JEE Main & Advanced Mathematics Differentiation Question Bank Partial Differentiation

  • question_answer
    If \[u={{\tan }^{-1}}\left( \frac{{{x}^{3}}+{{y}^{3}}}{x-y} \right)\], then \[x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\]                                                                                                                [EAMCET 1999]

    A)            \[\sin 2u\]

    B)            \[\cos 2u\]

    C)            \[\tan 2u\]

    D)            \[\sec 2u\]

    Correct Answer: A

    Solution :

                       \[\tan u\] is homogeneous in x, y of degree 2.                    \[\therefore \] \[x\frac{\partial }{\partial x}(\tan u)+y\frac{\partial }{\partial y}(\tan u)=2(\tan u)\]                    \[\therefore \] \[x{{\sec }^{2}}u\frac{\partial u}{\partial x}+y{{\sec }^{2}}u\frac{\partial u}{\partial y}=2\tan u\]                    Þ \[x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=2\frac{\tan u}{{{\sec }^{2}}u}\] = \[2\sin u\cos u=\sin 2u\].


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