JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \(x\,{\log _e}({\log _e}\,\,x)\, - \,{x^2} + {y^2} = 4\,(y\, > \,0),\) then \(\frac{{dy}}{{dx}}\) at \(x = e\) is equal to
- A \(\frac{{(1 + 2e)}}{{2\sqrt {4 + {e^2}} }}\)
- B \(\frac{{(2e - 1)}}{{2\sqrt {4 + {e^2}} }}\)
- C \(\frac{{(1 + 2e)}}{{\sqrt {4 + {e^2}} }}\)
- D \(\frac{e}{{\sqrt {4 + {e^2}} }}\)
Answer & Solution
Correct Answer
(B) \(\frac{{(2e - 1)}}{{2\sqrt {4 + {e^2}} }}\)
Step-by-step Solution
Detailed explanation
\(x.\frac{1}{{\ell nx}}.\frac{1}{x} + {\log _e}\left( {{{\log }_e}x} \right) - 2x + 2y.\frac{{dy}}{{dx}} = 0\) Put \(x=e\) \(1 - 2e + 2y\frac{{dy}}{{dx}} = 0\) \(\frac{{dy}}{{dx}} = \frac{{2e - 1}}{{2y}}\,\,\,\,\,\,\,\,\left( 1 \right)\) Put \(x=e\) in original equation…
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