Question
यदि $\ln \,(x + y) = 2xy,$ तब $y'(0) =$
$x$ के सापेक्ष दोनों पक्षों का अवकलन करने पर,
$\left( {\frac{1}{{x + y}}} \right)\,\left( {1 + \frac{{dy}}{{dx}}} \right) = 2\,\left( {x\frac{{dy}}{{dx}} + y} \right)$
==> $\frac{{dy}}{{dx}} = \frac{{1 - 2xy - 2{y^2}}}{{2{x^2} + 2xy - 1}}$
$\because \;x = 0$ पर, $y = 1$ ($\ln (x + y) = 2xy$ से)
अत: $y'(0) = \frac{{1 - 2}}{{ - 1}} = 1$.
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$(A)$ $e_1^2+e_2^2=\frac{43}{40}$
$(B)$ $e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
$(C)$ $\left|e_1^2-e_2^2\right|=\frac{5}{8}$
$(D)$ $e_1 e_2=\frac{\sqrt{3}}{4}$