A) 0
B) 1
C) 2
D) 3
Correct Answer: C
Solution :
Given, \[(e-1){{e}^{xy}}\,+{{x}^{2}}\,={{e}^{{{x}^{2}}+{{y}^{2}}}}\] \[(e-1).{{e}^{xy}}.\left( x.\frac{dy}{dx}+y \right)\,+2x={{e}^{{{x}^{2}}+{{y}^{2}}}}\,.\left( 2x+2y.\frac{dy}{dx} \right)\]put \[x=-1,\,\,x=0;\] we get \[(e-1).\,\left( \frac{dy}{dx} \right)+2=e(2+0)\Rightarrow \,{{\left. \frac{dy}{dx} \right|}_{(1,\,0)}}=2\]You need to login to perform this action.
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