan Shift 2)
For $x > 1$, if $(2x)^{2y} = 4e^{2x-2y}$, then $(1 + \log_e 2x)^2 \frac{dy}{dx}$ is equal to:
(1) $\frac{x\log_e 2x - \log_e 2}{x}$ (2) $\log_e 2x$ (3) $\frac{x\log_e 2x + \log_e 2}{x}$ (4) $x\log_e 2x$
Show Answer
Answer: (1) $\frac{x\log_e 2x - \log_e 2}{x}$
Solution
Taking log: $2y\ln(2x) = \ln 4 + 2x - 2y$. Differentiating and simplifying: $(1 + \ln 2x)^2\frac{dy}{dx} = \frac{x\ln(2x) - \ln 2}{x}$.
BETA
