A) 1
B) ?1
C) 2
D) 0
Correct Answer: A
Solution :
\[\ln (x+y)=2xy\] Differentiate both sides w.r.t x, \[\left( \frac{1}{x+y} \right)\,\left( 1+\frac{dy}{dx} \right)=2\,\left( x\frac{dy}{dx}+y \right)\] Þ \[\frac{dy}{dx}=\frac{1-2xy-2{{y}^{2}}}{2{{x}^{2}}+2xy-1}\] As at \[x=0,y=1\], (From \[\ln (x+y)=2xy\]) Hence \[y'(0)=\frac{1-2}{-1}=1\].
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