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

  • question_answer
    If \[A=\frac{{{2}^{x}}\cot x}{\sqrt{x}},\]then \[\frac{dA}{dx}=\]

    A)          \[\frac{{{2}^{x-1}}\left\{ -2x\,\text{cos}\text{e}{{\text{c}}^{2}}x+\cot x.\log \left( \frac{{{4}^{x}}}{e} \right) \right\}}{{{x}^{3/2}}}\]

    B)            \[\frac{{{2}^{x-1}}\left\{ -2x\cos \text{e}{{\text{c}}^{2}}x+\cot x.\log \left( \frac{{{4}^{x}}}{e} \right) \right\}}{x}\]

    C)            \[\frac{2x\left\{ -2x\text{cose}{{\text{c}}^{2}}x+\cot x.\log \left( \frac{{{4}^{x}}}{e} \right) \right\}}{{{x}^{\text{3/2}}}}\]

    D)            None of these

    Correct Answer: A

    Solution :

      \[\frac{dA}{dx}=\frac{\sqrt{x}\{{{2}^{x}}{{\log }_{e}}2\cot x-{{2}^{x}}\text{cose}{{\text{c}}^{2}}x\}-{{2}^{x}}\cot x\frac{1}{2\sqrt{x}}}{x}\]                          \[=\frac{{{2}^{x-1}}\left\{ -2x\,\text{cose}{{\text{c}}^{2}}x+\cot x.\log \left( \frac{{{4}^{x}}}{e} \right) \right\}}{{{x}^{3/2}}}\].


You need to login to perform this action.
You will be redirected in 3 sec spinner