BCECE Engineering BCECE Engineering Solved Paper-2002

  • question_answer
    If \[{{e}^{x}}\sin \,y-{{e}^{y}}\cos x=1,\] then \[\frac{dy}{dx}\]is equal to:

    A) \[\frac{{{e}^{x}}\sin y+{{e}^{y}}\sin x}{{{e}^{y}}\cos x-{{e}^{x}}\cos y}\]

    B)         \[\frac{{{e}^{x}}\sin x+{{e}^{y}}\sin y}{{{e}^{y}}\cos x-{{e}^{x}}\cos y}\]

    C)  \[\frac{{{e}^{x}}\sin y-{{e}^{y}}\sin x}{{{e}^{y}}\cos x-{{e}^{x}}\cos y}\]

    D)  none of the above

    Correct Answer: A

    Solution :

     \[{{e}^{x}}\sin y-{{e}^{y}}\cos x=1\] On differentiating both sides w.r.t. \[x,\] we get \[{{e}^{x}}\cos y\frac{dy}{dx}+{{e}^{x}}\sin y+{{e}^{y}}\sin x-\] \[\times \,\,{{e}^{y}}\cos x\frac{dy}{dx}=0\]                 \[\Rightarrow \]               \[\frac{dy}{dx}({{e}^{x}}\cos y-{{e}^{y}}\cos x)\]                 \[=-({{e}^{x}}\sin y+{{e}^{y}}\sin x)\]                 \[\Rightarrow \]               \[\frac{dy}{dx}=\frac{{{e}^{x}}\sin y+{{e}^{y}}\sin x}{({{e}^{y}}\cos x-{{e}^{x}}\cos y)}\]


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