RAJASTHAN ­ PET Rajasthan PET Solved Paper-2003

  • question_answer
    If\[y=\frac{1}{(t+2)(t+1)},\]then\[\frac{dy}{dx}\]is equal to

    A)  \[-\frac{1}{{{(t+1)}^{2}}}+\frac{1}{{{(t+2)}^{2}}}\]

    B)  \[\frac{1}{{{(t+1)}^{2}}}+\frac{1}{{{(t+2)}^{2}}}\]

    C)  \[-\frac{1}{{{(t+1)}^{2}}}-\frac{1}{{{(t+2)}^{2}}}\]

    D)  None of the above

    Correct Answer: A

    Solution :

     \[y=\frac{1}{(t+2)(t+1)}\] \[y=-\frac{1}{t+2}+\frac{1}{t+1}\] On differentiating w.r.t.\[t,\] \[\Rightarrow \] \[\frac{dy}{dt}=\frac{1}{{{(t+2)}^{2}}}-\frac{1}{{{(t+1)}^{2}}}\]


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