JEE Mains · Maths · STD 12 - 5. continuity and differentiation
For \(x > 1\), if \({\left( {2x} \right)^{2y}} = 4{e^{2x - 2y}}\), then \({\left( {1 + {{\log }_e}\,2x} \right)^2}\frac{{dy}}{{dx}}\) is equal to
- A \(\frac{{x{{\log }_e}\,2x - {{\log }_e}\,2}}{x}\)
- B \(log_e\, 2x\)
- C \(\frac{{x{{\log }_e}\,2x + {{\log }_e}\,2}}{x}\)
- D \(x log_e\, 2x\)
Answer & Solution
Correct Answer
(A) \(\frac{{x{{\log }_e}\,2x - {{\log }_e}\,2}}{x}\)
Step-by-step Solution
Detailed explanation
\(2y\ell n2x = \ell n4 + 2x - 2y\) \(2y\left( {1 + \ell n2x} \right) = \ell n4 + 2x\) \(y = \frac{{\ell n2x - \frac{{\ell n2}}{x}}}{{{{\left( {1 + \ell n2x} \right)}^2}}}.{\left( {1 + \ell n2x} \right)^2}\)…
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