JEE Main & Advanced Mathematics Differentiation Question Bank Derivative at a point Standard Differentiation

  • question_answer
    If \[y=\frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}\], then  \[\frac{dy}{dx}=\]             [AISSE 1986]

    A)          \[\frac{ay}{x\sqrt{{{a}^{2}}-{{x}^{2}}}}\]

    B)            \[\frac{ay}{\sqrt{{{a}^{2}}-{{x}^{2}}}}\]

    C)            \[\frac{ay}{x\sqrt{{{x}^{2}}-{{a}^{2}}}}\]

    D)            None of these

    Correct Answer: A

    Solution :

               \[y=\frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}\Rightarrow y=\frac{{{(\sqrt{a+x}-\sqrt{a-x})}^{2}}}{(a+x)-(a-x)}\]                    \[\Rightarrow y=\frac{(a+x)+(a-x)-2(\sqrt{{{a}^{2}}-{{x}^{2}}})}{2x}\]                            \[=\frac{2a-2\sqrt{{{a}^{2}}-{{x}^{2}}}}{2x}\] or \[y=\frac{a-\sqrt{{{a}^{2}}-{{x}^{2}}}}{x}\]  ?..(i)                    Differentiating w.r.t. x of y, we get                    \[\frac{dy}{dx}=\frac{x\left[ -\frac{1}{2\sqrt{{{a}^{2}}-{{x}^{2}}}}(-2x) \right]-(a-\sqrt{{{a}^{2}}-{{x}^{2}}})}{{{x}^{2}}}\]                        \[=\frac{{{x}^{2}}-a\sqrt{{{a}^{2}}-{{x}^{2}}}+{{a}^{2}}-{{x}^{2}}}{{{x}^{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}}=\frac{a(a-\sqrt{{{a}^{2}}-{{x}^{2}}})}{{{x}^{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}}\]                       \[=\frac{a}{x\sqrt{{{a}^{2}}-{{x}^{2}}}}\left[ \frac{a-\sqrt{{{a}^{2}}-{{x}^{2}}}}{x} \right]=\frac{ay}{x\sqrt{{{a}^{2}}-{{x}^{2}}}}\] [By (i)]


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