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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2008

  • question_answer
    The derivative of \[\sin ({{x}^{3}})\] w.r.t. \[cos({{x}^{3}})\] is

    A)  \[-\tan ({{x}^{3}})\]                       

    B)  \[\tan ({{x}^{3}})\]

    C)  \[-cot({{x}^{3}})\]                          

    D)  \[cot({{x}^{3}})\]

    Correct Answer: C

    Solution :

    Let \[u=\sin {{x}^{3}}\] and \[v=\cos {{x}^{3}}\]. On differentiating w.r.t. x, we get \[\frac{du}{dx}=\cos \,{{x}^{3}}.3{{x}^{2}}\] and \[\frac{dv}{dx}=-\sin {{x}^{3}}.3{{x}^{2}}\] \[\therefore \] \[\frac{du}{dv}=\frac{du/dx}{dv/dx}=\frac{3{{x}^{2}}\,\cos \,{{x}^{3}}}{-3{{x}^{2}}\,\sin {{x}^{3}}}=-\cot {{x}^{3}}\]


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