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JEE Main & Advanced Sample Paper JEE Main Sample Paper-24

  • question_answer
    If \[\ell n\left( e-1){{e}^{xy}}+{{x}^{2}} \right)={{x}^{2}}+{{y}^{2}}\], then \[\frac{dy}{dx}\] at (1, 0) is equal to

    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 \[(e-1).\,\left( \frac{dy}{dx} \right)+2=e(2+0)\Rightarrow \,{{\left. \frac{dy}{dx} \right|}_{(1,\,0)}}=2\]


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